problem solving for kinetic and potential energy

Potential and Kinetic Energy

Energy is the capacity to do work .

The unit of energy is J (Joule) which is also kg m 2 /s 2 (kilogram meter squared per second squared)

Energy can be in many forms! Here we look at Potential Energy (PE) and Kinetic Energy (KE).

Potential Energy and Kinetic Energy

• when raised up has potential energy (the energy of position or state)
• when falling down has kinetic energy (the energy of motion)

Potential energy (PE) is stored energy due to position or state

• a raised hammer has PE due to gravity.
• fuel and explosives have Chemical PE
• a coiled spring or a drawn bow also have PE due to their state

Kinetic energy (KE) is energy of motion

From PE to KE

For a good example of PE and KE have a play with a pendulum .

Gravitational Potential Energy

When the PE is due to an objects height then:

PE due to gravity = m g h

• m is the objects mass (kg)
• g is the "gravitational field strength" of 9.8 m/s 2 near the Earth's surface
• h is height (m)

Example: This 2 kg hammer is 0.4 m up. What is it's PE?

Kinetic energy.

The formula is:

KE = ½ m v 2

• m is the object's mass (kg)
• v is the object's speed (m/s)

Example: What is the KE of a 1500 kg car going at suburban speed of 14 m/s (about 50 km/h or 30 mph)?

KE = ½ × 1500 kg × (14 m/s) 2

KE = 147,000 kg m 2 /s 2

KE = 147 kJ

Let's double the speed!

Example: The same car is now going at highway speed of 28 m/s (about 100 km/h or 60 mph)?

KE = ½ × 1500 kg × (28 m/s) 2

KE = 588,000 kg m 2 /s 2

KE = 588 kJ

Wow! that is a big increase in energy! Highway speed is way more dangerous.

Double the speed and the KE increases by four times. Very important to know

A 1 kg meteorite strikes the Moon at 11 km/s. How much KE is that?

KE = ½ × 1 kg × (11,000 m/s) 2

KE = 60,500,000 J

KE = 60.5 MJ

That is 100 times the energy of a car going at highway speed.

When falling, an object's PE due to gravity converts into KE and also heat due to air resistance.

Let's drop something!

Example: We drop this 0.1 kg apple 1 m. What speed does it hit the ground with?

At 1 m above the ground it's Potential Energy is

PE = 0.1 kg × 9.8 m/s 2 × 1 m

PE = 0.98 kg m 2 /s 2

Ignoring air resistance (which is small for this little drop anyway) that PE gets converted into KE:

Swap sides and rearrange:

½ m v 2 = KE

v 2 = 2 × KE / m

v = √( 2 × KE / m )

Now put PE into KE and we get:

v = √( 2 × 0.98 kg m 2 /s 2 / 0.1 kg )

v = √( 19.6 m 2 /s 2 )

v = 4.427... m/s

Note: for velocity we can combine the formulas like this:

The mass does not matter! It is all about height and gravity. For our earlier example:

v = √( 2gh )

v = √( 2 × 9.8 m/s 2 × 1 m )

• Energy is the ability to do work

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Mechanics: Work, Energy and Power

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Kinetic Energy Problemset

SHOW ALL WORK!

1.  What is the kinetic energy of a jogger with a mass of 65.0 kg traveling at a speed of 2.5 m/s?

2.  What is the mass of a baseball that has a kinetic energy of 100 J and is traveling at 5 m/s?

Write down what you know:

3. What is the kinetic energy of a 0.5 kg soccer ball that is traveling at a speed of 3 m/s?

4  What is the kinetic energy of a 1 kg pie travelling at a speed of 4 m/s ?

5.  What is the kinetic energy of the pie if it is thrown at 10 m/s?

6. A student is hit with a 1 kg pumpkin pie. The kinetic energy of the pie 32 J. What was the speed of the pie?

GPE = mgh | g = 9.8 m/s 2

1.  Find the gravitational potential energy of a light that has a mass of 13.0 kg and is 4.8 m above the ground.

3.  A marble is on a table 2.4 m above the ground.  What is the mass of the marble if it has a GPE of 568 J. 

4.  A box with a mass of 12.5 kg sits on the floor.  How high would you need to lift it for it to have a GPE of 355J ?

5.  A cart at the top of a 300 m hill has a mass of  40 kg.  What is the cart’s gravitational potential energy?

6. Examine the graphic below.

What is the gravitational potential energy of the 6 kg cart as it sits the the top of the incline?  _______________

What is the KINETIC ENERGY of the cart if it is moving at a speed of 2 m/s at the bottom of the ramp? _____________

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Kinetic and Potential Energy

The mechanical energy of a system is the sum of its kinetic energy and potential energy.

Kinetic energy

Kinetic energy is the energy that an object possesses due to its motion. Kinetic energy of a marticle of mass $m$ moving with a velocity $v$ is given by \begin{align} K=\frac{1}{2}mv^2. \end{align} The unit of kinetic energy is Joule. Its dimensions are ML 2 T -2 . Kinetic energy K$ and linear momentum $p$ of a particle of mass $m$ are related by \begin{align} K=\frac{p^2}{2m}. \end{align} The work-energy theorem relates work done by conservative forces with the change in kinetic energy.

Potential energy

Potential energy is the energy that an object possesses due to its position or configuration. It is the energy stored in an object due to its position or configuration. The gravitational potential energy and elastic potential energy are two important forms of potential energy.

Gravitational potential energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. The gravitational potential energy of an object of mass $m$ positioned at a height $h$ (above some reference) is given by \begin{align} U_g = mgh \end{align} where $g$ is the acceleration due to gravity .

Elastic potential energy

Elastic potential energy is the energy stored in an object due to its configuration. The elastic potential energy of a stretched spring of spring constant $k$ is given by \begin{align} U_s = \frac{1}{2}kx^2 \end{align} where $x$ is the difference between stretched and natural (equilibrium) lengths of the spring.

Problems from IIT JEE

Problem (JEE Mains 2022): A body of mass 8 kg and another of mass 2 kg are moving with equal kinetic energy. The ratio of their respective momenta will be

Problem (JEE Mains 2003): A spring of spring constant $5\times10^3$ N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

Solution: The work required is equal to the change in spring potential energy, \begin{align} W&=U_f-U_i\\ &=\frac{1}{2}kx_f^2-\tfrac{1}{2}kx_i^2\nonumber\\ &=\frac{1}{2}(5\times{10}^{3}) (0.10)^2-\frac{1}{2}(5\times{10}^{3}) (0.05)^2 \nonumber\\ &=18.75\;\mathrm{J}.\nonumber \end{align}

• Work-Energy Theorem
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FREE K-12 standards-aligned STEM

curriculum for educators everywhere!

Find more at TeachEngineering.org .

• TeachEngineering
• Kinetic and Potential Energy of Motion

Lesson Kinetic and Potential Energy of Motion

Grade Level: 8 (7-9)

Time Required: 45 minutes

Lesson Dependency: None

Subject Areas: Physical Science, Physics

NGSS Performance Expectations:

• Print lesson and its associated curriculum

Curriculum in this Unit Units serve as guides to a particular content or subject area. Nested under units are lessons (in purple) and hands-on activities (in blue). Note that not all lessons and activities will exist under a unit, and instead may exist as "standalone" curriculum.

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TE Newsletter

Engineering connection, learning objectives, more curriculum like this, introduction/motivation, associated activities, lesson closure, vocabulary/definitions, user comments & tips.

Mechanical engineers are concerned about the mechanics of energy — how it is generated, stored and moved. Product design engineers apply the principles of potential and kinetic energy when they design consumer products. For example, a pencil sharpener employs mechanical energy and electrical energy. When designing a roller coaster, mechanical and civil engineers ensure that there is sufficient potential energy (which is converted to kinetic energy) to move the cars through the entire roller coaster ride.

After this lesson, students should be able to:

• Recognize that engineers need to understand the many different forms of energy in order to design useful products.
• Explain the concepts of kinetic and potential energy.
• Understand that energy can change from one form into another.
• Understand that energy can be described by equations.

Educational Standards Each TeachEngineering lesson or activity is correlated to one or more K-12 science, technology, engineering or math (STEM) educational standards. All 100,000+ K-12 STEM standards covered in TeachEngineering are collected, maintained and packaged by the Achievement Standards Network (ASN) , a project of D2L (www.achievementstandards.org). In the ASN, standards are hierarchically structured: first by source; e.g. , by state; within source by type; e.g. , science or mathematics; within type by subtype, then by grade, etc .

Ngss: next generation science standards - science, common core state standards - math.

View aligned curriculum

Do you agree with this alignment? Thanks for your feedback!

International Technology and Engineering Educators Association - Technology

State standards, colorado - math, colorado - science.

Begin by showing the class three items: 1) an item of food (such as a bagel, banana or can of soda water), 2) a battery, and 3) you, standing on a stool or chair. Ask the class what these three things have in common. The answer is energy. The food contains chemical energy that is used by the body as fuel. The battery contains electrical energy (in the form of electrical, potential or stored energy), which can be used by a flashlight or a portable CD player. A person standing on a stool has potential energy (sometimes called gravitational potential energy) that could be used to crush a can, smash the banana, or really hurt the foot of someone standing under you. Do a dramatic demonstration of jumping down on the banana or an empty soda can. (Be careful! Banana peels are slippery!) Explain the ideas of potential energy and kinetic energy as two different kinds of mechanical energy . Give definitions of each and present the equations, carefully explaining each variable, as discussed in the next section,

PE = mass x g x height

KE = 1/2 m x v 2

Lesson Background and Concepts for Teachers

Whenever something moves, you can see the change in energy of that system. Energy can make things move or cause a change in the position or state of an object. Energy can be defined as the capacity for doing work. Work is done when a force moves an object over a given distance. The capacity for work, or energy, can come in many different forms. Examples of such forms are mechanical, electrical, chemical or nuclear energy.

This lesson introduces mechanical energy , the form of energy that is easiest to observe on a daily basis. All moving objects have mechanical energy. There are two types of mechanical energy: potential energy and kinetic energy. Potential energy is the energy that an object has because of its position and is measured in Joules (J). Potential energy can also be thought of as stored energy. Kinetic energy is the energy an object has because of its motion and is also measured in Joules (J). Due to the principle of conservation of energy, energy can change its form (potential, kinetic, heat/thermal, electrical, light, sound, etc.) but it is never created or destroyed.

Within the context of mechanical energy, potential energy is a result of an object's position, mass and the acceleration of gravity. A book resting on the edge of a table has potential energy; if you were to nudge it off the edge, the book would fall. It is sometimes called gravitational potential energy ( PE ). It can be expressed mathematically as follows:

PE = mass x g x height or PE = weight x height

where PE is the potential energy, and g is the acceleration due to gravity. At sea level, g = 9.81 meters/sec 2 or 32.2 feet/sec 2 . In the metric system, we would commonly use mass in kilograms or grams with the first equation. With English units it is common to use weight in pounds with the second equation.

Kinetic energy ( KE ) is energy of motion. Any object that is moving has kinetic energy. An example is a baseball that has been thrown. The kinetic energy depends on both mass and velocity and can be expressed mathematically as follows:

Here KE stands for kinetic energy. Note that a change in the velocity will have a much greater effect on the amount of kinetic energy because that term is squared. The total amount of mechanical energy in a system is the sum of both potential and kinetic energy, also measured in Joules (J).

Total Mechanical Energy = Potential Energy + Kinetic Energy

Engineers must understand both potential and kinetic energy. A simple example would be the design of a roller coaster — a project that involves both mechanical and civil engineers. At the beginning of the roller coaster, the cars must have enough potential energy to power them for the rest of the ride. This can be done by raising the cars to a great height. Then, the increased potential energy of the cars is converted into enough kinetic energy to keep them in motion for the length of the track. This is why roller coaters usually start with a big hill. As the cars start down the first hill, potential energy is changed into kinetic energy and the cars pick up speed. Engineers design the roller coaster to have enough energy to complete the course and to overcome the energy-draining effect of friction.

Watch this activity on YouTube

Restate that both potential energy and kinetic energy are forms of mechanical energy. Potential energy is the energy of position and kinetic energy is the energy of motion. A ball that you hold in your hand has potential energy, while a ball that you throw has kinetic energy. These two forms of energy can be transformed back and forth. When you drop a ball, you demonstrate an example of potential energy changing into kinetic energy.

Explain that energy is an important engineering concept. Engineers need to understand the many different forms of energy so that they can design useful products. An electric pencil sharpener serves to illustrate the point. First, the designer needs to know the amount of kinetic energy the spinning blades need in order to successfully shave off the end of the pencil. Then, the designer must choose an appropriately-powered motor to supply the necessary energy. Finally, the designer must know the electrical energy requirements of the motor in order for the motor to properly do its assigned task.

conservation of energy: A principle stating that the total energy of an isolated system remains constant regardless of changes within the system. Energy can neither be created nor destroyed.

energy: Energy is the capacity to do work.

kinetic energy: The energy of motion.

mechanical energy: Energy that is composed of both potential energy and kinetic energy.

potential energy: The energy of position, or stored energy.

Pre-Lesson Assessment

Discussion Questions: Solicit, integrate and summarize student responses.

• What are examples of dangerous unsafe placement of objects? (Possible answers: Boulders on the edge of a cliff, dishes barely on shelves, etc.).

Post-Introduction Assessment

Question/Answer: Ask the students and discuss as a class:

• What has more potential energy: a boulder on the ground or a feather 10 feet in the air? (Answer: The feather because the boulder is on the ground and has zero potential energy. However, if the boulder was 1 mm off the ground, it would probably have more potential energy.)

Lesson Summary Assessment

Group Brainstorm: Give groups of students each a ball (example, tennis ball). Remind them that energy can be converted from potential to kinetic and vice versa. Write a question on the board and have them brainstorm the answer in their groups. Have the students record their answers in their journals or on a sheet of paper and hand it in. Discuss the student groups' answers with the class.

• How can you throw a ball and have its energy change from kinetic to potential and back to kinetic without touching the ball once it relases from your hand? (Answer: Throw it straight up in the air.)

Calculating: Have students practice problems solving for potential energy and kinetic energy:

• If a mass that weighs 8 kg is held at a height of 10 m, what is its potential energy? (Answer: PE = (8 kg)*(9.8 m/s 2 )*(10 m) = 784 kg*m 2 /s 2 = 784 J)
• Now consider an object with a kinetic energy of 800 J and a mass of 12 kg. What is its velocity? (Answer: v = sqrt(2*KE/m) = sqrt((2 * 800 J)/12 kg) = 11.55 m/s)

Lesson Extension Activities

There is another form of potential energy, not related to height, which is called spring potential or elastic potential energy . In this case, energy is stored when you compress or elongate a spring. Have the students search the Internet or library for the equation of spring potential energy and explain what the variables in the equation represent. The answer is

PE spring = ½ k∙x 2

where k is the spring constant measured in N/m (Newton/meters) and x is how far the spring is compressed or stretched measured in m (meters).

This activity shows students the engineering importance of understanding the laws of mechanical energy. More specifically, it demonstrates how potential energy can be converted to kinetic energy and back again. Given a pendulum height, students calculate and predict how fast the pendulum will swing ...

This activity demonstrates how potential energy (PE) can be converted to kinetic energy (KE) and back again. Given a pendulum height, students calculate and predict how fast the pendulum will swing by understanding conservation of energy and using the equations for PE and KE.

High school students learn how engineers mathematically design roller coaster paths using the approach that a curved path can be approximated by a sequence of many short inclines. They apply basic calculus and the work-energy theorem for non-conservative forces to quantify the friction along a curve...

Students explore the physics exploited by engineers in designing today's roller coasters, including potential and kinetic energy, friction and gravity. During the associated activity, students design, build and analyze model roller coasters they make using foam tubing and marbles (as the cars).

Argonne Transportation - Laser Glazing of Rails. September 29, 2003. Argonne National Laboratory, Transportation Technology R&D Center. October 15, 2003. http://www.anl.gov/index.html

Asimov, Isaac. The History of Physics. New York: Walker & Co., 1984.

Jones, Edwin R. and Richard L. Childers. Contemporary College Physics. Reading, MA: Addison-Wesley Publishing Co., 1993.

Kahan, Peter. Science Explorer: Motion, Forces, and Energy. Upper Saddle River, NJ: Prentice Hall, 2000.

Luehmann, April. Give Me Energy. June 12, 2003. Science and Mathematics Initiative for Learning Enhancement, Illinois Institute of Technology. October 15, 2003. http://www.iit.edu/~smile/ph9407.html

Nave, C.R. HyperPhysics. 2000. Department of Physics and Astronomy, Georgia State University. October 15, 2003. hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

The Atoms Family - The Mummy's Tomb – Raceways. Miami Museum of Science and Space Transit Planetarium. October 15, 2003. http://www.miamisci.org/af/sln/mummy/raceways.html

Other Related Information

Browse the NGSS Engineering-aligned Physics Curriculum hub for additional Physics and Physical Science curriculum featuring Engineering.

Contributors

Supporting program, acknowledgements.

The contents of this digital library curriculum were developed under a grant from the Fund for the Improvement of Postsecondary Education (FIPSE), U.S. Department of Education and National Science Foundation GK-12 grant no. 0338326. However, these contents do not necessarily represent the policies of the Department of Education or National Science Foundation, and you should not assume endorsement by the federal government.

Last modified: December 14, 2022

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Kinetic energy problems

When solving kinetic energy problems, you may be asked to find 3 variables. These variables are the kinetic energy, the mass, or the speed.

Problem # 1:

Suppose a car has 3000 Joules of kinetic energy. What will be its kinetic energy if the speed is doubled? What if the speed is tripled?

We already proved in kinetic energy lesson that whenever the speed is doubled, the kinetic energy is quadrupled or four times as big.

4 × 3000 = 12000

Therefore, the kinetic energy is going to be 12000 joules.

Let v be the speed of a moving object. Let speed =  3v after the speed is tripled. 

9 × 3000 = 27000

Therefore, the kinetic energy is going to be 27000 joules.

Problem # 2:

Calculate the kinetic energy of a 10 kg object moving with a speed of 5 m/s. Calculate the kinetic energy again when the speed is doubled.

Tricky kinetic energy problems

Problem # 3: 

Suppose a rat and a rhino are running with the same kinetic energy. Which one do you think is going faster?

The only tricky and hard part is to use the kinetic energy formula to solve for v.

Multiply both sides by 2

Problem # 4: 

The kinetic energy of an object is 8 times bigger than the mass. Is it possible to get speed of the object?

Think carefully and try to solve this problem yourself.

Potential energy

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7.2 Kinetic Energy and the Work-Energy Theorem

Learning objectives.

By the end of this section, you will be able to:

• Explain work as a transfer of energy and net work as the work done by the net force.
• Explain and apply the work-energy theorem.

Work Transfers Energy

What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in Figure 7.2 (a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up stairs in Figure 7.2 (d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in Figure 7.2 (e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work.

In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion.

Net Work and the Work-Energy Theorem

We know from the study of Newton’s laws in Dynamics: Force and Newton's Laws of Motion that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion.

Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work on an object. The net work can be written in terms of the net force on an object. F net F net . In equation form, this is W net = F net d cos θ W net = F net d cos θ where θ θ is the angle between the force vector and the displacement vector.

Figure 7.3 (a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an F cos θ F cos θ vs. d d graph. In this case, F cos θ F cos θ is constant. You can see that the area under the graph is F d cos θ F d cos θ , or the work done. Figure 7.3 (b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force ( F cos θ ) i ( ave ) ( F cos θ ) i ( ave ) . The work done is ( F cos θ ) i ( ave ) d i ( F cos θ ) i ( ave ) d i for each strip, and the total work done is the sum of the W i W i . Thus the total work done is the total area under the curve, a useful property to which we shall refer later.

Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in Figure 7.4 .

The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force F app F app and the horizontal friction force f f . Thus, as expected, the net force is parallel to the displacement, so that θ = 0º θ = 0º and cos θ = 1 cos θ = 1 , and the net work is given by

The effect of the net force F net F net is to accelerate the package from v 0 v 0 to v v . The kinetic energy of the package increases, indicating that the net work done on the system is positive. (See Example 7.2 .) By using Newton’s second law, and doing some algebra, we can reach an interesting conclusion. Substituting F net = ma F net = ma from Newton’s second law gives

To get a relationship between net work and the speed given to a system by the net force acting on it, we take d = x − x 0 d = x − x 0 and use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance d d if the acceleration has the constant value a a ; namely, v 2 = v 0 2 + 2 ad v 2 = v 0 2 + 2 ad (note that a a appears in the expression for the net work). Solving for acceleration gives a = v 2 − v 0 2 2 d a = v 2 − v 0 2 2 d . When a a is substituted into the preceding expression for W net W net , we obtain

The d d cancels, and we rearrange this to obtain

This expression is called the work-energy theorem , and it actually applies in general (even for forces that vary in direction and magnitude), although we have derived it for the special case of a constant force parallel to the displacement. The theorem implies that the net work on a system equals the change in the quantity 1 2 mv 2 1 2 mv 2 . This quantity is our first example of a form of energy.

The Work-Energy Theorem

The net work on a system equals the change in the quantity 1 2 mv 2 1 2 mv 2 .

The quantity 1 2 mv 2 1 2 mv 2 in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass m m moving at a speed v v . ( Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the translational kinetic energy,

is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together.

We are aware that it takes energy to get an object, like a car or the package in Figure 7.4 , up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy.

Example 7.2

Calculating the kinetic energy of a package.

Suppose a 30.0-kg package on the roller belt conveyor system in Figure 7.4 is moving at 0.500 m/s. What is its kinetic energy?

Because the mass m m and speed v v are given, the kinetic energy can be calculated from its definition as given in the equation KE = 1 2 mv 2 KE = 1 2 mv 2 .

The kinetic energy is given by

Entering known values gives

which yields

Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves.

Example 7.3

Determining the work to accelerate a package.

Suppose that you push on the 30.0-kg package in Figure 7.4 with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N.

(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force.

Strategy and Concept for (a)

This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See Figure 7.4 .) As expected, the net work is the net force times distance.

Solution for (a)

The net force is the push force minus friction, or F net = 120 N – 5 . 00 N = 115 N F net = 120 N – 5 . 00 N = 115 N . Thus the net work is

Discussion for (a)

This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force.

Strategy and Concept for (b)

The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work.

Solution for (b)

The applied force does work.

The friction force and displacement are in opposite directions, so that θ = 180º θ = 180º , and the work done by friction is

So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively,

The total work done as the sum of the work done by each force is then seen to be

Discussion for (b)

The calculated total work W total W total as the sum of the work by each force agrees, as expected, with the work W net W net done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach.

Example 7.4

Determining speed from work and energy.

Find the speed of the package in Figure 7.4 at the end of the push, using work and energy concepts.

Here the work-energy theorem can be used, because we have just calculated the net work, W net W net , and the initial kinetic energy, 1 2 m v 0 2 1 2 m v 0 2 . These calculations allow us to find the final kinetic energy, 1 2 mv 2 1 2 mv 2 , and thus the final speed v v .

The work-energy theorem in equation form is

Solving for 1 2 mv 2 1 2 mv 2 gives

Solving for the final speed as requested and entering known values gives

Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package.

Example 7.5

Work and energy can reveal distance, too.

How far does the package in Figure 7.4 coast after the push, assuming friction remains constant? Use work and energy considerations.

We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has removed all of the package’s kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing.

The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force, and it acts opposite to the displacement, so θ = 180º θ = 180º . To reduce the kinetic energy of the package to zero, the work W fr W fr by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Thus W fr = − 95 . 75 J W fr = − 95 . 75 J . Furthermore, W fr = f d ′ cos θ = – f d ′ W fr = f d ′ cos θ = – f d ′ , where d ′ d ′ is the distance it takes to stop. Thus,

This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the work done by friction is negative (the force is in the opposite direction of motion), so it removes the kinetic energy.

Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics and dynamics alone.

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Middle school physics - NGSS

Course: middle school physics - ngss   >   unit 3, potential energy.

• Understand: potential energy

Key points:

• Potential energy is energy that has the potential to become another form of energy . An object’s potential energy depends on its physical properties and position in a system.
• Potential energy comes in many forms, such as:

Gravitational potential energy due to an object’s mass and position in a gravitational field Magnetic potential energy due to a magnetic object’s position in a magnetic field Electric potential energy due to the size of an electric charge and its position in an electric field Elastic potential energy of a spring or rubber band that is stretched

• Changing an object’s position can change its potential energy. This depends on the forces between objects.

For example, if two objects attract each other, moving them apart will increase their potential energies. If two objects repel each other, moving them apart will decrease their potential energies.

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Potential Energy Problems & Solutions for high schools

Some simple problems about potential energy involving gravitational and spring potential are provided and solved. All of these questions are for high school students.

Potential Energy: Introduction

Potential energy is the energy that an object will have by virtue of its position relative to a reference point. Potential because it has the potential to do work when it is released from that position. 

Gravitational Potential Energy Problems

Problem: A rock with a mass of 2 kg is dropped from a height of 10 meters. Calculate the gravitational potential energy of the rock just before it hits the ground.

Solution: The gravitational potential energy due to Earth's gravity is defined as the product of an object's weight $mg$ and its height $y$ above some reference level (such as the ground). \begin{align*} \text{PE}_g &=mgh \\ &=2\times 9.8 \times 10 \\ &= 196\,\rm J \end{align*} Therefore, the gravitational potential energy of the rock just before it hits the ground is 196 Joules.

Problem:  A block of mass 2 kg is placed on an inclined plane with an angle of inclination of 30 degrees. Calculate the gravitational potential energy of the block when it is at a height of 5 meters above the ground.

Solution : As defined, the gravitational potential energy possessed by an object located near the Earth's surface is the object's weight $mg$ times its vertical position $h$ above the surface level, $\text{PE}_g=mgh$ . 

It does not matter how this object was elevated to that height. It might have been lifted by force $F$ directly straight up, pushed with force $F'$ up an incline of length $L$, or lifted with force $F$ up over a stair.

At a height of $h$ above the ground on the inclined plane, the object has the potential energy of $mgh$. 

Plugging in the given values into this equation, we get: \begin{align*} \text{PE}_g &=mg\times h\times \sin\theta \\\\ &=(2\times 9.8)(5) \times \sin 30^\circ \\\\ &=49\,\rm J\end{align*} 

Problem: When a block is pushed up an incline, how does its potential energy compare to being raised vertically to the same height?

Solution : We must do work to raise an object against Earth's gravity. Gravitational potential energy is defined as the amount of work done against gravity in lifting an object upward times the height $h$ through which this work is applied. 

On the other hand, work is defined as the required force $F$ to move an object times the distance it is moved $d$ (in this case, vertically). \[\text{work}=\text{force}\times \text{distance}\] When it comes to gravitational potential energy, the upward force is the weight of the object, $mg$. 

Consider a scenario, in which you are going to lift an object straight up to a height of $h$. The amount of work you do to overcome Earth's gravity is equal to the object's weight, $mg$, multiplied by the height $h$ relative to the ground or the floor of a building. 

In this case, the amount of gravitational potential energy possessed by the object is given by \[\text{PE}_g=mgh\] In the second scenario, imagine that the object needs to be pushed up an incline of length $L$. From problems involving inclined planes, you have learned that the force required to push a block up a frictionless incline is given by $F=mg\sin\theta$. 

Now, when a block is pushed up an incline instead of being lifted vertically, it requires less force, $mg\sin\theta$, compared to lifting it straight up, $mg$. This less force is exerted on the object over a longer distance $L$, along the incline (although, it reaches the same height), compared to exerting a force of $mg$ over height $h$. 

From these observations, we can conclude that the work done in pushing the object up the incline is equal to the work done in lifting it straight up to the same height $h$.  \[W_{incline}=W_{straight}\] Because, in both scenarios, the object is elevated to the same height, the gravitational potential in both cases is equal. \[\text{PE}_{g-incline}=\text{PE}_{g-straight}\] 

Problem: An object weighing 100 N slides down a frictionless ramp inclined at an angle of 40 degrees from the horizontal plane. If its initial height above the ground is measured at 6 meters, calculate its final gravitational potential energy when it reaches the bottom of the ramp.

Solution : What matters for calculating the gravitational potential energy is the position of that object relative to a reference point. Considering the reference point to be at the Earth's surface, the object is initially at height $6\,\rm m$ above the surface. Thus, its potential energy is found to be \begin{align*} \text{PE}_g&=mgh \\ &=100\times 6 \\ &=600\,\rm J\end{align*} Thus, the object initially has 600 joules of potential energy, and when it reaches the bottom of the plane, its position relative to the surface becomes zero, $y_f=0$. As a result, its final potential energy is zero. 

Problem A crate weighing 145 N is pushed on an inclined plane with an angle of inclination of 25 degrees for 2 meters. Calculate its gravitational potential energy.

Solution : We are not given the height of the incline or the amount of height by which the crate is vertically displaced. The only given values are the distance traveled over the incline $L$ and its angle $\theta$.

When an object is pushed a distance of $L$ over an incline (wedge, ramp, etc.) angled at $\theta$ to a final height of $h$ above the surface, its potential energy is calculated as $\text{PE}_g=mg\times \underbrace{L \times \sin\theta}_{h}$.

Actually, trigonometry gives us the height $h$ at which the crate is elevated relative to the ground on the incline. \begin{align*} \text{PE}_g &=mg\times\underbrace{L\sin\theta}_{h} \\\\ &= 145\times (2\sin 25^\circ) \\\\ &=122.5\,\rm J\end{align*} In fact, the crate has been elevated a distance of about $h=L\sin\theta=84\,\rm cm$ above the ground.  

Problem: On a single day, a mountain climber weighing 75 kg successfully scales a vertical cliff, starting from an elevation of 1500 m and reaching the summit at 2400 m. The following day, she descends from the summit to the cliff's base, situated at an elevation of 1350 m. What is the alteration in her gravitational potential energy during (a) the initial day and (b) the subsequent day?

Solution : To solve any potential energy problem, first consider a reference level as the origin and calculate the position of the object relative to it. 

The change in the gravitational potential energy is given by $\rm \Delta PE_g=mg\Delta h$, where $\Delta h$ is the change in the elevation of the object while going up or down.  

(a) Assuming the Earth's surface as the origin, The climber's initial and final positions are $y_i=1500\,\rm m$ and $y_f=2400\,\rm m$. The change in the potential energy of the climber during the ascending time interval is \begin{align*} \rm \Delta PE_g&=mg\underbrace{(y_f-y_i)}_{\Delta h} \\\\ &=75\times 9.8\times (2400-1500) \\\\ &=661500\,\rm J\end{align*} Therefore, during the initial day, there is an alteration of 661,500 Joules in her gravitational potential energy.

(b) On the subsequent day, the climber descends from the summit at an elevation of $y_i=2400\,\rm m$ to the cliff's base at an elevation of $y_f=1350\,\rm m$. The change in its height is \[\Delta h=y_f-y_i=1350-2400=-1050\,\rm m\] Notice that since the climber is moving down, $\Delta h$ will be negative. Therefore, the change in its potential energy is \begin{align*} \rm \Delta PE_g&=mg\Delta h \\\\ &=75\times 9.8\times (-1050) \\\\ &=-764250\,\rm J\end{align*} Therefore, during the subsequent day, there is an alteration of -764,250 Joules (or -764.25 kJ) in her gravitational potential energy. The negative sign indicates a decrease in potential energy as she descends.

The two next potential problems are a bit challenging: 

Problem: A pocket, weighing 1.25 kg, is raised by a person who is 1.75 m tall to a height of 2.20 m above the ground. Determine the potential energy of the pocket with respect to (a) the ground and (b) the top of the person's head. Additionally, explain how the work performed by the person is connected to the solutions in parts (a) and (b).

Solution : Remember that to find the potential energy of an object, we need a reference point to determine the position of the object relative to that point. The aim of this question is to illustrate the importance of choosing a reference level when solving potential problems. 

(a) In this part, we are asked to place the origin on the ground. Hence, the pocket is positioned at a height of $h=2.2\,\rm m$ from the ground, and its corresponding gravitational potential energy is weight times the vertically displaced position \begin{align*} \text{PE}_g &=mgh \\ &=1.25\times 9.8\times 2.2 \\ &=26.96\,\rm J\end{align*}  (b) Now, place the origin of the potential energy at the top of the person's head. In this situation, the pocket $2.2\,\rm m$ above the surface, rests at a height of $\Delta h=2.2-1.7=0.5\,\rm m$ from the person's head. Thus, the potential energy stored in the pocket relative to the person's head will be \begin{align*} \text{PE}_g &=mgh \\ &=1.25\times 9.8\times 0.5 \\ &=6.125\,\rm J\end{align*} The work done by the person $W_{ext}$ to lift the pocket is the change in its gravitational potential energy over the course of which the person applies the force \[W_{ext}=\Delta \text{PE}_g\] The person exerted a force to lift the pocket straight up a distance of $2.2\,\rm m$ from the ground. Therefore, in lifting this pocket to the final height of 2.2 meters, the person does $26.96\,\rm J$ work. 

The second part would not make sense. Since the total work done by an external agent (person) over a vertical distance depends on the initial and final heights, which in this case are the ground and 2.2 m height. 

Spring Potential Energy

A compressed or stretched spring can do (physical) work on the object attached to it by displacing it. Because of this ability, potential energy can be associated with the spring force, which is called spring or elastic potential energy. 

The amount of spring potential energy is directly proportional to the square of the distance $x$ the spring is stretched or compressed from equilibrium. \[\text{PE}_s=\frac 12 kx^2\] where $K$ is the spring constant.

Problem: A spring with a spring constant of 100 N/m is compressed by a distance of 0.2 meters. Calculate the elastic potential energy stored in the spring.

Solution : The given values are $k=100\,\rm N/$ and the amount of compression of the spring from equilibrium (or unstretched length) $x=0.2\,\rm m$. Plugging in the given numerical values into this, we get \begin{align*} \text{PE}_s &=\frac 12 kx^2 \\\\ &=\frac 12 \times 100 \times (0.2)^2 \\\\ &=2\,\rm J\end{align*} Therefore, the elastic potential energy stored in the spring is $2\,\rm N\cdot m$ or 2 joules. 

Problem: A rubber band has a spring constant of 50 N/m and is stretched by a distance of 0.1 meters. Calculate the elastic potential energy stored in the rubber band.

Solution : The rubber band acts like a spring since when it is stretched or compressed, it has the ability or potential to work on a body attached to it. Similar to spring, the potential energy stored in the rubber band is determined by $\text{PE}_s=\frac 12 kx^2$, where $x$ is the amount of distance that the band is displaced from its unstretched length. 

Plugging in the numerical values into this equation, we get \begin{align*} \text{PE}_s&=\frac 12 kx^2 \\\\ &=\frac 12 \times 50 \times (0.1)^2 \\\\ &=0.25\,\rm J \end{align*} 

There is another article that goes into the problems on spring potential energy in detail. 

Problem: A bowstring has a spring constant of 200 N/m and is pulled back by a distance of 0.3 meters to shoot an arrow. Calculate the elastic potential energy stored in the bowstring just before releasing it.

Solution : When a bow is stretched, energy is stored in it. Then, this stored energy can do work on an arrow by shooting it. The amount of potential energy stored in the bow is determined as follows \begin{align*} \text{PE}_s&=\frac 12 kx^2 \\\\ &=\frac 12 \times 200 \times 0.3 \\\\ &=30\,\rm J\end{align*}  

Conservation of Energy

Problem: A roller coaster starts at the top of a hill with an initial potential energy of 5000 J. As it descends, it encounters friction and air resistance, causing its mechanical energy to decrease. Calculate the final kinetic energy of the roller coaster when it reaches the bottom of the hill.

Solution : To solve this problem, we need to consider the conservation of mechanical energy. The total mechanical energy of the roller coaster is the sum of its potential energy (PE) and kinetic energy (KE). According to the law of conservation of mechanical energy, the total mechanical energy remains constant as long as no external forces are acting on the system.

In this question, we were told that the only forces acting on the roller coaster are friction and air resistance, which are internal forces. Thus, according to the definition of conservation of mechanical energy, these forces cannot change the system's mechanical energy.

Given that the initial potential energy (PE) is $5000\,\rm J$, we can assume that at the top of the hill, all of this potential energy is converted into kinetic energy (KE) when it reaches the bottom.

Therefore, at the bottom of the hill, all $5000\,\rm J$ of potential energy will be converted into kinetic energy. Hence, the final kinetic energy (KE) when it reaches the bottom is $5000\,\rm J$.

Author : Dr. Ali Nemati Published : Aug 12, 2023

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• School Guide
• Class 11 Syllabus
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• Maths Notes Class 11
• Physics Notes Class 11
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• NCERT Solutions Class 11 Maths
• RD Sharma Solutions Class 11
• Math Formulas Class 11
• CBSE Class 11 Physics Notes

Chapter 1: Physical World

• What is Physics? Definition, History, Importance, Scope
• How is Physics related to Other Sciences?
• Fundamental Forces

Chapter 2: Units and Measurement

• System of Units
• Length Measurement
• Measurement of Area, Volume and Density
• Rounding Numbers
• Dimensional Analysis
• Significant Figures
• Errors in Measurement

Chapter 3: Motion in a Straight Line

• What is Motion?
• Distance and Displacement
• Speed and Velocity
• Acceleration
• Uniform Acceleration
• Sample Problems on Equation of Motion
• Solving Problems Based on Free Fall
• Relative Motion
• Relative Motion in One Dimension
• Relative Motion in Two Dimension
• Calculating Stopping Distance and Reaction Time

Chapter 4: Motion in a Plane

• Scalar and Vector
• Vector Operations
• Product of Vectors
• Scalar Product of Vectors
• Dot and Cross Products on Vectors
• Position and Displacement Vectors
• Average Velocity
• Motion in Two Dimension
• Projectile Motion
• Uniform Circular Motion
• Centripetal Acceleration
• Motion in Three Dimensions

Chapter 5: Laws of Motion

• Contact and Non Contact Forces
• Inertia Meaning
• Law of Inertia
• What is Impulse?
• Solving Problems in Mechanics
• Linear Momentum of a System of Particles
• Newton's Second Law of Motion: Definition, Formula, Derivation, and Applications
• Laws of Conservation of Momentum
• What is Equilibrium? - Definition, Types, Laws, Effects
• Law of Action and Reaction
• Types of Friction - Definition, Static, Kinetic, Rolling and Fluid Friction
• Increasing and Reducing Friction
• Factors Affecting Friction
• Motion Along a Rough Inclined Plane
• Problems on Friction Formula
• Centripetal and Centrifugal Force
• Solved Examples on Dynamics of Circular Motion
• Dynamics of Circular Motion
• Motion in a Vertical Circle

Chapter 6: Work, Energy and Power

• Work Energy Theorem

Practice Problems on Kinetic Energy

• Work Done by a Variable Force
• What is Potential Energy?
• Potential Energy of a Spring
• Practice Problems on Potential Energy
• Law of Conservation of Energy
• Difference Between Work and Energy
• Types of Collisions
• Collisions in One Dimension
• Collisions in Two Dimensions

Chapter 7: Systems of Particles and Rotational Motion

• Rigid Body - Definition, Rotation, Angular Velocity, Momentum
• Motion of a Rigid Body
• Centre of Mass
• Center of Mass of Different Objects
• Motion of Center of Mass
• Torque and Angular Momentum
• What are Couples? Definition, Moment of Couple, Applications
• What is the Principle of Moments?
• Centre of Gravity
• Moment of Inertia
• Kinematics of Rotational Motion
• Dynamics of Rotational Motion
• Angular Momentum in Case of Rotation About a Fixed Axis
• Rolling Motion
• Relation between Angular Velocity and Linear Velocity

Chapter 8: Gravitation

• Kepler's Laws of Planetary Motion
• Universal Law of Gravitation
• Factors affecting Acceleration due to Gravity
• Variation in Acceleration due to Gravity
• Potential Energy
• Escape Velocity
• Binding Energy of Satellites
• Weightlessness

Chapter 9: Mechanical Properties of Solids

• Elastic Behavior of Materials
• Elasticity and Plasticity
• Stress and Strain
• Hooke's Law
• Stress-Strain Curve
• Young's Modulus
• Shear Modulus and Bulk Modulus
• Poisson's Ratio
• Elastic Potential Energy
• Stress, Strain and Elastic Potential Energy

Chapter 10: Mechanical Properties of Fluids

• Fluid Pressure
• Pascal's Law
• Variation of Pressure With Depth
• How to calculate Atmospheric Pressure?
• Hydraulic Machines
• Streamline Flow
• Bernoulli's Principle
• Bernoulli's Equation
• What is Viscosity?
• Stoke's Law
• Reynolds Number
• Surface Tension

Chapter 11: Thermal Properties of Matter

• Difference between Heat and Temperature
• Temperature Scales
• Ideal Gas Law
• Thermal Expansion
• Heat Capacity
• Calorimetry
• Change of State of Matter
• Latent Heat
• Thermal Conduction
• Sample Problems on Heat Conduction
• What is Radiation - Types, Scource, Ionizing and Non-Ionizing Radiation
• Greenhouse Effect
• Newton's Law of Cooling

Chapter 12: Thermodynamics

• Thermodynamics
• Zeroth Law of Thermodynamics
• Heat, Internal Energy and Work
• First Law of Thermodynamics
• Specific Heat Capacity
• Thermodynamic State Variables and Equation of State
• Thermodynamic Processes
• Second Law of Thermodynamics
• Reversible and Irreversible Processes

Chapter 13: Kinetic Theory

• Behavior of Gas Molecules - Kinetic Theory, Boyle's Law, Charles's Law
• Molecular Nature of Matter - Definition, States, Types, Examples
• Kinetic Theory of Gases
• Mean Free Path - Definition, Formula, Derivation, Examples

Chapter 14: Oscillations

• Oscillatory and Periodic Motion
• Simple Harmonic Motion
• Force Law for Simple Harmonic Motion
• Displacement in Simple Harmonic Motion
• Velocity and Acceleration in Simple Harmonic Motion
• Energy in Simple Harmonic Motion
• Some Systems executing Simple Harmonic Motion

Chapter 15: Waves

• Introduction to Waves - Definition, Types, Properties
• Speed of a Travelling Wave
• Reflection of Waves
• Properties of Waves
• Principle of Superposition of Waves
• Energy in Wave Motion
• Doppler Effect - Definition, Formula, Examples

When work is done by a force on an object. It acquires energy, it can be any form. Energy can take on many forms and can be converted from one form to another form. Potential energy, electric potential energy, kinetic energy, etc. are some examples of different types of energy. Kinetic energy comes when the object starts moving. This energy is due to motion. Although this energy is due to motion, this energy is not created. It is usually converted from one type of energy to another type. Let’s look at this concept in detail. 

Kinetic Energy

If an object is stationary, and we want to put that object into motion. We need to apply force. Any type of acceleration requires some force. When this force is applied, work is done on the object. When the work is done on an object, this means energy is getting transferred to the object is one form or another. Force can be removed once the object is in motion, but till the time force was applied on the object. The work that was done during that time is converted into energy. 

Kinetic energy is the energy an object acquires by virtue of its motion. 

This energy can be transferred from one object to another. For example, a moving ball hitting a stationary ball might cause the other ball to move. In this situation, some kinetic energy of the ball is transferred to another ball. 

Formula of Kinetic Energy 

To calculate the kinetic energy of the object, let’s consider a scenario where a force F, is acting on an object of mass M. In this case, the object starts moving with the acceleration “a” and covers a distance of “d”. 

Work done in this case will be, 

The acceleration “a” can be replaced using an equation of motion. 

v 2 = u 2 + 2a.d

⇒v 2 – u 2 = 2a.d

Substituting the value of “d” in the equation, 

So, this whole work done is converted into the K.E of the object. 

In case, initial velocity u = 0, 

One can also say, the network work done on the system is equal to the change in kinetic energy of the object. 

Note:  1. Kinetic energy depends on the velocity of the object squared. This means, when th velocity of the object is doubled, its kinetic energy becomes four times.  2. K.E must always have zero or positive values.  3. Kinetic energy is a scalar quantity, and it is expressed in Joules. 

Sample Problems

Question 1: A ball has a mass of 2Kg, suppose it travels at 10m/s. Find the kinetic energy possessed by it. 

Answer: 

Given: m = 2Kg, and v = 10m/s  The KE is given by,  K.E =  K.E =  ⇒ K.E =  ⇒ K.E = 100J

Question 2: A ball has a mass of 10Kg, suppose it travels at 100m/s. Find the kinetic energy possessed by it. 

Given: m = 10Kg, and v = 100m/s  The KE is given by,  K.E =  K.E =  ⇒ K.E =  ⇒ K.E = 50000J

Question 3: A spaceship has a mass of 20000Kg, suppose it travels at 10m/s. Find the kinetic energy possessed by it. 

Given: m = 20000Kg, and v = 10m/s  The KE is given by,  K.E =  K.E =  ⇒ K.E =  ⇒ K.E = 10 6 J

Question 4: Work done by a force on a moving object is 100J. It was traveling at a speed of 2 m/s. Find the new speed of the object if the mass of the object is 2Kg.

Given: W = 100J Work done by the force is equal to the change in kinetic energy.  W =  Given, u = 2 m/s and v = ?, m = 2kg.  Plugging the values in the given equation,  W =  ⇒  ⇒

Question 5: Work done by a force on a moving object is -50J. It was traveling at a speed of 10m/s. Find the new speed of the object if the mass of the object is 2Kg.

Given: W = -50J Work done by the force is equal to the change in kinetic energy.  W =  Given, u = 10m/s and v = ? . m = 2kg.  Plugging the values in the given equation,  W =  ⇒  ⇒  The speed is decreased because the work done was negative. This means that the force was acting opposite to the block and velocity was decreased. 

Question 6: Suppose a 1000Kg was traveling at a speed of 10m/s. Now, this mass transfers all its energy to a mass of 10Kg. What will be the velocity of the 10Kg mass after being hit by it? 

KE is given by the formula, K.E =    KE of the heavier object  M =1000Kg and v = 10m/s  K.E =    ⇒ K.E =  ⇒K.E = 50,000J  Now this energy is transferred to another ball.  m = 10Kg and v = ? 50,000 =  ⇒ 10,000 = v 2 ⇒ v = 100 m/s

Question 7: Suppose a 10Kg was traveling at a speed of 100m/s. Now, this mass transfers all its energy to a mass of 20Kg. What will be the velocity of the 20Kg mass after being hit by it? 

KE is given by the formula, K.E =    KE of the heavier object  M =10Kg and v = 100m/s  K.E =    ⇒ K.E =  ⇒K.E = 50,000J  Now this energy is transferred to another ball.  m = 20Kg and v = ? 50,000 =  ⇒ 5000 = v 2 ⇒ v = 50√2 m/s

Question 8: Suppose a 10Kg was traveling at a speed of 100m/s. Now, this mass transfers all its energy to a mass of 20Kg. What will be the velocity of the 20Kg mass after being hit by it? 

Question 9: Suppose a 10Kg was kept at 20m height. Now, this block is dropped. Find out the velocity of the block just before it hits the ground.

The block of 10Kg is kept at a height of 20m.  The potential energy of the block will be,  P.E = mgh  Here m = 10, g = 10m/s 2 and h = 20m.  P.E = mgh  ⇒ P.E = (10)(10)(20)  ⇒ P.E = 2000J  Now, this energy is converted completely into KE.  KE = PE  ⇒2000 =  Given m = 10Kg,  ⇒2000 =  ⇒400 = v 2 v = 20m/s 

Question 10: Suppose a rock of 100Kg was kept at 80m height. Now, this block is dropped. Find out the velocity of the block just before it hits the ground.

The block of 10Kg is kept at a height of 20m.  The potential energy of the block will be,  P.E = mgh  Here m = 100, g = 10m/s 2 and h = 80m.  P.E = mgh  ⇒ P.E = (100)(10)(80)  ⇒ P.E = 80000J  Now, this energy is converted completely into KE.  KE = PE  ⇒80000 =  Given m = 100Kg,  ⇒80000 =  ⇒1600 = v 2 v = 40m/s 

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9.6: Electric Potential and Potential Energy

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• Page ID 46934

Learning Objectives

• Define electric potential and electric potential energy.
• Describe the relationship between electric potential difference and electric field.
• Describe the relationship between electric potential and electrical potential energy.
• Explain electron volt and its usage in submicroscopic process.

When a free positive charge  \(q\)  is accelerated by an electric field, such as shown in  Figure  \(\PageIndex{1}\), it is given kinetic energy. The process is analogous to an object being accelerated by a gravitational field. It is as if the charge is going down an electrical hill where its electric potential energy is converted to kinetic energy. Let us explore the work done on a charge  \(q\)  by the electric field in this process, so that we may develop a definition of electric potential energy.

The electrostatic or Coulomb force is conservative, which means that the work done on  \(q\)  is independent of the path taken. This is exactly analogous to the gravitational force in the absence of dissipative forces such as friction. When a force is conservative, it is possible to define a potential energy associated with the force, and it is usually easier to deal with the potential energy (because it depends only on position) than to calculate the work done directly from force (\(W=\langle\boldsymbol{F}, \boldsymbol{d}\rangle\), where d is displacement and F is the force).

We use the letters PE to denote electric potential energy, which has units of joules (J). The change in potential energy, \(\triangle \mathrm{PE}\), is crucial, since the work done by a conservative force is the negative of the change in potential energy; that is, \(W=-\Delta \mathrm{PE}\). For example, work \(W\) done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative \(\triangle \mathrm{PE}\). There must be a minus sign in front of \(\triangle \mathrm{PE}\) to make \(W\) positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

Definition: POTENTIAL ENERGY

\(W=-\Delta \mathrm{PE}\). For example, work \(W\) done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative \(\Delta \mathrm{PE}\). There must be a minus sign in front of \(\Delta \mathrm{PE}\) to make \(W\) positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

Gravitational potential energy and electric potential energy are quite analogous. Potential energy accounts for work done by a conservative force and gives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. It is much more common, for example, to use the concept of voltage (related to electric potential energy) than to deal with the electric field (related to Coulomb force) directly.

Given some conservative force \(F\) and displacement \(d\) under the force, the work done and the change in potential energy can be calculated as, \(W=\langle\boldsymbol{F}, \boldsymbol{d}\rangle\) and \(\Delta \mathrm{PE}=-\mathrm{W}=\langle-\mathbf{F}, \boldsymbol{d}\rangle\). For electric force, the force is given by the product of electric charge and the electric field, \(\boldsymbol{F}=q \boldsymbol{E}\), where \(q\) is the charge experiencing the force and \(E\) is the electric field at the location of the charge. So the potential energy change due to work done by electric force is \(\Delta \mathrm{PE}=q(\langle-\mathbf{E}, \boldsymbol{d}\rangle)\). If we define change in  electric potential  \(V\) as \(\Delta V=\langle-\mathbf{E}, \boldsymbol{d}\rangle\), then the electric potential energy PEPE is simply expressed in terms of electric potential, \(\mathrm{PE}=q V\), or,

\[V=\frac{\mathrm{PE}}{q}, \nonumber \]

electric potential energy per charge.

Definition: ELECTRIC POTENTIAL

Electric potential is the electric potential energy per unit charge.

\[V=\frac{\mathrm{PE}}{q} \nonumber\]

With potential energy, the case often is that its value at a single point has no significant meaning but what is important is the difference in potential energy. From the difference in potential energy, we are able to calculate other quantities, such as change in kinetic energy (if no force other than the conservative force acts) or work needing to be done by other forces (if other forces act). So likewise, rather than the electric potential itself, we are often interested in difference in electric potential \(\Delta V\) between two points, where,

\[\Delta V=V_{\mathrm{B}}-V_{\mathrm{A}}=\frac{\Delta \mathrm{PE}}{q}. \nonumber \]

The  potential difference  between points A and B, \(V_{\mathrm{B}}-V_{\mathrm{A}}\), is thus defined to be the change in potential energy of a charge  \(q\)  moved from A to B, divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

\[1 \mathrm{~V}=1 \frac{\mathrm{J}}{\mathrm{C}} \nonumber \]

Definition: POTENTIAL DIFFERENCE

The potential difference between points A and B, \(V_{\mathrm{B}}-V_{\mathrm{A}}\), is defined to be the change in potential energy of a charge  q  moved from A to B, divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

\[1 \mathrm{~V}=1 \frac{\mathrm{J}}{\mathrm{C}} \nonumber\]

The familiar term  voltage  is the common name for potential difference. Keep in mind that whenever a voltage is quoted, it is understood to be the potential difference between two points. For example, every battery has two terminals, and its voltage is the potential difference between them. More fundamentally, the point you choose to be zero volts is arbitrary. This is analogous to the fact that gravitational potential energy has an arbitrary zero, such as sea level or perhaps a lecture hall floor.

In summary, the relationship between potential difference (or voltage) and electrical potential energy is given by

\[\Delta V=\frac{\Delta \mathrm{PE}}{q} \text { and } \Delta \mathrm{PE}=q \Delta V. \nonumber \]

POTENTIAL DIFFERENCE AND ELECTRICAL POTENTIAL ENERGY

The relationship between potential difference (or voltage) and electrical potential energy is given by

\[\Delta V=\frac{\Delta \mathrm{PE}}{q} \text { and } \Delta \mathrm{PE}=q \Delta V. \nonumber\]

The second equation is equivalent to the first.

Voltage is not the same as energy. Voltage is the energy per unit charge. Thus a motorcycle battery and a car battery can both have the same voltage (more precisely, the same potential difference between battery terminals), yet one stores much more energy than the other since \(\Delta \mathrm{PE}=q \Delta V\). The car battery can move more charge than the motorcycle battery, although both are 12 V batteries.

Example \(\PageIndex{1}\): Calculating Energy

Suppose you have a 12.0 V motorcycle battery that can move 5000 C of charge, and a 12.0 V car battery that can move 60,000 C of charge. How much energy does each deliver? (Assume that the numerical value of each charge is accurate to three significant figures.)

To say we have a 12.0 V battery means that its terminals have a 12.0 V potential difference. When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to \(\Delta \mathrm{PE}=q \Delta V\).

So to find the energy output, we multiply the charge moved by the potential difference.

For the motorcycle battery, \(q=5000 \ \mathrm{C}\) and \(\Delta V=12.0 \mathrm{~V}\). The total energy delivered by the motorcycle battery is

\[\begin{aligned} \Delta \mathrm{PE}_{\text {cycle }} &=(5000 \mathrm{C})(12.0 \mathrm{~V}) \\ &=(5000 \mathrm{C})(12.0 \mathrm{~J} / \mathrm{C}) \\ &=6.00 \times 10^{4} \mathrm{~J}. \end{aligned} \nonumber\]

Similarly, for the car battery, \(q=60,000 \ \mathrm{C}\) and

\[\begin{aligned} \Delta \mathrm{PE}_{\mathrm{car}} &=(60,000 \mathrm{C})(12.0 \mathrm{~V}) \\ &=7.20 \times 10^{5} \mathrm{~J}. \end{aligned} \nonumber\]

While voltage and energy are related, they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each is quite different. Note also that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as when headlights dim because of a low car battery. The energy supplied by the battery is still calculated as in this example, but not all of the energy is available for external use.

Note that the energies calculated in the previous example are absolute values. The change in potential energy for the battery is negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge—electrons in particular. The batteries repel electrons from their negative terminals (A) through whatever circuitry is involved and attract them to their positive terminals (B) as shown in  Figure  \(\PageIndex{2}\). The change in potential is \(\Delta V=V_{\mathrm{B}}-V_{\mathrm{A}}=+12 \mathrm{~V}\) and the charge \(q\) is negative, so that \(\Delta \mathrm{PE}=q \Delta V\) is negative, meaning the potential energy of the battery has decreased when \(q\) has moved from A to B.

Example \(\PageIndex{2}\): How Many Electrons Move through a Headlight Each Second?

When a 12.0 V car battery runs a single 30.0 \(W\) headlight, how many electrons pass through it each second?

To find the number of electrons, we must first find the charge that moved in 1.00 s. The charge moved is related to voltage and energy through the equation \(\Delta \mathrm{PE}=q \Delta V\). A 30.0 W lamp uses 30.0 joules per second. Since the battery loses energy, we have \(\Delta \mathrm{PE}=-30.0 \mathrm{~J}\) and, since the electrons are going from the negative terminal to the positive, we see that \(\Delta V=+12.0 \mathrm{~V}\).

To find the charge  \(q\)  moved, we solve the equation \(\Delta \mathrm{PE}=q \Delta V\):

\[q=\frac{\Delta \mathrm{PE}}{\Delta V}. \nonumber\]

Entering the values for \(\triangle \mathrm{PE}\) and \(\Delta V\), we get

\[q=\frac{-30.0 \mathrm{~J}}{+12.0 \mathrm{~V}}=\frac{-30.0 \mathrm{~J}}{+12.0 \mathrm{~J} / \mathrm{C}}=-2.50 \mathrm{C}. \nonumber\]

The number of electrons \(\mathrm{n}_{\mathrm{e}}\) is the total charge divided by the charge per electron. That is,

\[\mathrm{n}_{\mathrm{e}}=\frac{-2.50 \mathrm{C}}{-1.60 \times 10^{-19} \mathrm{C} / \mathrm{e}^{-}}=1.56 \times 10^{19} \text { electrons. } \nonumber\]

This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so many being present in ordinary systems. In fact, electricity had been in use for many decades before it was determined that the moving charges in many circumstances were negative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determine which is moving or whether both are moving.

The Electron Volt

The energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful X-rays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects.  Figure  \(\PageIndex{3}\) shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates as it might be in an old-model television tube or oscilloscope. The electron is given kinetic energy that is later converted to another form—light in the television tube, for example. (Note that downhill for the electron is uphill for a positive charge.) Since energy is related to voltage by \(\Delta \mathrm{PE}=q \Delta V\), we can think of the joule as a coulomb-volt.

Definition: ELECTRON VOLT

On the submicroscopic scale, it is more convenient to define an energy unit called the electron volt (eV), which is the energy given to a fundamental charge accelerated through a potential difference of 1 V .  In equation form,

\[\begin{aligned} 1 \mathrm{eV} &=\left(1.60 \times 10^{-19} \mathrm{C}\right)(1 \mathrm{~V})=\left(1.60 \times 10^{-19} \mathrm{C}\right)(1 \mathrm{~J} / \mathrm{C}) \\ &=1.60 \times 10^{-19} \mathrm{~J}. \end{aligned} \nonumber\]

An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electron accelerated through 50 V is given 50 eV. A potential difference of 100,000 V (100 kV) will give an electron an energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 V will be given 200 eV of energy. These simple relationships between accelerating voltage and particle charges make the electron volt a simple and convenient energy unit in such circumstances.

CONNECTIONS: ENERGY UNITS

The electron volt (eV) is the most common energy unit for submicroscopic processes. This will be particularly noticeable in the chapters on modern physics. Energy is so important to so many subjects that there is a tendency to define a special energy unit for each major topic. There are, for example, calories for food energy, and kilowatt-hours for electrical energy.

The electron volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron volts. For example, about 5 eV of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, it is given an energy of 30 keV (30,000 eV) and it can break up as many as 6000 of these molecules (\(30,000 \ \mathrm{eV} \div 5 \mathrm{eV} \text { per molecule }=6000 \text { molecules }\)). Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can, thus, produce significant biological damage.

Conservation of Energy

The total energy of a system is conserved if there is no net addition (or subtraction) of work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.

Mechanical energy  is the sum of the kinetic energy and potential energy of a system; that is, \(\mathrm{KE}+\mathrm{PE}=\text { constant }\). A loss of PE of a charged particle becomes an increase in its KE. Here PE is the electric potential energy. Conservation of energy is stated in equation form as

\[\mathrm{KE}+\mathrm{PE}=\text { constant } \nonumber \]

\[\mathrm{KE}_{\mathrm{i}}+\mathrm{PE}_{\mathrm{i}}=\mathrm{KE}_{\mathrm{f}}+\mathrm{PE}_{\mathrm{f}}, \nonumber \]

where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.

Example \(\PageIndex{3}\): Electrical Potential Energy Converted to Kinetic Energy

Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is accurate to three significant figures.)

We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we will check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms of energy to be \(\mathrm{KE}_{\mathrm{i}}=0, \mathrm{KE}_{\mathrm{f}}=1 / 2 m v^{2}, \mathrm{PE}_{\mathrm{i}}=q V, \text { and } \mathrm{PE}_{\mathrm{f}}=0\).

Conservation of energy states that

\[\mathrm{KE}_{\mathrm{i}}+\mathrm{PE}_{\mathrm{i}}=\mathrm{KE}_{\mathrm{f}}+\mathrm{PE}_{\mathrm{f}}. \nonumber\]

Entering the forms identified above, we obtain

\[q V=\frac{m v^{2}}{2}. \nonumber\]

We solve this for \(v\) :

\[v=\sqrt{\frac{2 q V}{m}}. \nonumber\]

Entering values for \(q\),\(V\), and \(m\) gives

\[\begin{aligned} v &=\sqrt{\frac{2\left(-1.60 \times 10^{-19} \mathrm{C}\right)(-100 \mathrm{~J} / \mathrm{C})}{9.11 \times 10^{-31} \mathrm{~kg}}} \\ &=5.93 \times 10^{6} \mathrm{~m} / \mathrm{s}. \end{aligned} \nonumber\]

Note that both the charge and the initial voltage are negative, as in  Figure  \(\PageIndex{3}\). We know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns.

Section Summary

• Electric potential is potential energy per unit charge.

\[.\Delta V=\frac{\Delta \mathrm{PE}}{q} \text { and } \Delta \mathrm{PE}=q \Delta V \nonumber\]

• Mechanical energy is the sum of the kinetic energy and potential energy of a system, that is, \(\mathrm{KE}+\mathrm{PE}\) This sum is a constant.