GCSE Physics (9-1)

1.1 Energy stores and transfers — 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

  1. Work out 0.5×1200×1520.5 \times 1200 \times 15^2.
  2. Write 2500 in standard form.
  3. Convert 500 g to kilograms.
Answers
  1. 0.5×1200×225=1350000.5 \times 1200 \times 225 = 135000.
  2. 2.5×1032.5 \times 10^3.
  3. 0.50 kg0.50\,\text{kg}.

If question 1 went wrong, look again at the order of operations: work out the power before you multiply. If question 3 went wrong, remember that there are 1000 g in 1 kg, so you divide by 1000.

Practise with help

1. A climber of mass 75 kg reaches a ledge 8.0 m above the ground. Take g=9.8 N/kgg = 9.8\,\text{N/kg}. Calculate the energy gained by the gravitational potential store.

Hint

The store that fills as the climber rises is the gravitational potential store. Which equation has mass, gravitational field strength and height in it?

Next step

Write Ep=mghE_p = m g h and put in m=75 kgm = 75\,\text{kg}, g=9.8 N/kgg = 9.8\,\text{N/kg} and h=8.0 mh = 8.0\,\text{m}.

Full working Ep=mgh=75×9.8×8.0=5880 J=5900 J(2 significant figures)\begin{aligned} E_p &= m g h \\ &= 75 \times 9.8 \times 8.0 \\ &= 5880\,\text{J} \\ &= 5900\,\text{J} \quad (2\ \text{significant figures}) \end{aligned}

2. A ball of mass 0.16 kg is kicked at 20 m/s. Calculate the energy in its kinetic store.

Hint

The speed enters the kinetic energy equation as a square. Work out the square first, then multiply.

Next step

Write Ek=12mv2E_k = \tfrac{1}{2} m v^2, then replace v2v^2 with 202=40020^2 = 400.

Full working Ek=12mv2=0.5×0.16×202=0.5×0.16×400=32 J\begin{aligned} E_k &= \tfrac{1}{2} m v^2 \\ &= 0.5 \times 0.16 \times 20^2 \\ &= 0.5 \times 0.16 \times 400 \\ &= 32\,\text{J} \end{aligned}

3. A block of copper of mass 1.2 kg is heated from 20 °C to 70 °C. The specific heat capacity of copper is 390 J/kg°C. Calculate the energy transferred to the block.

Hint

The symbol Δθ\Delta\theta means the change in temperature, not the final temperature. Find the change first.

Next step

Work out Δθ=70−20=50 ∘C\Delta\theta = 70 - 20 = 50\,^\circ\text{C}, then substitute into ΔE=mcΔθ\Delta E = m c \Delta\theta.

Full working ΔE=mcΔθ=1.2×390×50=23400 J=2.3×104 J(2 significant figures)\begin{aligned} \Delta E &= m c \Delta\theta \\ &= 1.2 \times 390 \times 50 \\ &= 23400\,\text{J} \\ &= 2.3 \times 10^4\,\text{J} \quad (2\ \text{significant figures}) \end{aligned}

4. A motor transfers 18000 J in 12 s. Calculate its power.

Hint

Power is the energy transferred each second, so it needs the energy and the time.

Next step

Write P=EtP = \dfrac{E}{t} and divide 18000 J by 12 s.

Full working P=Et=1800012=1500 W\begin{aligned} P &= \frac{E}{t} \\ &= \frac{18000}{12} \\ &= 1500\,\text{W} \end{aligned}

5. A ball is dropped and bounces back to a lower height each time. Explain, in terms of energy stores, why it never returns to its starting height.

Hint

Think about what happens at the moment the ball hits the floor, and while it is moving through the air.

Next step

Name the store that receives the energy that the ball loses on each bounce.

Full working

As the ball falls, energy moves from the gravitational potential store to the kinetic store. When the ball hits the floor, some of the kinetic energy is transferred to the thermal store of the ball, the floor and the air. Air resistance transfers more energy to the thermal store of the air on every trip. The energy in the gravitational potential store at the top of each bounce is therefore smaller than before, so the ball reaches a lower height. This answer is a model answer for you to compare with your own wording.

6. A car of mass 1000 kg is travelling at 12 m/s. The driver brakes until the car stops. Calculate the energy transferred to the thermal store of the brakes, and name the store that decreased.

Hint

The car loses one store of energy as it slows down. Identify that store before you calculate.

Next step

Use the kinetic energy equation with m=1000 kgm = 1000\,\text{kg} and v=12 m/sv = 12\,\text{m/s}.

Full working

The kinetic store decreases. Its energy is transferred to the thermal store of the brakes.

Ek=12mv2=0.5×1000×122=0.5×1000×144=72000 J=7.2×104 J(2 significant figures)\begin{aligned} E_k &= \tfrac{1}{2} m v^2 \\ &= 0.5 \times 1000 \times 12^2 \\ &= 0.5 \times 1000 \times 144 \\ &= 72000\,\text{J} \\ &= 7.2 \times 10^4\,\text{J} \quad (2\ \text{significant figures}) \end{aligned}

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