Work and the Work-Energy Theorem

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Work, Energy, and Power· 14 min read

1. Work Done by a Force (Constant and Variable)★★☆☆☆⏱ 4 min

📘 Definition

Work

$W$

Work is the measure of energy transfer caused by a force acting over a displacement. The sign of work indicates whether energy is added to (positive) or removed from (negative) the object.

Example:

Centripetal force does zero work for uniform circular motion, as it is perpendicular to displacement.

For a constant force $F$ acting on an object with displacement $Δ r$ , work is defined as the dot product:

W = \vec{F} \cdot \Delta\vec{r} = F\Delta r \cos\theta

where $θ$ is the angle between the force and displacement vectors. If $θ < 9 0^{\circ}$ , work is positive (energy added); if $θ > 9 0^{\circ}$ , work is negative (energy removed); if $θ = 9 0^{\circ}$ , work is zero (no energy transfer).

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 integral equals the net area between the force vs. position curve and the x-axis, a common AP exam interpretation.

📐 Worked Example

A 5 kg box is pulled 3 m up a $3 0^{\circ}$ 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
Normal force is perpendicular to displacement, so:
3
$W_N = 0$
4
Tension is parallel to displacement ( $θ = 0^{\circ}$ ), so:
5
$W_T = (40\ \text{N})(3\ \text{m})\cos 0^\circ = 120\ \text{J}$
6
Gravity acts at $12 0^{\circ}$ to displacement. Kinetic friction has magnitude $f_{k} = μ_{k} m g cos 3 0^{\circ} \approx 8.49 N$ , opposite displacement:
7
$W_g = mgd\cos 120^\circ = (5)(9.8)(3)(-0.5) = -73.5\ \text{J}$
8
$W_f = -f_k d \approx -25.5\ \text{J}$
9
Sum the individual works to get total net work:
10
$W_{\text{net}} = 120 - 73.5 - 25.5 = 21.0\ \text{J}$

Exam tip:

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

2. The Work-Energy Theorem: Derivation and Statement★★★☆☆⏱ 4 min

The work-energy theorem is derived directly from Newton's second law, giving a fundamental relation between net work done on an object and its change in kinetic energy. It is particularly powerful because it can be used for variable forces without integrating acceleration over time.

🔬 Derivation

Goal:

Derive the 1-dimensional work-energy theorem from Newton's second law

Starting from:

Newton's second law: $F_{net} = ma$

1
Rewrite acceleration using the chain rule:
2
$a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$
3
Substitute into Newton's second law:
4
$F_{\text{net}} = mv\frac{dv}{dx}$
5
Integrate both sides from initial position $x_{1}$ (velocity $v_{1}$ ) to final position $x_{2}$ (velocity $v_{2}$ ):
6
$\int_{x_1}^{x_2} F_{\text{net}} dx = \int_{v_1}^{v_2} mv\ dv$

Result:

The left-hand side is net work $W_{net}$ , and the right-hand side simplifies to the change in kinetic energy, giving the general theorem:

W_{\text{net}} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

A critical point: only net work (the 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 $Δ K E$ unless it is the only force acting.

📐 Worked Example

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

1
Apply the work-energy theorem: $W_{net} = Δ K E = K E_{f} - K E_{i}$ . Initial speed is zero, so $K E_{i} = 0$ .
2
Calculate net work by integrating the variable force:
3
$W_{\text{net}} = \int_0^3 (4x - x^2) dx = \left[2x^2 - \frac{x^3}{3}\right]_0^3 = 2(9) - \frac{27}{3} = 9\ \text{J}$
4
Set equal to final kinetic energy and solve for $v$ :
5
$9\ \text{J} = \frac{1}{2}(2\ \text{kg})v^2 \implies v^2 = 9 \implies v = 3\ \text{m/s}$

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 $Δ K E$ .

3. Net Work from Multiple Forces★★☆☆☆⏱ 2 min

When multiple forces act on an object, there are two equivalent ways to calculate net work: you can calculate work done by each force individually and sum them, or calculate the net force first then find the work done by the net force. Mathematically, this equivalence is written as:

W_{\text{net}} = \sum W_i = \left(\sum \vec{F}_i\right) \cdot \Delta\vec{r} = W_{\vec{F}_{\text{net}}}

Common special cases tested on the AP exam: work done by kinetic friction on a sliding object is always negative (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
Calculate work done by the pushing force; gravity and normal force are perpendicular to displacement, so their work is zero:
3
$W_{\text{push}} = Fd\cos 0^\circ = (50)(4) = 200\ \text{J}$
4
$W_g = W_N = 0$
5
By the work-energy theorem, constant speed means $Δ K E = 0$ , so net work must be zero:
6
$W_{\text{net}} = 0$
7
Solve for work done by friction from the sum of works:
8
$W_{\text{net}} = 200 + W_f = 0 \implies W_f = -200\ \text{J}$

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.

4. Exam-Style Worked Applications★★★★☆⏱ 4 min

✓ Quick check

Test your conceptual understanding with this AP-style multiple choice question:

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?
- Work done by gravity equals change in gravitational potential energy, net work equals change in kinetic energy
- Work done by gravity equals negative change in gravitational potential energy, net work equals change in kinetic energy
- Work done by gravity equals negative change in gravitational potential energy, net work equals negative change in kinetic energy
- Work done by gravity equals change in gravitational potential energy, net work equals negative change in kinetic energy
Reveal answer
1 —
The work-energy theorem always holds: net work equals change in kinetic energy. By definition, $Δ U_{g} = - W_{g}$ , so work done by gravity equals the negative change in gravitational potential energy.

📐 Worked Example

A 0.5 kg toy car is pushed along a horizontal track by a variable pushing force $F (x) = 2 e^{- 0.5 x}$ 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.

1
Part (a): Work done by a variable force is the integral of $F (x)$ over displacement:
2
$W_F = \int_0^2 2e^{-0.5x} dx = 2\left[-2e^{-0.5x}\right]_0^2 = -4\left(e^{-1} - e^0\right) = 4\left(1 - \frac{1}{e}\right) \approx 2.53\ \text{J}$
3
Part (b): Calculate work done by friction, find net work, then apply the work-energy theorem:
4
$W_f = -f d = -(0.3\ \text{N})(2\ \text{m}) = -0.6\ \text{J}$
5
$W_{\text{net}} = 2.53 - 0.6 = 1.93\ \text{J}$
6
$W_{\text{net}} = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2W_{\text{net}}}{m}} = \sqrt{\frac{2(1.93)}{0.5}} \approx 2.78\ \text{m/s}$
7
Part (c): Maximum speed occurs when acceleration is zero, which is when net force is zero. Set the pushing force equal to friction:
8
$2e^{-0.5x} = 0.3 \implies e^{-0.5x} = 0.15 \implies -0.5x = \ln(0.15) \approx -1.90 \implies x \approx 3.80\ \text{m}$
9
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 this position.

📐 Worked Example

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?

1
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:
2
$W_{\text{engine}} + W_{\text{friction}} = \Delta KE$
3
Calculate work done by friction and change in kinetic energy:
4
$W_{\text{friction}} = -fd = -(400\ \text{N})(300\ \text{m}) = -120000\ \text{J}$
5
$\Delta KE = \frac{1}{2}mv_f^2 - 0 = 0.5(1500\ \text{kg})(25\ \text{m/s})^2 = 468750\ \text{J}$
6
Solve for work done by the engine:
7
$W_{\text{engine}} = \Delta KE - W_{\text{friction}} = 468750 + 120000 = 588750\ \text{J} \approx 589\ \text{kJ}$

5. Common Pitfalls

Wrong move:

Calculating work as $W = F d$ when the force is not parallel to displacement, ignoring the $cos θ$ term.

Why:

Students memorize the simplified parallel form and forget the general dot product definition.

Correct move:

Always explicitly write $W = F \cdot Δ r$ and identify the angle between force and displacement before plugging in numbers.

Wrong move:

Setting the work done by a single force equal to $Δ K E$ 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 $W_{n e t} = Δ K E$ , 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 $Δ K E$ .

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. Quick Reference Cheatsheet

Category	Formula	Notes
Work (constant force)	$W = F \cdot Δ r = F Δ r cos θ$	$θ$ = angle between force and displacement; applies to rigid objects
Work (variable 1D force)	$W = \int_{x_{1}}^{x_{2}} F_{x} (x) d x$	Equals net area under force vs. position graph
Work (variable 3D force)	$W = \int_{r_{1}}^{r_{2}} F \cdot d r$	General form for any path and force
Net work (multiple forces)	$W_{net} = \sum W_{i} = W_{F_{net}}$	Sum work per force or calculate work of net force; same result
Work-Energy Theorem	$W_{net} = Δ K E = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$	Applies to translational KE; requires net work from all forces
Work done by normal force	$W_{N} = 0$	For objects moving along a fixed surface, normal is perpendicular to displacement
Work done by kinetic friction	$W_{f} = - f_{k} d$	$d$ = distance traveled; always negative for sliding motion
Average Power	$P_{avg} = \frac{W}{Δ t}$	Average rate of work transfer over a time interval

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Work done by variable force
2022 · FRQ
Net work and speed change

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.