A train of mass 3.3 × 10^6 kg is moving at a constant speed up a slope inclined at an angle of 0.64° to the horizontal. The engine of the train is producing a useful output power of 14 MW.
Assume that there are no frictional forces opposing the motion of the train.
What is the speed of the train?
A 0.43 m s–1 B 4.2 m s–1 C 39 m s–1 D 380 m s–1.
Ans is C how?
ref: Q14 - 9702_s18_qp_11

C. 39 m/s

Explanation:

First we need to calculate the total force required to move the train along the inclined plane. So, it is clear that the work done will be equal to the component of the weight that is parallel to the inclined plane, because there is no frictional force present:

Force = F = mg Sin θ

where,

m = mass of train = 3.3 x 10⁶ kg

g = 9.8 m/s²

θ = Angle of Inclination = 0.64°

Therefore,

F = (3.3 x 10⁶ kg)(9.8 m/s²)Sin 0.64°

F = 3.612 x 10⁵ N

Now, the formula for power is:

P = FV

V = P/F

where,

V = Velocity of Train = ?

P = Power of Engine = 14 MW = 1.4 x 10⁷ W

Therefore,

V = 1.4 x 10⁷ W/3.612 x 10⁵ N

V = 38.75 m/s

which is approximately equal to:

C. 39 m/s

grav potential energy

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