Work and the work-energy theorem — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Work done by constant and variable forces, dot product formulation, net work from multiple forces, derivation of the work-energy theorem, and application to find changes in speed and kinetic energy.

You should already know: Vectors and dot product operations, Newton's second law of motion, definition of translational kinetic energy.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Work and the work-energy theorem?

Work is the fundamental measure of energy transfer caused by a force acting over a displacement. The work-energy theorem generalizes the relation between force, displacement, and motion, stating that the net work done on a rigid object by all forces acting on it equals the object’s change in kinetic energy. Unlike kinematics approaches that rely on constant acceleration, the work-energy theorem can be applied to problems with variable forces and non-constant acceleration without requiring integration of acceleration over time, making it a powerful problem-solving shortcut. Per the AP Physics C: Mechanics CED, Unit 3 (Work, Energy, and Power) contributes 14-18% of your total exam score, and this topic forms the foundational framework for all other energy concepts in the unit. Work and work-energy theorem questions appear in both MCQ and FRQ sections: MCQs typically test conceptual understanding of work sign or work done by variable forces, while FRQs combine the theorem with other topics like forces or energy conservation for multi-step problems.

2. Work Done by a Force (Constant and Variable)

Work quantifies how much energy a force transfers to or from an object as it moves through a displacement. For a constant force $\vec{F}$ acting on an object that undergoes displacement $\Delta \vec{r}$, work is defined as the dot product: $$W=\vec{F}\cdot \Delta \vec{r}=F\Delta r\cos \theta$$ where $\theta$ is the angle between the force vector and displacement vector. The sign of work carries physical meaning: positive work means energy is added to the object ($\theta <90^{\circ }$), negative work means energy is removed from the object ($\theta >90^{\circ }$), and zero work means no energy is transferred ($\theta =90^{\circ }$, e.g., centripetal force does no work on uniform circular motion).

For a variable force that changes with position, we extend the definition by integrating over infinitesimal displacements. For 1-dimensional motion along the x-axis: $$W={\int }_{x_1}^{x_2}{F}_x(x)dx$$ Geometrically, this equals the net area between the force vs. position curve and the x-axis, a common exam interpretation.

Worked Example

A 5 kg box is pulled 3 m up a 30° incline by a rope with tension 40 N parallel to the incline. The coefficient of kinetic friction between the box and incline is 0.2. Find the net work done on the box by all forces.

1. Identify all forces acting on the box: tension, gravity, normal force, and kinetic friction.
2. Calculate work for each force: Normal force is perpendicular to displacement, so $W_N=0$. Tension is parallel to displacement ($\theta =0^{\circ }$), so $W_T=(40\text{N})(3\text{m})\cos 0^{\circ }=120\text{J}$.
3. Gravity acts at 120° to displacement, so $W_g=mgd\cos 120^{\circ }=(5)(9.8)(3)(-0.5)=-73.5\text{J}$. Kinetic friction has magnitude $f_k={\mu }_{k}mg\cos 30^{\circ }\approx 8.49\text{N}$, opposite displacement, so $W_f=-f_{k}d\approx -25.5\text{J}$.
4. Sum the works to get net work: $W_{\text{net}}=120-73.5-25.5=21.0\text{J}$.

Exam tip: Always confirm the angle between the force and displacement, not irrelevant angles from the problem geometry. Normal force never does work on an object moving along a fixed surface, so you can immediately set its work to zero.

3. The Work-Energy Theorem: Derivation and Statement

The work-energy theorem is derived directly from Newton's second law, making it a fundamental relation rooted in dynamics. For 1-dimensional motion, start with Newton's second law: $$F_{\text{net}}=ma=m\frac{dv}{dt}$$ Use the chain rule to rewrite acceleration as $a=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$, then substitute: $$F_{\text{net}}=mv\frac{dv}{dx}$$ Integrate both sides from initial position $x_1$ (velocity $v_1$) to final position $x_2$ (velocity $v_2$): $${\int }_{x_1}^{x_2}{F}_{\text{net}}dx={\int }_{v_1}^{v_2}mvdv$$ The left-hand side is net work $W_{\text{net}}$, and the right-hand side simplifies to $\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2=\Delta KE$, the change in translational kinetic energy. This generalizes to 3D motion to give the theorem: $$W_{\text{net}}=\Delta KE$$ A critical point to remember: only net work (sum of work from all forces acting on the object) equals the change in kinetic energy. Work done by a single force will not give $\Delta KE$ unless it is the only force acting.

Worked Example

A 2 kg block moves along the x-axis, acted on by a net force $F(x)=(4x-x^2)\text{N}$, where $x$ is in meters. If the block starts from rest at $x=0$, what is its speed at $x=3\text{m}$?

1. Apply the work-energy theorem: $W_{\text{net}}=\Delta KE=K{E}_f-K{E}_i$. Initial speed is zero, so $K{E}_i=0$.
2. Calculate net work by integrating the variable force: $$W_{\text{net}}={\int }_0^3(4x-x^2)dx={\left[2x^2-\frac{x^3}{3}\right]}_0^3=2(9)-\frac{27}{3}=9\text{J}$$
3. Set equal to final kinetic energy and solve for $v$: $$9\text{J}=\frac{1}{2}(2\text{kg})v^2⟹v^2=9⟹v=3\text{m/s}$$

Exam tip: If a problem asks for change in speed after work is done by one force, do not forget to add the work from all other forces (like gravity or friction) before setting net work equal to $\Delta KE$.

4. Net Work from Multiple Forces

When multiple forces act on an object, there are two equivalent ways to calculate net work: you can calculate the work done by each force individually and sum them, or you can calculate the net force first then find the work done by the net force. Mathematically, this is: $$W_{\text{net}}=\sum W_i=\left(\sum {\vec{F}}_i\right)\cdot \Delta \vec{r}=W_{{\vec{F}}_{\text{net}}}$$ This equivalence is a useful shortcut for many problems, especially when acceleration or net force is already known. Special cases that come up frequently on exams: work done by kinetic friction on a sliding object is always negative (since friction opposes displacement), work done by any force perpendicular to displacement is always zero, and if an object moves at constant speed, net force is zero so net work is zero.

Worked Example

A person pushes a 10 kg crate 4 m across a horizontal floor at constant speed. The pushing force is 50 N parallel to the floor. What is the work done by friction, and what is the net work done on the crate?

1. Constant speed means acceleration is zero, so net force on the crate is zero by Newton's second law.
2. Work done by the pushing force is $W_{\text{push}}=Fd\cos 0^{\circ }=(50)(4)=200\text{J}$. Gravity and normal force are perpendicular to displacement, so their work is $W_g=W_N=0$.
3. Net work is $W_{\text{net}}=W_{{\vec{F}}_{\text{net}}}d=0$, which matches the work-energy theorem: $\Delta KE=0$ for constant speed, so net work must be zero.
4. Sum of works gives $W_{\text{net}}=200+W_f=0⟹W_f=-200\text{J}$.

Exam tip: If an object moves at constant speed, you can immediately conclude net work is zero per the work-energy theorem, which is a huge shortcut for problems asking for work done by an unknown force.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating work as $W=Fd$ when the force is not parallel to displacement, ignoring the $\cos \theta$ term. Why: Students memorize the simplified parallel form and forget the general dot product definition. Correct move: Always explicitly write $W=\vec{F}\cdot \Delta \vec{r}$ and identify the angle between force and displacement before plugging in numbers.
• Wrong move: Setting the work done by a single force equal to $\Delta KE$ instead of net work. Why: Problems often highlight one force (like tension or pushing), leading students to forget other forces contribute to net work. Correct move: Before applying $W_{net}=\Delta KE$, list all forces acting on the object and confirm you have included the work from every force.
• Wrong move: Calculating the area under a force-time graph to get work. Why: Students mix up work (area under F-position) and impulse (area under F-time). Correct move: Label your graph axes before calculating area; area under F-x is work, area under F-t is impulse.
• Wrong move: Getting a positive work for kinetic friction on a sliding object, after correctly calculating the magnitude of friction. Why: Students forget friction points opposite displacement. Correct move: For any sliding displacement, kinetic friction work is always negative, so add the negative sign explicitly after calculating the magnitude.
• Wrong move: Splitting kinetic energy into x and y components to calculate $\Delta KE$. Why: Students used to vector components for velocity incorrectly extend this to kinetic energy. Correct move: Calculate the magnitude of initial and final velocity, then compute $\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$ directly; kinetic energy is a scalar and cannot be split into components.
• Wrong move: Integrating a variable force with respect to time instead of position to get work. Why: Students default to integrating over time from kinematics problems. Correct move: Work is force integrated over displacement, so always set up the integral with respect to position for variable force work.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A block slides down a curved ramp that has kinetic friction. Which of the following correctly describes the relation between work done by gravity and change in gravitational potential energy, and between net work and change in kinetic energy? A. Work done by gravity equals change in gravitational potential energy, net work equals change in kinetic energy B. Work done by gravity equals negative change in gravitational potential energy, net work equals change in kinetic energy C. Work done by gravity equals negative change in gravitational potential energy, net work equals negative change in kinetic energy D. Work done by gravity equals change in gravitational potential energy, net work equals negative change in kinetic energy

Worked Solution: The work-energy theorem is a fundamental result that always holds for translational motion: net work done on an object equals its change in kinetic energy. This eliminates options C and D. By definition, the change in gravitational potential energy is $\Delta U_g=-W_g$, where $W_g$ is work done by gravity. When the block slides down, gravity does positive work, and potential energy decreases, so $W_g=-\Delta U_g$. The correct answer is B.

Question 2 (Free Response)

A 0.5 kg toy car is pushed along a horizontal track by a variable pushing force $F(x)=2e^{-0.5x}$ N, where $x$ is position in meters from the starting point. The car experiences a constant kinetic friction force of 0.3 N. The car starts from rest at $x=0$. (a) Find the total work done by the variable pushing force when the car reaches $x=2$ m. (b) Find the speed of the car at $x=2$ m using the work-energy theorem. (c) At what position $x$ does the car reach its maximum speed after starting? Explain your reasoning.

Worked Solution: (a) Work done by a variable force is the integral of $F(x)$ over displacement: $$W_F={\int }_0^{2}2e^{-0.5x}dx=2{\left[-2e^{-0.5x}\right]}_0^2=-4\left(e^{-1}-e^0\right)=4\left(1-\frac{1}{e}\right)\approx 2.53\text{J}$$

(b) Work done by friction is $W_f=-fd=-(0.3\text{N})(2\text{m})=-0.6\text{J}$. Net work is $W_{\text{net}}=2.53-0.6=1.93\text{J}$. By the work-energy theorem: $$W_{\text{net}}=\frac{1}{2}m{v}^2⟹v=\sqrt{\frac{2W_{\text{net}}}{m}}=\sqrt{\frac{2(1.93)}{0.5}}\approx 2.78\text{m/s}$$

(c) Maximum speed occurs when acceleration is zero, which occurs when net force is zero. Set the pushing force equal to friction: $$2e^{-0.5x}=0.3⟹e^{-0.5x}=0.15⟹-0.5x=\ln (0.15)\approx -1.90⟹x\approx 3.80\text{m}$$ Before this position, net force is positive so speed increases; after this position, net force is negative so speed decreases, so maximum speed occurs at $x\approx 3.80\text{m}$.

Question 3 (Application / Real-World Style)

A 1500 kg electric car accelerates from rest to 25 m/s on a horizontal road. The combined force of air resistance and rolling friction is a constant 400 N over a total displacement of 300 m. What is the total work done by the car's engine to achieve this acceleration?

Worked Solution: Apply the work-energy theorem: net work equals change in kinetic energy. Net work is the sum of work done by the engine and work done by friction: $W_{\text{engine}}+W_{\text{friction}}=\Delta KE$. Work done by friction is $W_{\text{friction}}=-fd=-(400\text{N})(300\text{m})=-120000\text{J}$. Change in kinetic energy is $\Delta KE=\frac{1}{2}m{v}_f^2-0=0.5(1500\text{kg})(25\text{m/s}{)}^2=468750\text{J}$. Solve for $W_{\text{engine}}$: $W_{\text{engine}}=\Delta KE-W_{\text{friction}}=468750+120000=588750\text{J}\approx 589\text{kJ}$. In context, this means the engine must do 589 kJ of work: 469 kJ goes into the car's kinetic energy, and 120 kJ is lost to friction and air resistance over the 300 m acceleration interval.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational framework for all energy-based problem solving in AP Physics C: Mechanics. Up next, you will extend the work-energy relation to separate contributions from conservative and nonconservative forces, leading to the principle of conservation of mechanical energy, one of the most widely used problem-solving tools in the course. Without correctly understanding how to calculate net work and apply the work-energy theorem, you will not be able to correctly set up or solve energy conservation problems that include work done by nonconservative forces like friction. This topic also lays the groundwork for instantaneous and average power calculations, and later for rotational work and rotational kinetic energy in the rotation unit.

Conservation of energy Power and energy rate Rotational work and energy Forces and free-body diagrams