A force applied by the engine of a train of mass 2.05 × 10⁶ kg changes its velocity from 5ms^–1 to 25ms^–1 in 5 minutes. The power of the engine

Given,

mass of train m =$2.05\times {10}^{6}kg$

time t =5 minutes = 300s

v = 25m/s

u = 5m/s

We know that acceleration,

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{25-5}{300} \\ \Rightarrow \mathrm{a}=\frac{1}{15}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

By using the equation of motion,

$$\begin{gathered} {\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as} \\ \Rightarrow {25}^2-5^2=2\times \frac{1}{15}\times \mathrm{s} \\ \Rightarrow \mathrm{s}=4500\mathrm{m} \end{gathered}$$

Also power,

$\mathrm{P}=\frac{\mathrm{work} \mathrm{done}}{\mathrm{time}}=\frac{\mathrm{F}\times \mathrm{d}}{\mathrm{time}} (\because \mathrm{work} \mathrm{done}=\mathrm{F}\times \mathrm{d})$

$\mathrm{P}=\frac{\mathrm{m}\times \mathrm{a}\times \mathrm{s}}{\mathrm{t}}(\because \mathrm{force}=\mathrm{m}\times \mathrm{a})$

Substituting the values,

$\mathrm{P}=\frac{2.05\times {10}^6\times 1\times 4500}{15\times 300}=2.05\times {10}^6\mathrm{W}=2.05\mathrm{MW}$

Hence the correct answer is 2.05MW.