Unit 3 · Topic 3.2

3.2 Work

Work is energy transferred to or from a system by a force acting through a displacement: $W = \int_a^b\vec{F}\cdot d\vec{r}$ . Only the force component along the motion does work, so work can be positive, negative or zero. The work–energy theorem says the net work on an object equals its change in kinetic energy.

Key terms

work
dot product
work–energy theorem
conservative force
nonconservative force
force–position graph

Work by a constant force

When a constant force $\vec{F}$ acts while an object moves through a straight displacement $\vec{d}$ , the work is the dot product:

$W = \vec{F}\cdot\vec{d} = Fd\cos\theta$

θ is the angle between the force and the displacement. In components, the dot product is $F_xd_x + F_yd_y + F_zd_z$ . Work is a scalar, measured in joules.

Force along the motion (θ < 90°): positive work. The force adds energy.
Force against the motion (θ > 90°): negative work. The force removes energy, as kinetic friction does.
Force perpendicular to the motion (θ = 90°): zero work. A normal force on a level floor or the tension in a string keeping a ball in uniform circular motion does no work.

Work by a changing force

If the force changes with position, split the path into tiny steps and add up $\vec{F}\cdot d\vec{r}$ with an integral. In one dimension:

$W = \int_{x_1}^{x_2} F_x\,dx$

Graphically, that's the area under a graph of $F_x$ against x. Area below the axis is negative work.

The classic case is a spring. Stretching an ideal spring from $x_1$ to $x_2$ (measured from its relaxed length), the force you apply is kx, so the work you do is $\frac{1}{2}k(x_2^2 - x_1^2)$ . The spring itself does the negative of that.

The work–energy theorem

The net work done on an object, the work by all the forces together, equals its change in kinetic energy:

$W_{net} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

It comes straight from Newton's second law: $F = m\frac{dv}{dt} = mv\frac{dv}{dx}$ , so $\int F\,dx = \int mv\,dv$ , which is $\Delta(\frac{1}{2}mv^2)$ . Positive net work speeds an object up; negative net work slows it down; zero net work leaves its speed unchanged.

The theorem as written is for an object you can treat as a single point. For a system that changes shape, like a person pushing off a wall, the wall does no work (its push doesn't move through any distance), yet the person speeds up; the energy comes from inside the person. Most problems in this unit use point objects.

Conservative and nonconservative forces

The work done by a conservative force, like gravity or an ideal spring, depends only on the start and end points, not the path. Gravity does −mgh on an object raised by h whether you lift it straight up or carry it up a winding ramp. Around any closed loop, a conservative force does zero total work.

The work done by a nonconservative force depends on the path. Kinetic friction does more negative work over a longer path, so dragging a box the long way around costs more energy. Conservative forces are the ones that get a potential energy (3.3).

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Work with the dot product
(a) A 40 N pull at 30° above the horizontal drags a sled 10 m across level snow. How much work does the pull do? (b) A force $\vec{F} = (3.0\hat{i} + 4.0\hat{j})$ N acts during a displacement $\vec{d} = (5.0\hat{i} - 2.0\hat{j})$ m. Find the work.
Show the solutionHide the solution
1. Step 1: (a) $W = Fd\cos\theta = (40)(10)\cos 30^\circ \approx 346$ J. Only the horizontal component of the pull, about 34.6 N, does work; the vertical part is perpendicular to the motion.
2. Step 2: (b) $W = F_xd_x + F_yd_y = (3.0)(5.0) + (4.0)(-2.0) = 15 - 8.0 = 7.0$ J.
Answer: (a) about 346 J; (b) 7.0 J.
Example 2Calculator allowed
A position-dependent force and the work–energy theorem
A force $F(x) = 3x^2$ N (x in meters) acts along the x-axis on a 2.0 kg object, the only force doing work. The object moves at 1.0 m/s at x = 1.0 m. Find the work done from x = 1.0 m to x = 3.0 m and the object's speed at x = 3.0 m.
Show the solutionHide the solution
1. Step 1: $W = \int_1^3 3x^2\,dx = \left[x^3\right]_1^3 = 27 - 1 = 26$ J.
2. Step 2: Starting kinetic energy: $\frac{1}{2}(2.0)(1.0)^2 = 1.0$ J.
3. Step 3: Work–energy theorem: $K_f = 1.0 + 26 = 27$ J, so $v_f = \sqrt{\frac{2(27)}{2.0}} = \sqrt{27} \approx 5.2$ m/s.
Answer: 26 J; about 5.2 m/s.
Example 3Calculator allowed
Stretching a spring further (classic trap)
A spring with k = 200 N/m is already stretched 0.10 m. How much work do you do to stretch it from 0.10 m to 0.30 m?
Show the solutionHide the solution
1. Step 1: The force grows with x, so use the integral (or the area under the F–x graph): $W = \int_{0.10}^{0.30} 200x\,dx = \frac{1}{2}(200)(0.30^2 - 0.10^2)$ .
2. Step 2: $W = 100(0.090 - 0.010) = 8.0$ J.
3. Step 3: The trap is using the 0.20 m change in stretch: $\frac{1}{2}(200)(0.20)^2 = 4.0$ J. That's the work to stretch it 0.20 m starting from relaxed, which is easier because the force starts at zero.
Answer: 8.0 J.

Common mistakes

Using Fd when the force is at an angle to the motion. Use $Fd\cos\theta$ or the dot product.
Using $\frac{1}{2}k(\Delta x)^2$ with Δx as the change in stretch. Use $\frac{1}{2}k(x_2^2 - x_1^2)$ , with both stretches measured from the relaxed length.
Saying a force does work just because it's large. A force perpendicular to the motion does no work at all.
Using the work of one force in the work–energy theorem. It's the net work, from all forces, that equals ΔK.

On the exam

Expect force–position graphs where you find work as an area, including parts below the axis. Count squares or split into triangles and rectangles.
Questions often ask for the sign of the work each force does. Compare the force's direction with the displacement: along it, positive; against it, negative; perpendicular, zero.

Connected topics

Videos

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Check yourself

4 questions on 3.2 Work. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A force $F(x) = 3x^2$ acts along the x-axis on an object, with F in newtons and x in meters. How much work does it do as the object moves from x = 0 to x = 2.0 m?

Question 2 of 4Calculator allowed

A constant force $\vec{F} = (3\hat{i} + 4\hat{j})$ N acts on an object as it moves through a displacement $\Delta\vec{r} = (2\hat{i} - \hat{j})$ m. How much work does the force do?

Question 3 of 4Calculator allowed

A 2.0 kg block slides 5.0 m down a ramp tilted 30° above the horizontal. How much work does gravity do on the block? Use g = 10 m/s².

Question 4 of 4Calculator allowed

A block slides down a frictionless curved ramp. How much work does the normal force do on the block?

0 of 4 answered