A force applied by an engine on a train of mass $2.05 \times 10^6 \mathrm{~kg}$ changes its velocity from $5…

A force applied by an engine on a train of mass $2.05 \times 10^6 \mathrm{~kg}$ changes its velocity from $5 \mathrm{~m} / \mathrm{s}$ to $25 \mathrm{~m} / \mathrm{s}$ in 5 minutes. The power of the engine is
  1. $1.025 \mathrm{MW}$
  2. $2.05 \mathrm{MW}$
  3. $5 \mathrm{MW}$
  4. $6 \mathrm{MW}$

Solution

$\begin{aligned} m=2.05 & \times 10^6 \mathrm{~kg}, v_1=5 \mathrm{~m} / \mathrm{s} \\ v_2 & =25 \mathrm{~m} / \mathrm{s}, t=5 \mathrm{~min} \\ & =5 \times 60=300 \mathrm{sec} \\ \text { Power } & =\frac{W}{t}=\frac{\text { change in KE }}{t} \\ & =\frac{1}{2} \frac{m\left(v_2^2-v_1^2\right)}{300} \\ & =\frac{1}{2} \times \frac{2.05 \times 10^6\left[(25)^2-(5)^2\right]}{300} \\ & =\frac{1}{2} \times \frac{2.05 \times 10^6 \times 600}{300} \\ & =2.05 \times 10^6 \mathrm{~W} \\ & =2.05 \mathrm{MW}\end{aligned}$

Asked in: AP EAMCET 2001

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