Q#14 (Past Exam Paper – June 2015 Paper 13 Q18)
A loaded aeroplane has a total mass of 1.2×105 kg while climbing after take-off. It climbs at an angle of 23°to the horizontal with a speed of 50 m s–1. What is the rate at which it is gaining potential energy at this time?
A 2.3 × 106 J s-1
B 2.5 × 106 J s-1
C 2.3 × 107 J s-1
D 2.5 × 107 J s-1
Solution:
Answer: C.
In this question, we need to manipulate the formulae to obtain quantities with value already given in the question.
Rate of gaining potential energy = Gain in PE / time
GPE = mgh where h is the vertical height
So, to consider the gain in GPE, we need to consider the vertical component of speed
The aeroplane climbs at a speed of 50 m s-1 at an angle of 23° to the horizontal.
Vertical component of speed = 50 sin 23°
Rate of gaining potential energy = Gain in PE / time
Rate of gaining potential energy = mgh / t = mg(h/t)
(h/t) is the vertical speed at which the aeroplane climbs.
Rate of gaining potential energy = 1.2×105 × 9.81 × 50 sin23°
Rate of gaining potential energy = 2.3×107 J s-1
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