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:
[tex]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[/tex]
Then the power delivered in 22 milliseconds (which is the same as 0.022 sec) would be:
[tex]P=\frac{0.139264}{0.022} \,watts\\P=6.33\,\,watts[/tex]
