Coaching Centre
Work and kinetic energy
Work done by a force (W = Fs cos θ), positive, negative and zero work, work as the area under a force–displacement graph, kinetic energy, and the work–energy theorem W = F_net s = ΔK.
Hold a heavy shopping bag still at arm's length for a minute. Your arm burns and you feel exhausted, but a physicist would say you did no work on the bag at all. That isn't the physicist being difficult. "Work" has a precise meaning: energy transferred to an object by a force acting through a distance.
Forces change motion; energy tracks those changes. Work is the bridge between the two. Once you know the work done on an object, you can find its change in speed without using time or acceleration at all.
Work done by a force
When a constant force $F$ acts on an object that moves a displacement $s$, the work done by the force is
$$W = Fs\cos\theta$$where $\theta$ is the angle between the force and the displacement. Work is a scalar, measured in joules: 1 J = 1 N m. Only the component of the force along the displacement, $F\cos\theta$, does work.
| Angle θ | Work | Example |
|---|---|---|
| 0° | $W = Fs$, positive | pushing a trolley forwards |
| between 0° and 90° | positive, less than $Fs$ | pulling a suitcase by an angled handle |
| 90° | zero | the normal force on a car on a level road |
| 180° | $W = -Fs$, negative | friction on a sliding box, brakes on a car |
Positive work transfers energy to the object; negative work takes energy away from it. Holding the bag still does no work because $s = 0$. (Your muscles do use energy, because muscle fibres keep contracting and relaxing, but none of that energy goes into the bag.)
A child pulls a sled 25 m across flat snow with a rope at 35° to the horizontal. The tension is 60 N and the friction on the sled is 38 N. Find the work done by each force on the sled.
Work from a force–displacement graph
If the force changes during the motion, $W = Fs$ no longer works directly. But the work is always the area under the force–displacement graph (using the component of force along the motion). For a constant force, that area is a rectangle, $Fs$. For a force that grows steadily from zero, like stretching a spring, it is a triangle, $\tfrac12 Fs$.
Kinetic energy and the work–energy theorem
A moving object has kinetic energy
$$K = \tfrac12 mv^2$$Here is where it comes from. Suppose a constant net force $F_{net}$ acts on a mass $m$ along its motion for a displacement $s$, taking its speed from $u$ to $v$. Using $F_{net} = ma$ and $v^2 = u^2 + 2as$:
$$W_{net} = F_{net}s = ma\,s = m\left(\frac{v^2 - u^2}{2}\right) = \tfrac12 mv^2 - \tfrac12 mu^2$$This is the work–energy theorem: the net work done on an object equals its change in kinetic energy.
$$W_{net} = F_{net}\,s = \Delta K$$Notice that $K$ depends on $v^2$. Double your speed and your kinetic energy is four times as large, so stopping takes four times the work. That is the physics behind lower speed limits near schools.
A 4.0 kg cart starts from rest on a smooth floor. A student pulls it 5.0 m with a constant 20 N force at 30° above the horizontal. Find its final speed.
A 1500 kg car travelling at 25 m s⁻¹ brakes with a constant 9000 N braking force. How far does it travel while stopping? How far would it travel from 50 m s⁻¹?
- With Steady pull at an angle, 20 N at 30° over 5.0 m on a 4.0 kg cart: the shaded area is 86.6 J and the cart reaches 6.58 m/s.
- Turn on friction. It takes 14.6 J (pulling upward reduces the normal force to 29.2 N), and the cart reaches 6.00 m/s.
- Set the angle to 0° and choose Stretching spring: 20 N maximum over 5.0 m does 50.0 J, half what a steady 20 N force would.
- Set the angle to 90°. Whatever the force, the area is zero: a force perpendicular to the motion does no work.
Test yourself
A 50 N force pulls a box 8.0 m along a floor. The force is at 40° above the horizontal. The work done by the force is
$W = 50 \times 8.0 \times \cos40° = 306$ J.
A satellite moves in a circular orbit. The work done on it by gravity during one orbit is
In a circular orbit the gravitational force points to the centre, perpendicular to the motion at every instant, so it does no work and the speed stays constant.
A 0.15 kg cricket ball is bowled at 40 m s⁻¹. Its kinetic energy is
$K = \tfrac12(0.15)(40)^2 = 120$ J.
The speed of a car increases from 30 km/h to 60 km/h. Its kinetic energy
$K \propto v^2$, and $(60/30)^2 = 4$.
Label the quantities in W = Fs cos θ.
What work does the named force do on the object?
Work is measured in and equals force times displacement times the of the angle between them. For a force that varies, work is the under the force–displacement graph. The net work on an object equals its change in energy. Doubling an object's speed makes its kinetic energy times as large.
A 65 kg skateboarder rolls onto a level, rough concrete path at 6.0 m s⁻¹ and slows to 4.0 m s⁻¹ over 12 m. (a) Calculate the average resistance force. (3 marks) (b) A friend predicts that, since the speed dropped by 2.0 m s⁻¹ in 12 m, the skateboarder will need another 24 m to stop from 4.0 m s⁻¹. Use the work–energy theorem to show that this prediction is wrong. (2 marks)
(a) $\Delta K = \tfrac12(65)(4.0^2 - 6.0^2) = \tfrac12(65)(-20) = -650$ J. The resistance force does this negative work: $-F(12) = -650$, so $F = 54$ N (opposite to the motion).
(b) The work done by a constant resistance force equals the loss of kinetic energy, and kinetic energy depends on $v^2$, not $v$. Slowing from 4.0 m s⁻¹ to rest removes $\tfrac12(65)(4.0^2) = 520$ J, so the distance needed is $520/54 = 9.6$ m, not 24 m. The first 12 m removed more energy (650 J) even though the speed dropped by less, because the stopping distance depends on the change in $v^2$, not the change in speed.