An aircraft of mass 1200 kg climbs upwards with a constant velocity of 45 m s-1, as shown in Fig. 3.1.

Question 27

(a) (i) Define power. [1]

(ii) State what is meant by gravitational potential energy. [1]

(b) An aircraft of mass 1200 kg climbs upwards with a constant velocity of 45 m s^-1, as shown in Fig. 3.1.

Fig. 3.1 (not to scale)

The aircraft’s engine produces a thrust force of 2.0 × 10³ N to move the aircraft through the air. The rate of increase in height of the aircraft is 3.3 m s^-1.

(i) Calculate the power produced by the thrust force. [2]

(ii) Determine, for a time interval of 3.0 minutes,

1. the work done by the thrust force to move the aircraft, [2]

2. the increase in gravitational potential energy of the aircraft, [2]

3. the work done against air resistance. [1]

(iii) Use your answer in (b)(ii) part 3 to calculate the force due to air resistance acting on the aircraft. [1]

(iv) With reference to the motion of the aircraft, state and explain whether the aircraft is in equilibrium. [2]

[Total: 12]

Reference: Past Exam Paper – November 2018 Paper 21 Q3

Solution:

(a) (i) Power is defined as the rate of doing work.

(ii) Gravitational potential energy is the energy possessed by a mass due to its position in a gravitational field.

(b)

(i)

P = Fv

{We need to use velocity of the aircraft, which is in the direction of the thrust (i.e. v = 45 m s^-1), not the rate of increase in height (i.e. not 3.3 m s^-1).}

P = 2.0 × 10³ × 45

P = 9.0 × 10⁴ W

(ii)

1.

{Power = work done / time

Work done = F × d                              or Work done = Power × time

Distance d = Speed × time = (45 × 3.0 × 60)}

W = (2.0 × 10³) × (45 × 3.0 × 60)       or W = 9.0×10⁴ × 3.0 × 60

W = 1.6 × 10⁷ J          

2.

(Δ)EP = mg(Δ)h

{(Δ)h is the increase in height.

Rate of increase in height (= speed) = 3.3 m s^-1

Speed = distance / time                      So, Distance = Speed × time

Distance (increase in height) = 3.3 × 3.0×60 m}

(Δ)EP = 1200 × 9.81 × 3.3 × 3.0 × 60

(Δ)EP = 7.0 × 10⁶ J

3.

{Work done by thrust force = Gain in GPE + Work done against air resistance

Work done against air resistance = Work done by thrust – Gain in GPE}

W = 1.6×10⁷ – 7.0×10⁶

W = 9.0 × 10⁶ J

(iii)

{Against air resistance,

Work done = Force × distance moved in direction of force

Distance moved in direction of force = Speed × time = (45 × 3.0 × 60) m}

Force = (9.0 × 10⁶) / (45 × 3.0 × 60)

F = 1.1 × 10³ N

(iv) Since the velocity of the aircraft is constant, there is no resultant force on it. So, it is in equilibrium as the resultant force is zero.