Figure 14 shows the vertical forces on an aeroplane - Edexcel - GCSE Physics - Question 7 - 2018 - Paper 1

In Figure 15, the resultant of the 400 N and 300 N forces can be represented as a vector that combines both forces. The resultant vector can be drawn from the tail of the 300 N force to the head of the 400 N force, forming a right triangle. The resultant can be calculated using the Pythagorean theorem:

$ext{Resultant} = \\sqrt{(400)^2 + (300)^2} = 500 ext{ N}$

The direction can be determined using trigonometry (tan inverse) to find the angle.

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

$\text{GPE} = m \cdot g \cdot h$

where:

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

Thus,

$\text{GPE} = 750 \cdot 10 \cdot 1300 = 9750000 ext{ J}$

Therefore, the change in energy is 9750000 J.

The power output can be calculated using the efficiency and energy supplied by the fuel. First, we calculate the useful output energy, which is:

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

Now, convert kJ to J:

$4550 \text{ kJ} = 4550 \times 10^3 ext{ J}$

Power is then calculated using the formula:

$\text{Power} = \frac{\text{Energy}}{\text{Time}}$

Given that energy is supplied every minute (60 seconds):

$\text{Power} = \frac{4550 \times 10^3}{60} \approx 75833.33 ext{ W} \approx 75.83 ext{ kW}$

The efficiency of an engine is less than 100% because not all the energy supplied by the fuel is converted into useful work. Some energy is lost to the surroundings in forms such as heat or sound. This means that a portion of the energy is dissipated, contributing to inefficiencies in the engine's operation. For example, only 70% of the input energy is transformed into useful work while the remaining 30% is wasted.