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 (GPE) can be calculated using the formula:

$GPE = m imes g imes h$

where $m = 750 ext{ kg}$, $g = 10 ext{ N/kg}$, and $h = 1300 ext{ m}$. Thus,

$GPE = 750 imes 10 imes 1300 = 9,750,000 ext{ J}$

So, the change in gravitational potential energy is 9,750,000 J.

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

$ext{Efficiency} = \frac{ ext{Useful output}}{ ext{Input}}$

Rearranging gives:

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

The input energy supplied by the fuel in kilojoules is:

$\text{Input} = 6500 \text{ kJ}$

Now we find:

$\text{Useful output} = 0.70 \times 6500 = 4550 \text{ kJ}$

Next, to convert kJ to kW, we remember that power is energy per time. Since the energy is supplied every minute,

$\text{Power} = \frac{4550 \text{ kJ}}{60 \text{ s}} = 75.83 \text{ kW}$

Rounding gives approximately 76 kW.

The efficiency of the engine is less than 1 because not all the energy supplied by the fuel is converted into useful work. Some energy is lost or transferred to less useful forms, such as heat or noise. In this case, only 70% of the input energy is utilized to produce useful output; the remaining energy is dissipated, resulting in lower efficiency.