Work done by a force — F times d times cosθ — Physics

Before this

Work done — energy transferred by a force · coming soon
SOH CAH TOA — sin, cos and tan · coming soon
Impulse — force acting over time

This leads to

Kinetic energy and the work-energy theorem
Gravitational potential energy
What is a vector · coming soon

Ten minutes of effort, and no work done

Hold a heavy suitcase at arm's length, perfectly still, for ten minutes. Your arm shakes. You sweat. You are exhausted. Now here is the physics verdict: you did no work on the suitcase. Zero joules.

That sounds wrong, even unfair. But it is not a trick. It tells you that "work" in physics means something precise, and different from effort. Effort is about what happens inside your body. Work is about what a force does to the object it acts on.

So the question of this lecture is: when does a force do work, and how much? The answer needs one new idea, and that idea is hiding in a small piece of trigonometry, $cos θ$ .

Work is energy moved by a force

A force does work when it acts on an object and the object moves in the direction of the force. The work done is the energy that the force transfers to the object, measured in joules (J).

The simplest case is a force pointing exactly along the motion. Then the work is the force multiplied by the displacement:

W = F s

Here $F$ is the force in newtons and $s$ is the displacement in metres, using the same letter $s$ as in the SUVAT equations. Many books write this as $W = F d$ . It is the same formula.

Look at the units. A newton multiplied by a metre is a newton-metre, and this is what a joule is: $1 J = 1 N m$ . One joule is the work done by a force of 1 N pushing something 1 m along its own direction. Lifting an apple of about 100 g through 1 m takes about 1 J.

The lecture on impulse said that a force multiplied by the time it acts gives a change in momentum. It also said that a force multiplied by the distance it acts over gives the work done. This is that second idea, made exact.

Now return to the suitcase. Your upward force is real, and it is large. But the suitcase does not move, so $s = 0$ . Then $W = F \times 0 = 0$ . Your muscles are using energy, because muscle fibres keep contracting and relaxing to hold the weight. That energy becomes heat in your body. None of it is transferred to the suitcase. The suitcase ends with exactly the energy it started with.

A pull at an angle

Now pull a suitcase along an airport floor by a strap. The strap is not horizontal. It points up and forwards at an angle $θ$ above the floor. The suitcase moves horizontally. How much of your pull does work?

The pull

F

splits into two parts.

F cos θ

points along the motion and does work.

F sin θ

points straight up, at right angles to the motion, and does none.

Split the force into two components, as in the figure. This is the same right-angled triangle as in trigonometry. The force is the hypotenuse. The part along the floor is the adjacent side, $F cos θ$ . The part straight up is the opposite side, $F sin θ$ .

The horizontal part pulls the suitcase in the direction it moves. So it does work: $F cos θ \times s$ .

The vertical part only lifts slightly on the suitcase, which stays on the floor. The suitcase never moves up, so the vertical part moves nothing in its own direction. It does no work, exactly like your arm holding the suitcase still.

Add the two parts and you get the general rule.

Work done by a constant force

A constant force $F$ acts on an object that moves through a straight-line displacement $s$ . The work done by the force is

$W = F s cos θ$

Here $θ$ is the angle between the force and the displacement. Work is measured in joules: $1 J = 1 N m$ .

There is a second way to read the same formula. Write it as $F \times (s cos θ)$ . Now $s cos θ$ is the part of the displacement that lies along the force. Both readings give the same number. Use whichever picture is clearer in a problem. In both, the rule is the same: only motion along the force, or force along the motion, counts.

Change the angle yourself below, and watch what happens to the useful part of the force.

Explore — the angle decides the work

A 50 N force acts on a box while the box moves 4 m to the right. Change the angle between the force and the motion from 0° to 180°. Watch the component along the motion and the work done. Find where the work becomes zero, and what happens after that.

Angle θ: 30°

Try:

The table shows a 50 N force acting while a box moves 4 m, for different angles between the force and the motion.

Angle θ	cos θ	Component along the motion	Work done
0°	1	50 N	200 J
30°	≈ 0.866	≈ 43.3 N	≈ 173 J
60°	0.5	25 N	100 J
90°	0	0 N	0 J
120°	−0.5	−25 N	−100 J
180°	−1	−50 N	−200 J

What did you see? At 0°, the whole force works: 200 J. As the angle grows, the useful component shrinks, slowly at first and then faster. At 90° it disappears completely. Past 90°, the component turns round and points against the motion, and the work becomes negative. At 180° the force pushes straight backwards, and the work is $- 200$ J.

That is it. That is the whole idea. Only the part of a force along the motion counts. The $cos θ$ in the formula is simply the tool that picks out that part.

Zero work: three different reasons

A force can do zero work in three different ways. Learning to see all three is most of the skill in this topic.

Nothing moves. Holding the suitcase still: $s = 0$ .
The force is at right angles to the motion. Carry the suitcase across a level room at a steady height. Your upward force is perpendicular to the horizontal motion, so $cos 9 0^{\circ} = 0$ . Gravity does no work on the suitcase either, for the same reason. The normal force from the floor on a sliding box does no work, also for the same reason.
The force keeps turning to stay at right angles. In the lectures on circular motion and orbits, the force on a satellite in a circular orbit always points to the centre. The motion is always along the circle. So at every moment the angle is 90°, and gravity does no work. That is exactly why the satellite's speed never changes.

Stop and think

A comet moves in a long, stretched ellipse around the Sun. Does the Sun's gravity do work on the comet?

Show the answer

Yes, for most of the orbit. In an ellipse, the pull towards the Sun is not at right angles to the motion, except at the nearest and farthest points. As the comet falls inwards, the pull has a component along its motion, so gravity does positive work and the comet speeds up. As it moves outwards, the pull has a component against its motion, so gravity does negative work and the comet slows down. Over one full orbit, the positive and negative work cancel. A circular orbit is the special case where the angle is 90° everywhere, so the work is zero at every moment.

Negative work: a force that takes energy away

When $θ$ is between 90° and 180°, $cos θ$ is negative, so the work is negative. A negative sign in physics usually means a direction. Here it means a transfer: the force is taking energy out of the object instead of putting it in.

Friction on a sliding box is the standard example. The box moves forwards and friction acts backwards, so $θ = 18 0^{\circ}$ and $W = - F s$ . The energy taken from the box becomes heat in the box and the floor. When you catch a ball, your hand pushes against the ball's motion, so your hand does negative work on it. That is how the ball's energy is removed.

Here is a point that is easy to get wrong. Positive work by one force does not mean that the object speeds up. Push a crate across a rough floor at a steady speed. Your push does positive work. Friction does an equal amount of negative work. The total is zero, and the speed stays the same. What decides whether an object speeds up is the total work done by all the forces on it. The next lecture shows exactly how that total is connected to the speed.

Worked examples

Worked Example 1 — A push angled downwards

Mei pushes a crate along a level floor with a force of 50 N, directed 30° below the horizontal. The crate moves 4.0 m. How much work does her push do?

The angle in the formula is between the force and the displacement. The displacement is horizontal and the force is 30° below horizontal, so $θ = 3 0^{\circ}$ . Being below rather than above the horizontal does not change the angle between the two directions.

$W = F s cos θ = 50 \times 4.0 \times cos 3 0^{\circ} \approx 50 \times 4.0 \times 0.866 \approx 173$ J.

Check: the answer is less than $50 \times 4.0 = 200$ J, as it must be, because only part of the push is along the motion. Notice the other part: $50 sin 3 0^{\circ} = 25$ N presses the crate into the floor. That does no work itself, but it increases the friction. This is why pushing down at an angle feels harder than pulling up at an angle.

Worked Example 2 — Carrying a book

Tomás lifts a 2.0 kg book from a table through 1.2 m, then carries it 3.0 m across the room at a steady height. Take $g = 9.8$ m/s². Find the work done on the book by gravity in each stage.

The weight is $m g = 2.0 \times 9.8 = 19.6$ N, downwards.

Lifting. The displacement is 1.2 m upwards and gravity points downwards, so $θ = 18 0^{\circ}$ . $W = 19.6 \times 1.2 \times cos 18 0^{\circ} = - 23.52$ J, so about $- 23.5$ J. Gravity takes energy from the book. (Tomás's lifting force does about $+ 23.5$ J. That energy is stored, and the lecture on gravitational potential energy follows it.)

Carrying. The displacement is horizontal and gravity is vertical, so $θ = 9 0^{\circ}$ . $W = 19.6 \times 3.0 \times 0 = 0$ J. Gravity does no work during the carry.

Worked Example 3 — Every force on a car

A car moves 500 m along a straight, level road. The driving force is 1200 N forwards, and the resistive forces (air resistance and rolling resistance) total 400 N backwards. Find the work done by each force, and the total.

Take forwards as the direction of the displacement.

Driving force: $θ = 0^{\circ}$ , so $W_{1} = 1200 \times 500 \times 1 = 600000$ J $= 600$ kJ.

Resistive forces: $θ = 18 0^{\circ}$ , so $W_{2} = 400 \times 500 \times (- 1) = - 200000$ J $= - 200$ kJ.

Weight and the normal force: both are vertical, at 90° to the motion, so each does 0 J.

Total work: $600 - 200 = 400$ kJ. This total is positive, so the car gains energy of motion. The next lecture shows how to turn that 400 kJ into a speed.

Worked Example 4 — A sledge in the snow

Amara pulls a sledge 30 m across level snow. Her rope pulls with 80 N at 25° above the horizontal. Friction on the sledge is 60 N. Find the work done by the rope, by friction, and in total.

Rope: $W = 80 \times 30 \times cos 2 5^{\circ} \approx 80 \times 30 \times 0.906 \approx 2175$ J.

Friction: $W = 60 \times 30 \times cos 18 0^{\circ} = - 1800$ J.

Total: $2175 - 1800 \approx 375$ J, positive, so the sledge speeds up a little over the 30 m.

Check the size: if the rope were horizontal, it would do $80 \times 30 = 2400$ J. The angle costs about 225 J, which is reasonable for a small angle, because $cos 2 5^{\circ}$ is close to 1.

The most common slip in these problems is using the wrong angle. Find it in the working below.

Spot the error — the angle from the wrong line

A rope pulls a box 10 m along level ground with a force of 200 N. The rope makes an angle of 60° with the vertical. A student finds the work done by the rope. Tap the line that is wrong.

The working below contains one error, in line 3.

1. The box moves horizontally, so the displacement is 10 m along the ground.
2. The angle in the formula is the angle between the force and the displacement.
3. The rope makes 60° with the vertical, so θ = 60°. Wrong: the displacement is horizontal, so the angle needed is 90° − 60° = 30°.
4. W = 200 × 10 × cos 60° = 1000 J. (The correct answer is 200 × 10 × cos 30° ≈ 1732 J.)
5. Gravity and the normal force do no work, because they are at right angles to the displacement.

A quick check catches this error. A rope at 60° to the vertical is closer to horizontal than vertical, so most of the force should be along the motion. The work should be more than half of $200 \times 10 = 2000$ J. The answer 1000 J is exactly half, which is too small. The correct 1732 J passes the check.

Three classic mistakes

Trap 1 — Measuring θ from the wrong line

θ is the angle between the force and the displacement. It is not the angle to the vertical, or to a slope. Draw both arrows from one point and mark the angle between them before you calculate anything.

Trap 2 — Confusing effort with work

Holding, pressing and supporting can be exhausting and still do no work on the object. Work needs a displacement with a component along the force. If you write "work is done because the person is tired", the answer is wrong.

Trap 3 — Using displacement for friction on a round trip

Slide a box 5 m to the right and 5 m back. Its displacement is zero, but friction did not do zero work. Friction turned round with the motion, so it acted against the box on both legs, and the work is $- F \times 5$ twice. $W = F s cos θ$ is for a constant force and one straight displacement. When the force changes direction, split the journey into parts.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Take $g = 9.8$ m/s².

Ravi lifts a 5.0 kg box at steady speed from the floor onto a shelf 1.6 m high. Find the work done by Ravi and the work done by gravity on the box.

Hint 1
At steady speed, Ravi's upward force equals the box's weight.

Hint 2
The weight is $m g$ . Ravi's force is along the displacement ( $θ = 0^{\circ}$ ); gravity is opposite to it ( $θ = 18 0^{\circ}$ ).

Full answer
Weight: $5.0 \times 9.8 = 49$ N. Ravi: $W = 49 \times 1.6 \times cos 0^{\circ} = 78.4$ J. Gravity: $W = 49 \times 1.6 \times cos 18 0^{\circ} = - 78.4$ J. The total work is zero, which matches the box rising at a steady speed: its energy of motion does not change. Ravi's 78.4 J is not lost. It is stored, because the box is now higher.
A traveller pulls a suitcase 250 m through an airport. The strap pulls with 120 N at 35° above the horizontal. How much work does the strap do? What horizontal pull would do the same work?

Hint 1
θ is 35°, the angle between the strap and the horizontal motion.

Hint 2
A horizontal pull doing the same work must equal the component $F cos θ$ .

Full answer
$W = 120 \times 250 \times cos 3 5^{\circ} \approx 120 \times 250 \times 0.819 \approx 24600$ J, about 24.6 kJ. A horizontal pull of $120 cos 3 5^{\circ} \approx 98.3$ N would do the same work over 250 m. The extra 21.7 N in the strap goes into the upward component, which does no work but does take some weight off the wheels.
A 0.15 kg ball is thrown straight up, rises 12 m, and falls back to the hand. Ignoring air resistance, find the work done by gravity on the way up, on the way down, and in total. Then suppose air resistance is a constant 0.3 N against the motion. What work does it do over the whole trip?

Hint 1
Gravity always points down. Ask whether it is along or against the motion on each leg.

Hint 2
Air resistance turns round when the ball turns round. It is against the motion on both legs, like friction in Trap 3.

Full answer
Weight: $0.15 \times 9.8 = 1.47$ N. Up: $W = 1.47 \times 12 \times cos 18 0^{\circ} = - 17.64$ J. Down: $W = 1.47 \times 12 \times cos 0^{\circ} = + 17.64$ J. Total: 0 J. Gravity takes energy on the way up and gives it all back on the way down. Air resistance: $- 0.3 \times 12$ on each leg, so $- 3.6 - 3.6 = - 7.2$ J in total. It never gives energy back. This is why a real ball returns to your hand a little slower than you threw it.
A 40 kg crate is pushed 5.0 m up a ramp that rises 1.5 m. Find the work done by gravity on the crate in two ways: with $W = F s cos θ$ , and by thinking only about the height. What do you notice?

Hint 1
The ramp makes an angle $α$ with the horizontal, where $sin α = 1.5 \div 5.0 = 0.3$ . Gravity points straight down, so the angle between gravity and the displacement up the ramp is $9 0^{\circ} + α$ .

Hint 2
$cos (9 0^{\circ} + α) = - sin α$ . For the second way, use the second reading of the formula: the part of the displacement along the force is the drop in height.

Full answer
Weight: $40 \times 9.8 = 392$ N. First way: $α \approx 17. 5^{\circ}$ , and $cos (9 0^{\circ} + α) = - sin α = - 0.3$ . So $W = 392 \times 5.0 \times (- 0.3) = - 588$ J. Second way: the part of the displacement along gravity is the vertical change, 1.5 m upwards, which is against gravity. So $W = - 392 \times 1.5 = - 588$ J. The answers agree. Gravity's work depends only on the change in height, not on the length or shape of the path. A steep 1.5 m climb and a gentle ramp give the same $- 588$ J. This fact is the reason gravitational potential energy can be defined at all, as a later lecture in this topic shows.

The bigger idea

Force and displacement are both vectors: each has a size and a direction. Work is not. It is a single number, a scalar, with a sign but no direction. The formula $W = F s cos θ$ is the way physics combines two vectors into one number, by asking how much of one lies along the other. In later mathematics this operation has its own name, the dot product, and it appears everywhere: in electric circuits, in magnetism, in computer graphics and in machine learning.

It also explains something you may have noticed. Some forces always act at right angles to the motion, such as the pull on a satellite in a circular orbit, or the magnetic force on a moving charge. They can bend a path without ever changing the speed. They steer. They never pay or take energy. Work is the part of a force that changes energy, and $cos θ$ is how you find it.

Key concepts in this lesson

Work is the energy transferred by a force, $W = F s cos θ$ , measured in joules, where $1 J = 1 N m$ .
The angle $θ$ is between the force and the displacement, and $cos θ$ picks out the part of the force that lies along the motion.
A force does zero work when nothing moves, when it is at right angles to the motion, or when it keeps turning to stay at right angles, as in a circular orbit.
A force with a component against the motion, such as friction, does negative work and takes energy away from the object.
Whether an object speeds up depends on the total work done by all the forces on it, not on the work done by any one force.