The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21.0 ms. The total mass of the train is 875 g.

Question

(a) Find the minimum power delivered to the train by electrical transmission from the metal rails during the acceleration. (b) Why is it the minimum power?

Answer 1

Let

v = 0.620 m/s be the train's final velocity,

t = 21.0 ms = 0.021 s be the time it takes to accelerate it,

m = 875 g = 0.875 kg be the train's mass.

After the train accelerates, its kinetic energy is

E = mv²/2

The motor delivered this energy in time t.

That gives average power of

P = E/t = mv²/2t

Substituting actual numbers

P = 0.875×0.62²/(2×0.021) = 8W

This is the minimum average power in the sence that actual average power must be greater,

because it must also overcome friction, air resistance, etc.

This is not the minimum instantaneous power; at some time the instantaneous power could drop to 0,

as long as it is made up later.