Question

The electric motor of a model train accelerates the train from rest to 0.665 m/s in 20.5 ms. the total mass of the train is 875 g. (a) find the minimum power delivered to the train by electrical transmission from the metal rails during the acceleration.

Answer 1

The power is defined as the work done (W) by the transmission per unit of time (t) to accelerate the train:
$P= \frac{W}{t}$

To find the work done, we can use the work-energy theorem, which states that the work done is equal to the variation of kinetic energy of the train. Since the trains starts from rest, its initial kinetic energy is zero, so the work done is simply equal to the final kinetic energy:
$W= \frac{1}{2}mv_f^2$
where m is the mass of the train and $v_f$ is its final velocity.
The mass of the train is m=0.875 kg while its final velocity is $v_f=0.665 m/s$, the work done by the electrical transmission is
$W= \frac{1}{2}(0.875 kg)(0.665 m/s)^2=0.19 J$

And since this work is done in a time of t=20.5 ms=0.0205 s, the minimum power delivered is
$P= \frac{W}{t}= \frac{0.19 J}{0.0205 s}=9.27 W$