4 (a) (i) Figure 5 shows the vertical forces on an aeroplane - Edexcel - GCSE Physics Combined Science - Question 4 - 2018 - Paper 1

The change in gravitational potential energy can be calculated using the formula:

$ext{GPE} = mgh$

Where:

• $m = 750 \text{ kg}$ (mass of the aeroplane)
• $g = 10 \text{ N/kg}$ (gravitational field strength)
• $h = 1300 \text{ m}$ (height)

Substituting the values:

$ext{GPE} = 750 \times 10 \times 1300 = 9,750,000 \text{ J}$

Thus, the change in gravitational potential energy is 9.8 x 10^6 J.

To find the power output, we use the efficiency equation:

$\text{Efficiency} = \frac{\text{Useful Output}}{\text{Input}}$

Rearranging gives:

$\text{Useful Output} = \text{Efficiency} \times \text{Input}$

Here, the input energy, given that the fuel supplies 6500 kJ per minute, is:

$\text{Input} = 6500 \text{ kJ} = 6500 \times 1000 = 6,500,000 \text{ J}$

Thus, the useful output energy is:

$\text{Useful Output} = 0.70 \times 6,500,000 = 4,550,000 \text{ J}$

Power is defined as energy per time. Given that this energy is supplied in one minute (60 seconds):

$\text{Power} = \frac{4,550,000 \text{ J}}{60 \text{ s}} \approx 75.83 \text{ kW}$

The efficiency of the engine is less than 1 because not all input energy is converted into useful work. Some energy is inevitably lost to the surroundings as heat or in less useful forms.

Specifically, this could be due to:

• Energy being dissipated as thermal energy.
• Mechanical losses in the components of the engine. Thus, only 70% of the input energy contributes to useful work.