Unit 3.2 | Work and the Work-Energy Theorem

- 16 minutes
- 4 formulas
- 19 questions
- 1 videos
Work is how forces add energy to an object or take it away. This lesson connects everything you learned about forces to the energy ideas in this unit.
Unit 3 Breakdown
In this lesson:
- You will learn what work means in physics (it is not the same as effort)
- How to calculate work, including forces applied at an angle
- When work is positive, negative, or zero
- Use the work-energy theorem to connect forces and energy
- The 4 work problem types: the sign of work, forces at an angle, force vs. position graphs, and stopping distances
Introduction
This lesson covers work and the work-energy theorem: how forces add energy to an object or take it away, and how the work-energy theorem links Unit 2 forces to energy.
In Unit 3.1 you learned that objects can have energy: kinetic, gravitational, and spring. But how does an object get that energy in the first place?
The answer is work. Work is how a force moves energy into an object or out of it. It is the bridge between the forces you mastered in Unit 2 and the energy ideas in this unit.
What Is Work?
In everyday language, “work” means effort. Holding a heavy backpack for an hour feels like a lot of work. In physics, you did zero work on that backpack, because it never moved. Work is only done when a force moves an object over a distance.
[katex] W = Fd [/katex]
Here [katex] F [/katex] is the force in newtons and [katex] d [/katex] is the displacement in meters. Work is measured in joules, the same unit as energy. That is not a coincidence: work is energy being transferred.
Push a box across the floor with [katex] 50\,\text{N} [/katex] for [katex] 4\,\text{m} [/katex] and you do [katex] W = (50)(4) = 200 \, J [/katex] of work on the box. You moved [katex] 200\,\text{J} [/katex] of energy from your muscles into the box.
Forces at an angle
What if you pull a wagon by a handle tilted upward? Only the part of the force along the motion does work. The part pointing up is not moving the wagon forward.
[katex] W = Fd\cos\theta [/katex]
Here [katex] \theta [/katex] is the angle between the force and the direction of motion.
A student pulls a wagon [katex] 5\,\text{m} [/katex] across a flat sidewalk. The handle makes a [katex] 37^\circ [/katex] angle with the ground and the student pulls with [katex] 50\,\text{N} [/katex]. How much work does the student do on the wagon?
- Identify the force and displacement: [katex] F = 50 \, N [/katex], [katex] d = 5 \, m [/katex], [katex] \theta = 37^\circ [/katex].
- Apply the equation: [katex] W = (50)(5)\cos 37^\circ [/katex].
- Solve: [katex] W = (50)(5)(0.8) = 200 \, J [/katex].
Notice the wagon’s mass was never needed. Work only cares about the force, the distance, and the angle between them.
Positive, Negative, and Zero Work
Work is a scalar, but it can still be positive or negative. The sign tells you which way the energy flows.
| Work | When | Example |
| Positive | Force has a part along the motion ([katex] \theta \lt 90^\circ [/katex]) | Pushing a stalled car forward |
| Negative | Force has a part against the motion ([katex] \theta \gt 90^\circ [/katex]) | Friction on a sliding box |
| Zero | Force is perpendicular to the motion ([katex] \theta = 90^\circ [/katex]), or nothing moves | Normal force on a box sliding across a floor |
Friction almost always does negative work. It pulls energy out of the object and turns it into heat. That’s the “lost” energy from Unit 3.1.
Work From a Force vs. Position Graph
In Unit 1.3 you found displacement as the area under a velocity vs. time graph. Force graphs work the same way.
The work done by a force equals the area under a force vs. position graph. This is how you find work when the force changes as the object moves.
The spring force is the classic example. It grows as you stretch the spring ([katex] F = kx [/katex]), so its graph is a straight line and the area under it is a triangle:
[katex] W = \frac{1}{2}(x)(kx) = \frac{1}{2}kx^2 [/katex]
That is exactly the spring potential energy from Unit 3.1. The work you do stretching a spring is the energy the spring stores.
The Work-Energy Theorem
Add up the work done by every force on an object (the net work) and you get the change in its kinetic energy:
[katex] W_{net} = \Delta KE = KE_f – KE_i [/katex]
Net positive work speeds an object up, net negative work slows it down, and zero net work leaves its speed unchanged. In Unit 2, a net force caused an acceleration. Now a net force acting over a distance causes a change in kinetic energy. Same physics, different lens.
Keep this idea in your back pocket: work is how energy enters or leaves. In Unit 3.3 it becomes the [katex] W [/katex] term in the conservation of energy framework, [katex] E_i + W = E_f [/katex], the one equation you will use for almost every energy problem.
Every work question on the AP exam is one of these four types:
| Problem type | The clue | The setup |
| 1. The sign of work | “positive, negative or zero work?” | Compare the force with the motion |
| 2. Forces at an angle | a handle, rope or push at an angle | [katex] W = Fd\cos\theta [/katex] |
| 3. Force vs. position graphs | a graph of [katex] F [/katex] against [katex] x [/katex] | Area = work |
| 4. Stopping distances | “how far does it slide?” | [katex] W_{net} = \Delta KE [/katex] |
Problem Type 1: The sign of work
How to spot it: Which force does positive, negative or zero work? Often no numbers at all.
Setup: Compare each force with the direction of motion: along it is positive, against it is negative, perpendicular is zero.
A ball on a string moves in a horizontal circle at constant speed (like in circular motion). How much work does the tension do on the ball?
Answer: zero. Tension points into the center of the circle while the ball moves along the circle, so the force is always perpendicular to the motion and [katex] \cos 90^\circ = 0 [/katex]. That is why the ball’s speed never changes.
AP trap: Saying a force does work just because it is there. The normal force on a box sliding across the floor is perpendicular to the motion, so it does zero work.
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Problem Type 2: Forces at an angle
How to spot it: A handle, rope or push tilted at an angle to the motion.
Setup: [katex] W = Fd\cos\theta [/katex], with [katex] \theta [/katex] the angle between the force and the motion. The wagon above is this type.
AP trap: Using the angle from the vertical, or sine instead of cosine. Measure [katex] \theta [/katex] from the direction of motion.
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Problem Type 3: Work from a force vs. position graph
How to spot it: A graph of force against position.
Setup: The area under the graph is the work. Split it into rectangles and triangles.
AP trap: Mixing up the graphs. Area under force vs. position is work; area under force vs. time is impulse (Unit 4).
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Problem Type 4: Stopping distances (the work-energy theorem)
How to spot it: A force speeds up or stops an object over a distance: “how far does it skid?”, “how fast is it going after?”
Setup: Find [katex] \Delta KE [/katex], write the net work as force times distance with its sign, and set [katex] W_{net} = \Delta KE [/katex].
A [katex] 1000\,\text{kg} [/katex] car is traveling at [katex] 20\,\text{m/s} [/katex] when the driver slams on the brakes. Friction exerts a constant [katex] 8000\,\text{N} [/katex] force on the car. How far does the car skid before stopping?
1st step: Find the change in kinetic energy ([katex] 20\,\text{m/s} [/katex] to [katex] 0\,\text{m/s} [/katex]): [katex] \Delta KE = 0 – \frac{1}{2}(1000)(20)^2 = -200{,}000 \, J [/katex]
2nd step: Find the net work. Only friction does work (gravity and the normal force are perpendicular to the motion), and it points against the motion: [katex] W_{net} = -8000d [/katex]
3rd step: Apply the work-energy theorem: [katex] -8000d = -200{,}000 \rightarrow d = 25 \, m [/katex]
Answer: [katex] 25 \, m [/katex].
Show Answer: What if the car were going twice as fast?
Kinetic energy depends on [katex] v^2 [/katex], so twice the speed means four times the kinetic energy ([katex] 800{,}000\,\text{J} [/katex]). With the same friction force, the car skids four times as far: [katex] 100 \, m [/katex].
AP trap: Dropping the minus sign on friction’s work. Friction points against the motion, so its work is negative, and so is the change in kinetic energy.
This question is the “twice as fast” idea from above, with arrows.
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Lesson 3.3 Preview
In the next lesson we put it all together with the conservation of energy framework: choose a system, mark two points, and solve roller coaster, pendulum, and spring problems without an FBD.
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This article was written by Jason
