1.2 Conservation, dissipation and energy resources — 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 lamp transfers 2000 J of energy in 50 s. Calculate its power.
Name the store that increases when a kettle heats water.

Answers

$P = \dfrac{E}{t} = \dfrac{2000}{50} = 40\,\text{W}$ .
The thermal store of the water and the kettle.

If question 1 went wrong, go back to Section 1.1 and the equation $P = \dfrac{E}{t}$ .

Practise with help

1. A ball has 30 J of energy in its gravitational potential store at the top of a drop. Air resistance is ignored. Calculate the energy in its kinetic store just before it lands.

Hint

The drop has no outside energy going in or out, so the total energy stored does not change.

Next step

The gravitational potential store empties by exactly the amount the kinetic store fills.

Full working

Energy is conserved, so the kinetic store gains the 30 J that the gravitational potential store loses.

E_k = 30\,\text{J}

2. In a real room, the ball from question 1 lands with 26 J in its kinetic store. Calculate the energy dissipated, and name the store that receives it.

Hint

The missing energy has not gone away. It has moved somewhere else.

Next step

Subtract the kinetic energy at the bottom from the energy at the top.

Full working

\text{dissipated} = 30 - 26 = 4\,\text{J}

The 4 J is transferred to the thermal store of the ball and the air. Air resistance is the cause.

3. A motor takes in 600 J of energy and transfers 450 J usefully. Calculate its efficiency as a percentage.

Hint

Efficiency compares the useful output with the total input, so use 600 J, not 150 J.

Next step

Work out $\dfrac{450}{600}$ , then multiply by 100 to turn the decimal into a percentage.

Full working

\begin{aligned} \text{efficiency} &= \frac{450}{600} \\ &= 0.75 \\ &= 75\% \end{aligned}

4. A machine is 25% efficient and its useful output is 150 J. Calculate the total input.

Hint

Write 25% as 0.25 before you use the equation.

Next step

Rearrange $\text{efficiency} = \dfrac{\text{useful output}}{\text{total input}}$ to make the total input the subject.

Full working

\begin{aligned} 0.25 &= \frac{150}{\text{total input}} \\ \text{total input} &= \frac{150}{0.25} \\ &= 600\,\text{J} \end{aligned}

5. A hot drink is poured into a thin plastic cup. The same drink in a thick insulated cup cools more slowly. Explain why.

Hint

Compare the two cups by the thickness and the thermal conductivity of the material between the drink and the air.

Next step

Say which cup transfers energy to the surroundings faster, and why.

Full working

The thin plastic cup lets energy transfer from the hot drink to the cooler surroundings quickly. The thick insulated cup has a layer of trapped air, which has a low thermal conductivity. The thicker layer and the lower thermal conductivity both slow the rate of energy transfer, so the drink cools more slowly. This answer is a model answer for you to compare with your own wording.

6. A wind turbine removes $4.0 \times 10^6\,\text{J}$ of energy from the kinetic store of the air every second. Its useful electrical output power is 1.6 MW. Calculate the efficiency of the turbine, and give one reason why the efficiency is not 100%.

Hint

$4.0 \times 10^6\,\text{J}$ each second is a power of $4.0 \times 10^6\,\text{W}$ , which is 4.0 MW.

Next step

Divide the useful output power by the total input power.

Full working

\begin{aligned} \text{efficiency} &= \frac{1.6}{4.0} \\ &= 0.40 \\ &= 40\% \end{aligned}

The air still leaves the turbine with kinetic energy, and some energy is transferred by friction in the moving parts to the thermal store of the surroundings. That energy is dissipated, so it does not reach the generator as useful electrical energy.