Q#523: [Work, Energy, Power] (Past Exam Paper – N95 / I / 6)
A space vehicle of mass m re-enters the Earth’s atmosphere at an angle θ to the horizontal. Because of air resistance, the vehicle travels at a constant speed v.
The heat-shield of the vehicle dissipates heat at a rate P, so that the mean temperature of the vehicle remains constant.
Taking g as the relevant value of the acceleration of free fall, which expression is equal to P?
A mgv
B mgv sin θ
C ½ mv2
D ½ mv2 sin2 θ
Solution 523:
Answer: B.
Since the vehicle is travelling at constant speed, the net acceleration (and thus net force) is zero. This means that the gravitational pull is balanced by the air resistance.
A constant speed also means that the kinetic energy of the vehicle stays constant throughout the travel. Heat is dissipated at a rate P. From the conservation of energy, this heat (energy) dissipation comes only (since KE is constant here) from the loss in gravitational potential energy of the vehicle as it approaches the Earth’s surface, losing its height.
The vehicle enters the Earth’s atmosphere at an angle θ to the horizontal with a speed v. The motion of the vehicle is of course towards the Earth’s surface (downwards at an angle θ to the horizontal), not upwards.
Horizontal component of speed = v cosθ
Vertical component of speed = v sinθ
As said before, the rate of heat (energy) dissipated, P is due to the rate of loss of G.P.E of the vehicle.
Rate of loss of G.P.E = Δ(mgh) / Δt = mg (Δh / Δt)
[since m and g are constant]
But, the rate of change of the vertical height of the vehicle, (Δh / Δt), is the vertical component of speed, v sinθ.
Rate of heat dissipated, P = Rate of loss of G.P.E = mgv sinθ
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