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

Draw the resultant force vector starting from the tail of the 300 N vector and pointing towards the direction that is the vector sum of the 300 N and 400 N forces.

Using the formula for gravitational potential energy (GPE) given by:

$GPE = m imes g imes h$

where,

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

Substituting the values:

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

Thus, energy = 9,750,000 J.

First, calculate the useful output energy of the engine:

$ext{Useful output energy} = ext{Efficiency} imes ext{Input energy}$

Input energy = 6500 kJ/min, so converting it to joules:

$ext{Input energy} = 6500 imes 10^3 ext{ J} = 6,500,000 ext{ J}$

Then,

$ext{Useful output energy} = 0.70 imes 6,500,000 ext{ J} = 4,550,000 ext{ J}$

Now, to find power output, we use:

ext{Power} = rac{ ext{Energy}}{ ext{Time}}

Power in kilowatts (kW), since time = 1 minute = 60 seconds:

The efficiency of the engine is less than 1 (100%) because not all input energy is converted into useful output energy. A significant portion is lost as heat due to friction and other inefficiencies, meaning that some energy is wasted rather than being harnessed for useful work.