Question

The electric motor of a model train accelerates the train from rest to 0.640 m/s in 22.0 ms. The total mass of the train is 680 g. Find the average power delivered to the train during its acceleration

Answer 1

Answer:

F=19.79N

Explanation:

Force F=m×a, a being the acceleration.

a=dv/dt (speed variation/interval of time)

a=(0.640-0)/22ms=

0.640/0.022=640/22=29.1m/s²

Mass: m=680g=0.68kg

Force: F=0.68×29.1=19.79N

Answer 2

Answer:

The average power delivered to the train is approximately 6.33 watts

Explanation:

Recall that power is defined as the work done (W) per unit of time (t). That is: Power= W/t

in our case the work done is equal to the change in kinetic energy (no potential energy change involved) based on the Kinetic Energy - Work theorem.

The initial kinetic energy is 0 J (the train was at rest, so its velocity is zero), and the final kinetic energy of the system is:

$K_f=\frac{1}{2} \,m\,v_f^2\\K_f=\frac{1}{2} \,0.680\ kg\,0.64^2\,\frac{m^2}{s^2} \\K_f=0.139264\,J$

Then the power delivered in 22 milliseconds (which is the same as 0.022 sec) would be:

$P=\frac{0.139264}{0.022} \,watts\\P=6.33\,\,watts$