sample problem solving in potential and kinetic energy

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Potential And Kinetic Energy Example Problem – Work and Energy Examples

Potential energy is energy attributed to an object by virtue of its position. When the position is changed, the total energy remains unchanged but some potential energy gets converted into kinetic energy . The frictionless roller coaster is a classic potential and kinetic energy example problem.

The roller coaster problem shows how to use the conservation of energy to find the velocity or position or a cart on a frictionless track with different heights. The total energy of the cart is expressed as a sum of its gravitational potential energy and kinetic energy. This total energy remains constant across the length of the track.

Potential And Kinetic Energy Example Problem

A cart travels along a frictionless roller coaster track. At point A, the cart is 10 m above the ground and traveling at 2 m/s. A) What is the velocity at point B when the cart reaches the ground? B) What is the velocity of the cart at point C when the cart reaches a height of 3 m? C) What is the maximum height the cart can reach before the cart stops?

The total energy of the cart is expressed by the sum of its potential energy and its kinetic energy.

Potential energy of an object in a gravitational field is expressed by the formula

where PE is the potential energy m is the mass of the object g is the acceleration due to gravity = 9.8 m/s 2 h is the height above the measured surface.

Kinetic energy is the energy of the object in motion. It is expressed by the formula

KE = ½mv 2

where KE is the kinetic energy m is the mass of the object v is the velocity of the object.

The total energy of the system is conserved at any point of the system. The total energy is the sum of the potential energy and the kinetic energy.

Total E = KE + PE

To find the velocity or position, we need to find this total energy. At point A, we know both the velocity and the position of the cart.

Total E = KE + PE Total E = ½mv 2  + mgh Total E = ½m(2 m/s) 2  + m(9.8 m/s 2 )(10 m) Total E = ½m(4 m 2 /s 2 ) + m(98 m 2 /s 2 ) Total E = m(2 m 2 /s 2 ) + m(98 m 2 /s 2 ) Total E = m(100 m 2 /s 2 )

We can leave the mass value as it appears for now. As we complete each part, you will see what happens to this variable.

The cart is at ground level at point B, so h = 0 m.

Total E = ½mv 2  + mgh Total E = ½mv 2  + mg(0 m) Total E = ½mv 2

All of the energy at this point is kinetic energy. Since total energy is conserved, the total energy at point B is the same as the total energy at point A.

Total E at A = Total Energy at B m(100 m 2 /s 2 ) = ½mv 2

Divide both sides by m 100 m 2 /s 2 = ½v 2

Multiply both sides by 2 200 m 2 /s 2 = v 2

v = 14.1 m/s

The velocity at point B is 14.1 m/s.

At point C, we know only a value for h (h = 3 m).

Total E = ½mv 2 + mgh Total E = ½mv 2 + mg(3 m)

As before, the total energy is conserved. Total energy at A = total energy at C.

m(100 m 2 /s 2 ) = ½mv 2 + m(9.8 m/s 2 )(3 m) m(100 m 2 /s 2 ) = ½mv 2 + m(29.4 m 2 /s 2 )

Divide both sides by m

100 m 2 /s 2 = ½v 2 + 29.4 m 2 /s 2 ½v 2 = (100 – 29.4) m 2 /s 2 ½v 2 = 70.6 m 2 /s 2 v 2 = 141.2 m 2 /s 2 v = 11.9 m/s

The velocity at point C is 11.9 m/s.

The cart will reach its maximum height when the cart stops or v = 0 m/s.

Total E = ½mv 2 + mgh Total E = ½m(0 m/s) 2 + mgh Total E = mgh

Since total energy is conserved, the total energy at point A is the same as the total energy at point D.

m(100 m 2 /s 2 ) = mgh

100 m 2 /s 2  = gh

100 m 2 /s 2  = (9.8 m/s 2 ) h

The maximum height of the cart is 10.2 m.

A) The velocity of the cart at ground level is 14.1 m/s. B) The velocity of the cart at a height of 3 m is 11.9 m/s. C) The maximum height of the cart is 10.2 m.

This type of problem has one main key point: total energy is conserved at all points of the system. If you know the total energy at one point, you know the total energy at all points.

Related Posts

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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Animated Physics Lessons

Mechanical Energy Problem Solutions

Mechanical energy problems and solutions.

See examples of mechanical energy problems involving kinetic energy, potential energy, and the conservation of energy. Check your work with ours.

1. How much gravitational potential energy do you have when you lift a 15 N object 10 meters off the ground?

2. How much gravitational potential energy is in a 20 kg mass when 0.6 meters above the ground?

3. How much gravitational potential energy does a 35 kg boulder have when 30 meters off the ground?

4. How many times greater is an objects potential energy when three times higher?

If you need help on ratio problems click the link below:

Rule of Ones: analyzing equations to determine how other variables change

5.  How much kinetic energy does a 0.15 kg ball thrown at 24 m/s have?

6.  How many times greater is the kinetic energy of a ball that is going five times faster?

7.  How much kinetic energy does a 1.2 kg ball have the moment it hits the ground 3.5 meters below when it starts from rest?

I cancelled out the initial kinetic energy because:

• KE i = ½ mv f 2
• KE i = (½)(3.5)(0 2 ) = 0 J

I cancelled out the final potential energy because:

• PE f = mgh f
• PE f = (3.5)(9.8)(0) = 0 J

8.  How fast is a 1.2 kg ball traveling the moment it hits the ground 3.5 meters below when it starts from rest?

(Note: In many of these problems I could cancel out mass but did not since it was provided)

Since I did not cancel out mass I could answer the following questions if asked:

• How much mechanical energy did you have at the beginning? (41.6 J)
• How much kinetic energy did you have at the beginning? (0 J)
• How much potential energy did you have at the beginning? (41.6 J)
• How much potential energy do you have at the end? (0 J)

If I cancelled out mass in my work it would not show the actual initial potential energy since PE i = mgh and not just gh.

9.  A 3.5 kg ball fell from a height of 12 meters.  How fast is it traveling when its still 5 meters off the ground?

10. An 85kg roller coaster cart is traveling 4 m/s at the top of a hill 50 meters off the ground.  How fast is it traveling at top of a second hill 20 meters off the ground?

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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:

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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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Mechanics: Work, Energy and Power

Name:___________________________________________

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? _____________

Practice Problems for Kinetic and Potential Energy

Students also viewed

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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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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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. 

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

⇒ [Tex]\frac{v^2 – u^2}{2a}[/Tex]  = d

Substituting the value of “d” in the equation, 

⇒ W =  [Tex]m.d.\frac{v^2 – u^2}{2d}[/Tex]

⇒W =  [Tex]m.\frac{v^2 – u^2}{2}[/Tex]

⇒W =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex]

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

In case, initial velocity u = 0, 

K.E =  [Tex]\frac{1}{2}mv^2[/Tex]

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 =  [Tex]\frac{1}{2}mv^2[/Tex] K.E =  [Tex]\frac{1}{2}mv^2[/Tex] ⇒ K.E =  [Tex]\frac{1}{2}(2)(10)^2[/Tex] ⇒ 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 =  [Tex]\frac{1}{2}mv^2[/Tex] K.E =  [Tex]\frac{1}{2}mv^2[/Tex] ⇒ K.E =  [Tex]\frac{1}{2}(10)(100)^2[/Tex] ⇒ 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 =  [Tex]\frac{1}{2}mv^2[/Tex] K.E =  [Tex]\frac{1}{2}mv^2[/Tex] ⇒ K.E =  [Tex]\frac{1}{2}(20000)(10)^2[/Tex] ⇒ 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.

Answer:  Given: W = 100J Work done by the force is equal to the change in kinetic energy.  W =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex] Given, u = 2 m/s and v = ?, m = 2kg.  Plugging the values in the given equation,  W =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex] ⇒  [Tex]100 = \frac{1}{2}(2)(v^2 – 2^2)[/Tex] ⇒ [Tex]100 = v^2 – 2^2 \\ = 104 = v^2 \\ = v = \sqrt{104}  \text{ m/s}[/Tex] [Tex]v = \sqrt{104} \text{ m/s}[/Tex]

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 =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex] Given, u = 10m/s and v = ? . m = 2kg.  Plugging the values in the given equation,  W =  [Tex]\frac{1}{2}m(v^2 – u^2)[/Tex] ⇒  [Tex]-50 = \frac{1}{2}(2)(v^2 – 10^2)[/Tex] ⇒  [Tex]-50 = v^2 – 10^2 \\ = 50 = v^2 \\ = v = \sqrt{50}  \\ = v = 5\sqrt{2} \text{ m/s}[/Tex] 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 =  [Tex]\frac{1}{2}mv^2     [/Tex]   KE of the heavier object  M = 1000Kg and v = 10m/s  K.E =  [Tex]\frac{1}{2}mv^2     [/Tex]   ⇒ K.E =  [Tex]\frac{1}{2}(1000)(10)^2[/Tex] ⇒K.E = 50,000J  Now this energy is transferred to another ball.  m = 10Kg and v = ? 50,000 =  [Tex]\frac{1}{2}(10)v^2[/Tex] ⇒ 10,000 = v 2 ⇒ v = 100 m/s

Question 7: Suppose a 10Kg mass 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? 

To find the velocity of the 20 kg mass after being hit by the 10 kg mass, we use the principle of conservation of momentum. According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system. m 1 = 10kg v 1 = 100m/s m 2 =20kg v 2 = ? The total momentum before the collision (P initial) is : = m 1 x v 1 = 10 x 100 = 1000 Kgm/s According to the conservation of momentum: P initial = P final m 1 x v 1 = m 2 x v 2 10 x 100 = 20 x v 2 1000 = 20 x v 2 v 2 = 1000/20 = 50m/s So, the velocity of the 20 kg mass after being hit by the 10 kg mass is 50m/s.

Question 8: Suppose a 10Kg block 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 =  [Tex]\frac{1}{2}mv^2[/Tex] Given m = 10Kg,  ⇒2000 =  [Tex]\frac{1}{2}10v^2[/Tex] ⇒400 = v 2 v = 20m/s 

Question 9: 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 80m.  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 =  [Tex]\frac{1}{2}mv^2[/Tex] Given m = 100Kg,  ⇒80000 =  [Tex]\frac{1}{2}100v^2[/Tex] ⇒1600 = v 2 v = 40m/s 

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High School Physics : Calculating Potential Energy

Study concepts, example questions & explanations for high school physics, all high school physics resources, example questions, example question #1 : calculating potential energy.

At the top of the hill the skier has purely potential energy. At the bottom, she has purely kinetic energy. 

We can solve by understanding the conservation of energy. The skier's energy at the top of the hill will be equal to her energy at the bottom of the hill.

Using the equations for potential and kinetic energy, we can solve for the height of the hill.

The masses cancel, and we can plug in our final velocity and gravitational acceleration.

This formula solves for the change in height. The negative sign implies she travelled in a downward direction. Because the question is asking how tall the hill is, we use an absolute value.

Example Question #2 : Calculating Potential Energy

We can use conservation of energy to solve. The potential energy when the astronaut is holding the rock will be equal to the kinetic energy just before it strikes the surface.

Plug in the given values and solve.

Plug in the values, and solve for the potential energy.

The units for energy are Joules.

Example Question #4 : Calculating Potential Energy

Use the conservation of energy equation to solve for the potential energy at the top of the hill.

Plug in the values given to you and solve for the final height.

The maximum height will be when the ball has only potential energy, and no kinetic energy. Initially, the ball has only kinetic energy and no potential energy. We can set these values equal to each other due to conservation of energy.

The masses will cancel out.

Plug in the values that were given and solve for the height.

We are given the height and mass of the ball. Using the given values, we can solve for the potential energy.

Keep in mind that the displacement will be negative because the ball is traveling in the downward direction.

Since we know the mass, height, and acceleration from gravity, we can simply multiply to find the potential energy.

Example Question #9 : Calculating Potential Energy

Given the values for the mass, height, and gravity, we can solve using multiplication. Note that the height is negative because the book falls in the downward direction.

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Middle school physics

Course: middle school physics   >   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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Exercise \(\PageIndex{1}\)

A ball of mass \(m\) is dropped onto a vertical spring with spring constant \(k\) . The spring will compress until the ball comes to rest. How much will it compress if the ball is dropped from a height \(h\) above the spring?

The two forces acting on the ball are gravity and the spring force. Both are conservative, so we can use conservation of mechanical energy. We will find the energy of the ball when it is at a height \(h\) above the spring, and the energy of the ball when the spring is fully compressed. Then, we will use conservation of mechanical energy to determine the compression of the spring.

Remember that the total mechanical energy is the sum of the total potential energy and the kinetic energy, \(E=U+K\) . Let’s call the initial position of the ball \(A\) and the final position of the ball \(B\) . You will notice that we set up our coordinate system so that \(y\) is positive upwards, with \(y=0\) at the point where the ball comes into contact with the spring. We choose to define both the gravitational potential energy and spring potential energy so that they are zero at \(y=0\) .

Since the ball starts from rest, its kinetic energy is zero at position \(A\) . At this point, the ball is not touching the spring, so the potential energy from the spring force is zero. The mechanical energy of the ball at position \(A\) is simply equal to its gravitational potential energy:

\[\begin{aligned} E_A&=U_A+K_A\\[4pt] E_A&=mgh\end{aligned}\]

At position \(B\) , the ball is again at rest, so the kinetic energy of the ball is zero. Now that the ball is in contact with the spring, it will experience a force from the spring that can be modeled with a potential energy \(U(y)=\frac{1}{2}ky_1^2\) , where \(y_1\) is the distance between the rest position of the spring and its compressed length. At point \(B\) ( \(y=-y_1\) ), the ball will have both spring and gravitational potential energy, so its mechanical energy at position \(B\) is given by:

\[\begin{aligned} E_B&=U_B+K_B=U_B\\[4pt] U_B&=mg(-y_1)+\frac{1}{2}ky_1^2\\[4pt] E_B&=-mgy_1+\frac{1}{2}ky_1^2\end{aligned}\]

Since mechanical energy is conserved in this system (no non-conservative forces are doing work), we can now set \(E_A=E_B\) and solve for \(y_1\) :

\[\begin{aligned} E_A&=E_B\\[4pt] mgh&=-mgy_1+\frac{1}{2}ky_1^2\\[4pt] 0&=\frac{1}{2}ky_1^2-mgy_1-mgh\\[4pt]\end{aligned}\]

where in the last line we rewrote the expression as a quadratic equation. We can solve for \(y_1\) with the quadratic formula:

\[\begin{aligned} y_1=\frac{mg\pm\sqrt{(mg)^2-4(1/2k)(-mgh)}}{k}\\[4pt] y_1=\frac{mg\pm\sqrt{mg(mg+2kh)}}{k}\end{aligned}\]

We now have an expression for the amount the spring is compressed, \(y_1\) , in terms of our known values.

Exercise \(\PageIndex{2}\)

A simple pendulum consists of a mass \(m\) connected to a string of length \(L\) . The pendulum is released from an angle \(\theta_0\) from the vertical. Use conservation of energy to find an expression for the velocity of the mass as a function of the angle.

We are going to find a general expression for the energy of the system, and then use this expression to find the velocity at any point. There are two forces acting on the mass:

The force of tension (from the string). This force is perpendicular to the direction of motion at any point, so it does no work on the mass.

The force of gravity, which has a potential energy function given by \(U(y)=mgy\) . We choose the gravitational potential energy to be zero when the pendulum hangs vertically (when \(\theta=0\) and \(y=0\) ).

The mechanical energy of the mass is conserved, and at any point is given by the sum of its kinetic and its gravitational potential energies:

\[\begin{aligned} E=mgy+\frac{1}{2}mv^2\end{aligned}\]

We want to find the velocity as a function of \(\theta\) , so we need to write \(y\) in terms of \(\theta\) . As you may recall from Problem 7.6.2 , we saw that from the geometry of the problem, we can express the height of the mass as \(y=L-L\cos\theta\) , or \(L(1-\cos\theta)\) , where \(y\) is the height as measured from the bottom point of the motion. You can refer to Figure 7.6.4 to refresh your memory. The energy at any point is then:

\[\begin{aligned} E=mgL(1-\cos\theta)+\frac{1}{2}mv^2\end{aligned}\]

Conservation of energy tells us that the total energy at any point must be the same as the initial energy. So, we can use our initial conditions to find the total energy of the system. The mass starts from rest (initial kinetic energy is zero) an angle \(\theta_0\) above the vertical:

\[\begin{aligned} E&=mgL(1-\cos\theta)+\frac{1}{2}mv^2\\[4pt] E_{initial}&=mgL(1-\cos\theta_0)\end{aligned}\]

Now that we have found the total energy of the system, we can write our general expression for the energy of the system at any point:

\[\begin{aligned} E&=mgL(1-\cos\theta)+\frac{1}{2}mv^2\\[4pt] mgL(1-\cos\theta_0)&=mgL(1-\cos\theta)+\frac{1}{2}mv^2\end{aligned}\]

All that’s left to do is simplify the expression and rearrange for \(v\) :

\[\begin{aligned} mgL(1-\cos\theta_0)&=mgL(1-\cos\theta)+\frac{1}{2}mv^2\\[4pt] gL(1-\cos\theta_0)-gL(1-\cos\theta)&=\frac{1}{2}v^2\\[4pt] gL-gL\cos\theta_0-gL+gL\cos\theta&=\frac{1}{2}v^2\\[4pt] gL(\cos\theta-\cos\theta_0)&=\frac{1}{2}v^2\\[4pt] \therefore v&=\sqrt{2gl(\cos\theta-\cos\theta_0)}\end{aligned}\]

We can see from this expression that the speed will be maximized when \(\cos\theta\) is maximized, which will occur when \(\theta=0\) (when the pendulum is vertical). This is as we expected. We can also see that we will get an imaginary number if the magnitude of \(\theta\) is greater than \(\theta_0\) , showing that the motion is constrained between \(-\theta_0\) and \(\theta_0\) . Finally, we showed that the velocity of the pendulum does not depend on the mass!

Exercise \(\PageIndex{3}\)

A block of mass \(m\) sits on a frictionless horizontal surface. It is attached to a wall by a spring with a spring constant \(k\) . The mass is pushed so as to compress the spring and then it is released (Figure \(\PageIndex{3}\)). Use the Lagrangian formalism to find an equation of motion for the mass/spring system (i.e. use the Lagrangian to determine the acceleration of the mass).

We are going to find an equation of motion of the system using the Lagrangian method. We choose to use a one dimension coordinate system, with the \(x\) axis defined to be co-linear with the spring, positive in the direction where the spring is extended, and set the origin to be located at the rest position of the spring. The kinetic energy and potential energy of the mass are given by

\[\begin{aligned} K&=\frac{1}{2}mv_x^2\\[4pt] U&=\frac{1}{2}kx^2\end{aligned}\]

since the only force exerted on the mass that can do work is the force from the spring. We have chosen the potential energy to be zero at \(x=0\) . The Lagrangian for this system is:

\[\begin{aligned} L&=K-U\\[4pt] L&=\frac{1}{2}mv_x^2-\frac{1}{2}kx^2\end{aligned}\]

The Euler-Lagrange equation in one dimension is:

\[\begin{aligned} \frac{d}{dt}\left(\frac{\partial L}{\partial v_{x}}\right)-\frac{\partial L}{\partial x} = 0\end{aligned}\]

We can calculate the terms of the Euler-Lagrange equation:

\[\begin{aligned} \frac{\partial L}{\partial v_{x}}&=\frac{\partial}{\partial v_{x}}\left(\frac{1}{2}mv_x^2-\frac{1}{2}kx^2\right)\\[4pt] &=mv_x\\[4pt] \therefore \frac{d}{dt}\left(\frac{\partial L}{\partial v_{x}}\right)&=\frac{d}{dt}(mv_x)\\[4pt] &=ma_x\\[4pt] \textrm{and}\qquad \frac{\partial L}{\partial x}&=\left(\frac{1}{2}mv_x^2-\frac{1}{2}kx^2\right)\\[4pt] &=-kx\end{aligned}\]

and then put them together to get:

\[\begin{aligned} \frac{d}{dt}\left(\frac{\partial L}{\partial v_{x}}\right)-\frac{\partial L}{\partial x} &= 0\\[4pt] \therefore ma_x&=-kx\\[4pt]\end{aligned}\]

We can see that this equation of motion is equivalent to Newton’s Second Law.

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Potential and Kinetic Energy Problems

9th - 12th grade.

10 questions

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No student devices needed.   Know more

The kinetic energy of a 7 kg cat moving at 4 m/s is

The gravitational potential energy of an object on the ground is

equal to the kinetic energy

The potential energy of a 550 kg missile flying at 800 meters is

4,312,000 Joules

440,000 Joules

176,000,000 Joules

A 80 kg runner going at 2.8 m/s has a kinetic energy of

2195.2 Joules

313.6 Joules

627.2 Joules

A ball that has a mass of 3kg and has a potential energy of 5,000 J is how high in the air.

A soccer ball has a gravitational potential energy of 80%. What is the percent of kinetic energy?

A student raises a 0.3 kg phone to a height of 0.8 meters. The potential energy of the phone is

.096 Joules

2.35 Joules

0.24 Joules

A plane with a mass 1200kg and a kinetic energy of 3,500,000 J is flying with a velocity of ?

5,833.3 m/s

What is the mass of a football that feel from a height of 28m with a potential energy of 4,150J?

A cannonball is launched with 500 Joules of kinetic energy and a velocity of 10m/s. What is its mass?

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Potential Energy Examples

Potential energy (PE) is the energy that is stored in an object due to its position charge, stress etc. Here are a few potential energy examples with solutions. These potential energy practice problems will help you learn how to calculate PE, mass, height.

Potential Energy Practice Problems

Let us consider a potential energy practice problem : An object of mass 30 kgs is placed on a hill top of height 80m. What is the potential energy possessed by the object?

We can calculate the potential energy of the object using the given formula

Substituting the values in the formula,

PE = m x g x h      = 30 x 9.8 x 80      = 23520 J

Note: We know that the acceleration due to gravity is constant and is always equal to 9.8 m/s 2

Therefore, the potential energy of the object is 23520 J .

Refer the below potential energy sample problem and calculate mass based on the potential energy, height and gravity. A fruit hangs from a tree and is about to fall to the ground of 10 meters height. It has a potential energy of 22.5 J. Calculate the mass of the fruit

Substitute the values using the given formula, M = PE / (g x h)      = 22.5 / (9.8 x 10)      = 0.229 kg

Therefore, the value of Mass is 0.229 kg

Refer the below example of potential energy and learn how to calculate height from PE and mass.

Let us consider a potential energy worked example problem: 5 kg weighing cat climbing at the top of the tree has a potential energy of 1176 kg. Find the height of the tree?

Substitute the values using the given formula, H = PE / (m x g)      = 1176 / (9.8 x 5)      = 24 cm

Therefore, the height of the tree is 24 cm

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