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Saturday, May 18, 2019

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.

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