Question

The electric motor of a model train accelerates the train from rest to $\mathbf{0}\mathbf{.}\mathbf{620}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$ in $\mathbf{21}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{s}$. The total mass of the train is $\mathbf{875}\mathbf{}\mathit{g}$. (a) Find the minimum power delivered to the train by electrical transmission from the metal rails during the acceleration. (b) Why is it the minimum power?

Short answer

(a) The minimum power delivered to the train by electrical transmission from the metal rails during the acceleration is $P_{av}=8.01W$.

(b) Some of the energy transferred into the system of the train goes into internal energy in warmer track and moving parts and some leaves the system by sound.

Step-by-step solution

Concept and formulae

The work done is given by

$W=\Delta U=U_f-U_i$

The average power is given by

$P_{av}=\frac{W}{\Delta t}$

where

$W=$Work done

$\Delta U=$Change in potential energy

$\Delta t=$ Change in time in second.

Calculation for the minimum power

(a)

Byusingtheconceptfromstep(1),wehave

$$\begin{gathered} P_{av}=\frac{W}{\Delta t} \\ =\frac{k_f}{\Delta t} \\ =\frac{m{v}^2}{2\Delta t} \\ {P}_{av}=\frac{\left(0.875kg\right){\left(0.620m/s\right)}^2}{2\left(21\times {10}^{-3}s\right)} \\ =8.01W \end{gathered}$$

Reason for the minimum power

(b)

Some of the energy transferred into the system of the train goes into internal energy in warmer track and moving parts and some leaves the system by sound. To account for this as well as the stated increase in kinetic energy, energy must be transferred at a rate higher than $P_{av}=8.01W$.