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

The formula for gravitational potential energy (GPE) is given by:

$GPE = m imes g imes h$

where:

• m = mass (750 kg)
• g = gravitational field strength (10 N/kg)
• h = height change (1300 m)

Substituting the values into the formula:

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

Thus, the change in gravitational potential energy is 9,800,000 J.

To determine the power output, we can use the efficiency equation:

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

Rearranging for useful output, we have:

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

Substituting the given values:

Input energy = 6500 kJ/min = 6500 kJ/60 s = 108.33 kJ/s = 108.33 kW

Thus, the useful output energy is:

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

Now applying the power formula:

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

The efficiency of the engine is less than 1 (100%) because not all the energy from the fuel is converted to useful output energy. Some of the energy is lost or dissipated in forms that are less useful, such as heat, sound, or vibrations. In this case, the engine only converts 70% of the input energy into useful work, while 30% is lost as waste energy.