Question

The electric motor of a model train accelerates the train from rest to 0.540 m/s in 27.0 ms. The total mass of the train is 610 g. Find the average power delivered to the train during its acceleration.

Answer 1

Answer:

3.29 Watts

Explanation:

Given that

Initial velocity, u = 0 m/s

Final velocity, v = 0.54 m/s

Change in time, Δt = 27 ms

Mass of the train, m = 610 g = 0.61 kg

Power of any object is given as

P = Work done / time

P = W / t

Work done in this question, W is

W = 1/2mv²

W = 1/2 * 0.61 * 0.54²

W = 305 * 0.2916

W = 0.0889 kgm²/s²

Plugging this value into the equation we'd stated earlier, we have

P = W / t

P = 0.0889 / 27*10^-3

P = 3.293 Watts

Therefore, the average power delivered to the train during its acceleration is 3.29 Watts

Answer 2

Answer:

The value  is  $P =3.294 \ W$

Explanation:

From the question we are that

     The velocity  $v = 0.540 \ m/s$

      The  time taken is  $t = 27.0 ms = 27.0 *10^{-3} \ s$

     The  total mass of the train is  $m = 610 \ g = 0.610 \ kg$

Generally the average power delivered is mathematically represented as

     $P =\frac{KE }{t}$    

     $P =\frac{ \frac{1}{2} * m * v^2 }{t}$    

=>  $P =\frac{ \frac{1}{2} * 0.610 * 0.540 ^2 }{ 27.0 *10^{-3}}$    

=>  $P =3.294 \ W$