Work
Holding a heavy backpack still for ten minutes is tiring, yet a physicist would say you did no work on it at all. This lesson explains what work means in physics, how to calculate it, and why it can be negative.
By the end of this lesson you can
- tell the physics meaning of work from the everyday meaning;
- calculate the work done by a constant force, \(W = Fd\cos\theta\);
- decide whether a force does positive, negative or zero work;
- find the net work done on an object by several forces.
Work in physics
Think first
Which of these involve work done on the object, in the physics sense? (a) Holding a box still above your head. (b) Lifting the box from the floor to a shelf. (c) Pushing hard on a wall that does not move.
Compare your answer
Only (b). In (a) and (c) you push, but the object does not move, so no work is done on it. Your muscles get tired because they burn energy internally, not because they do work on the box or the wall.
In physics, work is done on an object when a force makes it move. Two things are needed: a force, and a displacement of the object. If either is missing, the work is zero.
The amount of work depends on how big the force is and how far the object moves. Push a stalled car 10 m to the side of the road and then a friend pushes it another 20 m with the same force: your friend has done twice as much work.
Only part of the force counts
Pull a suitcase by its handle at an angle. Part of your pull drags it forward. The rest pulls it upward, and since the suitcase does not move up, that part achieves nothing. What counts is the piece of the force that lies in the direction the object moves, \(F\cos\theta\).
Work done by a constant force
$$W = Fd\cos\theta$$\(F\) is the size of the force, \(d\) how far the object moves, and \(\theta\) the angle between the direction of the force and the direction of motion. Work is measured in joules (J): \(1\ \text{J} = 1\ \text{N}\cdot\text{m}\).
When the force points exactly along the motion, \(\theta = 0^\circ\) and \(\cos\theta = 1\), so \(W = Fd\). When the force is perpendicular to the motion, \(\cos 90^\circ = 0\) and the force does no work at all.
Change the angle of the pull
InteractiveScroll sideways to see the whole demo, or turn your phone sideways.
Try this
- Keep the force and distance fixed and change the angle from 0° to 90°. How does the red component change, and what happens to the work?
- Go past 90°. What sign does the work have now, and what does the message say?
- Press Worked example values to load the numbers of Worked example 1 below and check the answer.
Worked example 1
A traveler pulls a suitcase 4.0 m along a level floor with a force of 40 N directed 25° above the horizontal. How much work does the traveler do on the suitcase?
- Identify the quantities. \(F = 40\ \text{N}\), \(d = 4.0\ \text{m}\), and the angle between the pull and the horizontal displacement is \(\theta = 25^\circ\).
- Use the work equation. $$W = Fd\cos\theta = (40\ \text{N})(4.0\ \text{m})\cos 25^\circ$$
- Calculate.
\(W = 145\ \text{J}\)
Worked example 2
A crane lifts a 50.0 kg crate 6.0 m straight up at constant speed. How much work does the crane do on the crate? How much work does gravity do?
- Find the crane's force. At constant speed the crate is in equilibrium, so the crane pulls up with a force equal to the weight: \(F = mg = (50.0)(9.81) = 491\ \text{N}\).
- Work by the crane. Force and displacement both point up, so \(\theta = 0^\circ\): $$W_{\text{crane}} = (491\ \text{N})(6.0\ \text{m}) = 2.94 \times 10^3\ \text{J}$$
- Work by gravity. Gravity points down while the crate moves up, so \(\theta = 180^\circ\) and \(\cos\theta = -1\): $$W_g = -2.94 \times 10^3\ \text{J}$$
Crane: \(+2.94 \times 10^3\ \text{J}\). Gravity: \(-2.94 \times 10^3\ \text{J}\). The two add to zero, which fits the constant speed.
Positive, negative or zero work
Work is a scalar, but it has a sign. The sign tells you whether the force helps the motion or opposes it.
| Angle between force and motion | Sign of work | Example |
|---|---|---|
| \(0^\circ \le \theta < 90^\circ\) | Positive | You push a shopping cart forward. |
| \(\theta = 90^\circ\) | Zero | Gravity on a ball rolling along a flat floor. |
| \(90^\circ < \theta \le 180^\circ\) | Negative | Friction on a sliding box; brakes on a moving car. |
Common mistake
"Carrying a book across the room at constant speed is work." Your hand pushes up on the book, but the book moves sideways. The angle is 90°, so your upward force does no work on the book. (Speeding the book up and slowing it down at the ends of the trip does involve work, but it adds up to zero.)
Net work
When several forces act on an object, the net work is the total of the work done by each force, signs included:
$$W_{\text{net}} = W_1 + W_2 + W_3 + \cdots$$You can also find it from the net force: \(W_{\text{net}} = F_{\text{net}}\, d\cos\theta\). Both ways give the same answer.
The sign of the net work is a preview of the next lesson. If the net work is positive, the object speeds up. If it is negative, the object slows down. If it is zero, the speed stays the same.
Worked example 3
A worker pushes a 25.0 kg box 8.0 m across a level floor with a horizontal force of 120 N. The coefficient of kinetic friction is 0.35. Find the work done by the worker, by friction, by gravity and by the normal force, and the net work.
- Worker. The push points along the motion: \(W_{\text{push}} = (120\ \text{N})(8.0\ \text{m}) = 960\ \text{J}\).
- Friction. \(F_k = \mu_k mg = 0.35 \times 245 = 85.8\ \text{N}\), opposite the motion: \(W_f = -(85.8)(8.0) = -687\ \text{J}\).
- Gravity and the normal force are both perpendicular to the motion, so each does zero work.
- Add. \(W_{\text{net}} = 960 - 687 + 0 + 0 = 273\ \text{J}\).
\(W_{\text{net}} = +273\ \text{J}\), so the box is speeding up.
Check yourself
Question 1
A tugboat pulls a barge 2.5 km through a harbor with a constant forward force of \(4.0 \times 10^3\ \text{N}\). How much work does it do? (You can type 1.0e7 for \(1.0 \times 10^7\).)
Show solution
\(W = Fd = (4.0 \times 10^3\ \text{N})(2.5 \times 10^3\ \text{m}) = 1.0 \times 10^7\ \text{J}\)
Question 2
How much work does it take to lift a 250 g book 1.60 m at constant speed?
Show solution
Convert to kilograms first: \(m = 0.250\ \text{kg}\). \(W = mgd = (0.250)(9.81)(1.60) = 3.92\ \text{J}\).
Question 3
A gardener pushes a lawnmower 15.0 m with a force of 60.0 N directed along the handle, 40.0° below the horizontal. How much work does the gardener do?
Show solution
The angle between the push and the horizontal motion is 40.0°: \(W = (60.0)(15.0)\cos 40.0^\circ = 689\ \text{J}\). The downward part of the push does no work.
Question 4
A satellite moves in a circular orbit around Earth. Gravity always points toward Earth's center. How much work does gravity do on the satellite?
- Positive work, because gravity is the only force
- Zero, because gravity is always perpendicular to the motion
- Negative work, because gravity pulls inward
- It depends on the satellite's mass
Show solution
In a circular orbit the velocity is always along the circle, at 90° to the inward pull. \(\cos 90^\circ = 0\), so gravity does no work and the satellite's speed stays the same.
Question 5
If 5.00 J of work is done to lift a 0.400 kg ball straight up at constant speed, how high is it lifted?
Show solution
\(d = \dfrac{W}{mg} = \dfrac{5.00}{(0.400)(9.81)} = 1.27\ \text{m}\)
Question 6
A sled slides down a snowy hill. What is the sign of the work done on the sled by friction, and by gravity?
- Friction positive, gravity positive
- Friction positive, gravity negative
- Friction negative, gravity positive
- Both zero
Show solution
Friction points up the slope, against the motion: negative work. Gravity has a component down the slope, along the motion: positive work.
Summary
- A force does work on an object only if the object moves while the force acts.
- \(W = Fd\cos\theta\): only the part of the force pointing along the motion counts. The unit is the joule, \(1\ \text{J} = 1\ \text{N}\cdot\text{m}\).
- Work is positive when the force helps the motion, negative when it opposes it, and zero when it is perpendicular.
- The net work is the sum of the work done by every force.
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