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

To calculate the change in gravitational potential energy (GPE), we use the formula:

$ext{GPE} = m imes g imes h$

where:

• $m$ is the mass (750 kg),
• $g$ is the gravitational field strength (10 N/kg),
• $h$ is the height change (1300 m).

Substituting in the values:

$ext{GPE} = 750 ext{ kg} imes 10 ext{ N/kg} imes 1300 ext{ m} = 9,750,000 ext{ J}$

Efficiency is calculated as follows:

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

Given that the efficiency is 0.70 and the input energy over one minute is 6500 kJ:

Converting kJ to J:

• Input = $6500 ext{ kJ} = 6500 imes 1000 ext{ J} = 6,500,000 ext{ J}$.

Now, rearranging the efficiency equation to find useful output:

• Useful output = Efficiency × Input = $0.70 × 6,500,000 ext{ J} = 4,550,000 ext{ J}$.

To find power (in watts): Power = rac{ ext{Energy}}{ ext{Time}}; with time being 60 seconds:

• Power = rac{4,550,000 ext{ J}}{60 ext{ s}} \approx 75.83 ext{ kW}.

The efficiency of the engine is less than 1 because not all the input energy is converted into useful output energy. Some energy is lost as heat or dissipated to the surroundings, meaning only a portion (70% in this case) is utilized for work, while the remaining 30% is wasted.