Physics Help Question

Question

A locomotive with a (constant) power capability of 1.5 MW can accelerate a train from a speed of 14 m/s to 30m/s in 7.2 min. (a) Calculate the mass of the train. Find (b) the speed of the train and (c) the force accelerating the train at 5.6 min after it started accelerating from the speed of 14 m/s. (d) Find the distance moved by the train during the 7.2 min interval that it took to reach the speed of 30 m/s. Note: This Problem involves constant power, not constant force.

Answer 1

Given P = 1.5MW

u = 14m/s (initial velocity)

v = 30 m/s (final velocity)

t = 7.2min = 7.2 x 60= 432sec

Required to find; a) mass of train, m

According to Newton's first w=equation of motion, v = u + at

So, re-writing the eqaution with a as subject we have, a = v -u /t

a = (30 - 140m/s / 432s = 0.037m/s^2

We know further that, power is equal to force times velocity, that is, P = Fx v

So, Force = power/velocity, that is F = P/v

F = 1.5MW / 30 = 50,000N (50KN)

Now, from Newton's second law of motion, we know that force = mass x acceleration.

Therefore, F = ma; then rearranging this formula we can find mass of train, m.

m = F/a = 50000N / 0.037 = 1,351,351Kg

Force at 5.6min = mass of train x (u-0/ 5.6 x 60) = 1,351,351 x (14-0/336) = 56,306.3N

From Newton third equation of motion s = ut + 1/2at^2

So, since s is the distance moved, then s = 14(432) + 1/2(0.037)(432)^2 = 6048 +3452.5 = 9500.5m