Work-Energy Theorem (College Board AP® Physics 1: Algebra-Based): Study Guide

Leander Oates

Written by: Leander Oates

Reviewed by: Caroline Carroll

Updated on

Work-energy theorem

  • The work-energy theorem states that:

The change in an object's kinetic energy is equal to the sum of the (net) work being done by all forces exerted on the object

  • This can be expressed by the following equation:

increment K space equals space sum for i of W subscript i space equals space sum for i of F subscript parallel to i end subscript d

  • Where:

    • increment K = change in kinetic energy, measured in straight J

    • stack sum W subscript i with i below = the sum of the net work being done by all the forces on the object, measured in straight N times straight m

    • sum for i of F subscript parallel to i end subscript d = the sum of all the forces acting on the object parallel to its displacement over a distance, measured in straight N times straight m

  • The change in kinetic energy of an object can be expressed as:

increment K space equals space K subscript f space minus space K subscript i space equals space 1 half m v subscript f superscript 2 space minus space 1 half m v subscript i superscript 2

  • The work-energy theorem means that if net work is done to a system, the kinetic energy of the system will change

  • This change in kinetic energy is a measure of the net work done and vice versa

    • Where 1 space straight J space equals space 1 space straight N times straight m

  • External forces may change the configuration of a system; that is, the objects within the system may change position

  • Only the component of the external force acting parallel to the displacement of the object contributes to the net work done on the object

W space equals space F subscript parallel to d space equals space F d space cos space theta

  • The displacement of the object is measured from the point at which the force is exerted on the object to give the change in kinetic energy of the object, or the net work done

Center of mass

  • Calculations involving the work-energy theorem can be simplified by considering the motion of a system’s center of mass

  • If the system’s center of mass and the point of application of the force move the same distance when a force is exerted on a system, then the system may be modeled as a single object

    • In this case, only the system’s kinetic energy can change

  • If the system cannot be modeled as a single object, then the point of application of each force, and the distance through which each force acts must be considered

    • In this case, the total energy of the system (kinetic and potential energies) can change

Worked Example

An apple of mass 75 g is thrown vertically upwards with an initial velocity of 18 m/s. According to the work-energy theorem, which of the following is the maximum height the apple can reach assuming air resistance is negligible?

A: 12.6 space straight m

B: 14.2 space straight m

C: 15.6 space straight m

D: 16.2 space straight m

The correct answer is D

Analyze the scenario

  • The question mentions the work-energy theorem, so this is the method that should be used

  • The force of gravity acts on the apple as the ball is displaced and acts parallel to the displacement

  • The force and the displacement are in opposite directions, theta space equals space 180 degree, therefore:

W space equals space F subscript g open parentheses cos space theta close parentheses d space equals space minus F subscript g d

  • As the apple reaches its maximum height, its velocity is zero as it changes direction; therefore, its kinetic energy is also zero

Step 1: List the known quantities

  • Mass, m space equals space 75 space straight g space equals space 0.075 space kg

  • Initial velocity, v subscript 0 space equals space 18 space straight m divided by straight s

  • Final kinetic energy, K subscript f space equals space 0

  • Acceleration due to gravity at Earth's surface, g space equals space 10 space straight m divided by straight s squared

Step 2: State the work-energy theorem equation in terms of the apple

W space equals space increment K space equals space K subscript f space minus space K subscript i

negative F subscript g d space equals space 0 space minus space 1 half m v subscript i superscript 2

Step 3: Rearrange the equation to solve for the maximum height, d

d space equals space fraction numerator 1 half m v subscript i superscript 2 over denominator F subscript g end fraction space equals space fraction numerator 1 half m v subscript i superscript 2 over denominator m g end fraction space equals space fraction numerator 1 half up diagonal strike m v subscript i superscript 2 over denominator up diagonal strike m g end fraction space equals space fraction numerator 1 half v subscript i superscript 2 over denominator g end fraction space

Step 4: Calculate the maximum height

d space equals space fraction numerator 1 half open parentheses 18 close parentheses squared over denominator 10 end fraction space equals space 16.2 space straight m

  • This is option D

Friction & energy dissipation

  • The energy dissipated, or work done, by friction is typically equated to the force of friction times the length of the path over which the frictional force is exerted

  • This can be expressed by the following equation:

increment E subscript m e c h end subscript space equals space F subscript f d space cos space theta

  • Where:

    • increment E subscript m e c h end subscript = the change in mechanical energy, measured in straight J

    • F subscript f = force of friction, measured in straight N

    • d = distance moved by object, measured in straight m

    • theta = angle at which the force of friction is applied, measured in degree

Worked Example

A block with a mass of 15 space straight g is used to compress a spring, with a spring constant of 12 space straight N divided by straight m, a distance of 2.4 space cm. After the block is released, it slides across a horizontal surface which has a coefficient of kinetic friction of 0.26. What is the maximum distance the block can reach?

Answer:

Step 1: Sketch out the scenario

Diagram of a block attached to a compressed spring at initial position xi, moving to final position xf with displacement Δxs shown by an arrow.

Step 2: List the known quantities:

  • Mass of block, m space equals space 15 space straight g space equals space 0.015 space kg

  • Spring constant, k space equals space 12 space straight N divided by straight m

  • Displacement of spring compression, increment x subscript s space equals space 2.4 space cm space equals space 0.024 space straight m

  • Coefficient of kinetic friction, mu subscript k space equals space 0.26

Step 3: Determine the initial and final positions of the block

  • The initial position of the block is taken to be when the spring is compressed to its maximum displacement

  • The final position of the block is at its maximum height up the slope

Step 4: State the initial and final kinetic energy of the block

  • The block is stationary at its initial position and its final position

  • Therefore, both initial and final kinetic energy are zero

W subscript n e t end subscript space equals space increment K space equals space 0

Step 5: Consider the work done by the force of the spring on the block

  • As the spring is compressed, it gains elastic potential energy

  • When the spring is released, the loss of elastic potential energy is equal to the work done by the force applied by the spring on the block

increment U subscript s space equals space 1 half k open parentheses increment x subscript s close parentheses squared

increment U subscript s space equals space 1 half open parentheses 12 close parentheses open parentheses 0.024 close parentheses squared space equals space 3.46 cross times 10 to the power of negative 3 end exponent space straight J

W subscript s space equals space 3.46 cross times 10 to the power of negative 3 end exponent space straight N times straight m

Step 6: Consider the work done on the block on the horizontal surface by the force of kinetic friction

  • The energy dissipated by the frictional force is:

increment E subscript m e c h end subscript space equals space F subscript f space d space cos space theta

  • The force of kinetic friction is F subscript f comma space k end subscript space equals space mu subscript k F subscript n

increment E subscript m e c h end subscript space equals space mu subscript k F subscript n d space cos space theta

  • The force acts perpendicular to the displacement but in the opposite direction, so theta space equals space 180 degree, and cos space theta space equals space minus 1, therefore:

increment E subscript m e c h end subscript space equals space minus mu subscript k m g d

  • The net work is equal to the work done in compressing the spring plus the work done by friction:

W subscript n e t end subscript space equals space W subscript s space plus space W subscript f space equals space 0

  • Therefore, the change in mechanical energy is equal to the work done by the spring

W subscript s space equals space minus W subscript f

W subscript s space equals space minus increment E subscript m e c h end subscript

W subscript s space equals space mu subscript k m g d

  • Rearrange to solve for the distance traveled

d space equals space fraction numerator W subscript s over denominator mu subscript k m g end fraction

  • Substitute the known values into the equation to calculate the distance traveled by the block

d space equals space fraction numerator 3.46 cross times 10 to the power of negative 3 end exponent over denominator open parentheses 0.26 close parentheses open parentheses 0.015 times 10 close parentheses end fraction

d space equals space 0.089 space straight m space equals space 8.9 space cm

Examiner Tips and Tricks

A problem like this could be solved with kinematic equations, but if the question specifically asks you to use a certain method, you must stick with that.

You've read 0 of your 5 free study guides this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Leander Oates

Author: Leander Oates

Expertise: Physics Content Creator

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Physics & Chemistry Subject Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.