Calculating Work Done (College Board AP® Physics 1: Algebra-Based)

Study Guide

Leander Oates

Written by: Leander Oates

Reviewed by: Caroline Carroll

Work done equation

  • When an applied force displaces an object over a distance, work is done

    • If a force is applied but the object is not displaced, then no work is done

  • For a constant force applied parallel to the displacement of the object, work is calculated using the equation:

W space equals space F d

  • Where:

    • W = work, measured in straight N times straight m

    • F = applied force parallel to displacement, measured in straight N

    • d = distance, measured in straight m

  • For example, if a bucket of water is raised from a well by pulling a rope with a force of 200 space straight N through a height of 2 space straight m, then the work done is:

W space equals space 200 space times space 20 space equals space 4000 space straight N times straight m

A rope is used to lift a bucket of water from a well. Labels identify the bucket and the well.
Work is done when a bucket of water is lifted through a height by an applied force

Work done at an angle

  • Only the components of the applied force that are parallel to the displacement contribute to the work done

  • When a constant force is applied at an angle to the displacement of the object, work can be calculated using:

W space equals space F subscript parallel to d space equals space F d open parentheses cos theta close parentheses

  • Where:

    • W = work, measured in straight N times straight m

    • F subscript parallel to = component of the force that is parallel to the displacement of the object, measured in straight N

    • d = distance, measured in straight m

    • theta = angle at which the force is applied, measured in degree from the displacement vector

A person in winter clothes pulls a sled with another person on it. The image illustrates the direction of the applied force (F) at angle θ and Fcosθ component of that force that acts parallel to the displacement of the sled
When a force is applied at an angle to the displacement of the object, only the component parallel to the displacement contributes to the work done
  • The point of application of the force will determine the system's total energy

  • When the angle of the applied force is 0 space less or equal than space theta space less or equal than space 90 degree:

    • cos space theta is positive

    • work done is positive

    • the speed of the object increases

    • the kinetic energy of the object increases

  • When the angle of the applied force is perpendicular to the displacement, theta space equals space 90 degree

    • cos space theta is zero

    • work done is zero

    • the speed of the object does not change

    • the kinetic energy of the object does not change

  • When the angle of the applied force is 90 space less or equal than space theta space less or equal than space 180 degree:

    • cos space theta is negative

    • work done is negative

    • the speed of the object decreases

    • the kinetic energy of the object decreases

  • When a force is applied perpendicularly to the direction of the displacement of the object, it is possible for the object to change direction without changing the kinetic energy of the object

  • Examples include:

Force-versus-displacement graph

  • If the force applied on an object is not constant, then a force-versus-displacement graph can be used to determine the work done

  • The force applied to an object may not always be constant

    • This is more representative of real-world scenarios

  • In this case, the parallel component of the variable force can be plotted as a function of the displacement of the object and the work done is then equal to the area under the graph

  • The work done is equivalent to the area under the graph, whether there is:

    • a small force applied over a large displacement 

    • a large force applied over a small displacement

A graph illustrating work done (in joules) with force (in newtons) on the Y-axis and displacement (in meters) on the X-axis. The area under the graph is shaded in blue.
Graph of work done by an applied force displacing an object

Worked Example

Line graph showing force as a function of displacement. A variable force acts over a distance of 80 m. The initial force applied is 100 N and increases to 250 N.

The graph above shows the parallel component of a variable force exerted on an object as a function of displacement. Which of the following is the correct value for the work done on the object?

A: 1.4 space cross times 10 cubed space straight N times straight m

B: 6.0 space cross times 10 cubed space straight N times straight m

C: 8.0 space cross times 10 cubed space straight N times straight m

D: 1.4 space cross times 10 to the power of 4 space straight N times straight m

The correct answer is D

Step 1: Split the area into sections

  • The area below the curve of a force-versus-displacement graph is equal to the work done

  • The total area is the sum of the areas of sections A and B

Force Displacement Graph Worked Example (2)

Step 2: Calculate the area of section A

A subscript A space equals space 1 half b h

A subscript A space equals space 1 half open parentheses 80 close parentheses open parentheses 250 space minus space 100 close parentheses space equals space 6000 space straight N times straight m

Step 3: Calculate the area of section B

A subscript B space equals space b h

A subscript B space equals space 80 space times space 100 space equals space 8000 space straight N times straight m

Step 4: Calculate the total work done

W space equals space A subscript A space plus space A subscript B

W space equals space 6000 space plus space 8000 space equals space 14 space 000 space straight N times straight m space equals space 1.4 cross times 10 to the power of 4 space straight N times straight m

  • The answer is therefore D

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Leander Oates

Author: Leander Oates

Expertise: Physics

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

Author: Caroline Carroll

Expertise: Physics 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.