3.3 Work, energy and power — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

A car accelerates uniformly from $8.0\,\text{m}\,\text{s}^{-1}$ to $24\,\text{m}\,\text{s}^{-1}$ in 4.0 s. Find its acceleration and the distance travelled.
A person of mass 60 kg stands in a lift that accelerates downwards at $2.0\,\text{m}\,\text{s}^{-2}$ . Find the normal reaction. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A force of 100 N acts at 35° to the horizontal. Find its horizontal and vertical components.

Answers

$a = 4.0\,\text{m}\,\text{s}^{-2}$ and $s = 64\,\text{m}$ .
$N = 60(9.81 - 2.0) = 469\,\text{N}$ .
$100\cos 35^\circ = 81.9\,\text{N}$ and $100\sin 35^\circ = 57.4\,\text{N}$ .

If question 1 goes wrong, use $a = \dfrac{v - u}{t}$ and $s = \dfrac{(u + v)t}{2}$ . If question 2 goes wrong, the acceleration is downwards, so $mg - N = ma$ . If question 3 goes wrong, the angle is measured from the horizontal, so that component uses the cosine.

Practise with help

1. A force of 150 N moves a crate 8.0 m along the direction of the force. Find the work done.

Hint

The force and the displacement are in the same direction, so the angle between them is zero.

Next step

Use $W = Fs$ , because $\cos 0 = 1$ .

Full working

W = Fs = 150 \times 8.0 = 1200\,\text{J}

The work done is 1.2 kJ.

2. A force of 200 N acts at 40° to the horizontal and drags a box 15 m along the ground. Find the work done by the force.

Hint

Only the component of the force along the displacement does work.

Next step

Use $W = Fs\cos\theta$ with $\theta = 40^\circ$ , and put your calculator in degree mode.

Full working

\begin{aligned} W &= Fs\cos 40^\circ \\ &= 200 \times 15 \times 0.766 \\ &= 2300\,\text{J} \end{aligned}

The vertical component of the force does no work, because the box moves horizontally.

3. A stone of mass 4.0 kg is released from rest and falls 12 m. Find the gravitational potential energy it loses and its speed just before it lands. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

The potential energy lost becomes kinetic energy. You can write $mg\Delta h = \tfrac{1}{2}mv^2$ .

Next step

Find $\Delta E_p = mg\Delta h$ first, then use $v = \sqrt{2g\Delta h}$ .

Full working

\begin{aligned} \Delta E_p &= mg\Delta h = 4.0 \times 9.81 \times 12 = 471\,\text{J} \\ v &= \sqrt{2g\Delta h} = \sqrt{2 \times 9.81 \times 12} \\ &= 15.3\,\text{m}\,\text{s}^{-1} \end{aligned}

The mass cancels in the speed, so a heavier stone would land at the same speed.

4. A ball of mass 0.50 kg is thrown vertically upwards at $12\,\text{m}\,\text{s}^{-1}$ . Use an energy method to find the maximum height it reaches. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

At the highest point the kinetic energy is zero, so all the kinetic energy has become gravitational potential energy.

Next step

Write $\tfrac{1}{2}mv^2 = mgh$ and cancel the mass before you substitute.

Full working

\begin{aligned} \tfrac{1}{2}mv^2 &= mgh \\ h &= \frac{v^2}{2g} = \frac{12^2}{2 \times 9.81} \\ &= 7.3\,\text{m} \end{aligned}

The energy method gives the same height as $v^2 = u^2 + 2as$ from Section 3.1, because both come from the same conservation law.

5. A motor lifts a load of mass 250 kg through 20 m in 12 s. Find the work done against gravity and the useful power of the motor. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

The work done against gravity is $mg\Delta h$ . Power is work divided by time.

Next step

Find the work first, then divide by 12 s.

Full working

\begin{aligned} W &= mg\Delta h = 250 \times 9.81 \times 20 = 49050\,\text{J} \\ P &= \frac{W}{t} = \frac{49050}{12} = 4088\,\text{W} \end{aligned}

The useful power is about 4.1 kW. A real motor draws more power than this, because some energy is transferred to thermal stores.

6. A cyclist of total mass 80 kg freewheels from rest down a hill, descending 20 m. The cyclist reaches $15\,\text{m}\,\text{s}^{-1}$ at the bottom. Find the energy transferred to thermal stores. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

Compare the gravitational potential energy lost with the kinetic energy gained. The difference has gone somewhere.

Next step

Find $\Delta E_p = mg\Delta h$ and $E_k = \tfrac{1}{2}mv^2$ separately, then subtract.

Full working

\begin{aligned} \Delta E_p &= 80 \times 9.81 \times 20 = 15696\,\text{J} \\ E_k &= \tfrac{1}{2} \times 80 \times 15^2 = 9000\,\text{J} \\ \text{thermal energy} &= 15696 - 9000 = 6696\,\text{J} \end{aligned}

About 6700 J has been transferred to thermal stores by friction and air resistance, so the cyclist is slower than the frictionless prediction.