Question

The electric motor of a model train accelerates the train from rest to 0.685 m/s in 21.5 ms. The total mass of the train is 875 g. Find the minimum power delivered to the train by electrical transmission from the metal rails during the acceleration.

Answer 1

Answer:

P=9.58 W

Explanation:

According to Newton's second law, and assuming friction force as zero:

$F_m=m.a\\F_m=0.875kg*a$

The acceleration is given by:

$a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2$

So the force exerted  by the motor is:

$F_m=0.875kg*31.9m/s^2\\F_m=27.9N$

The work done by the motor is given by:

$W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m$

$W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J$

And finally, the power is given by:

$P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W$