Q. A force applied by an engine of a train of mass $2.05 \times 10^{6}\ kg$ changes its velocity from $5\ m/s$ to $25\ m/s$ in 5 minutes. The power of the engine is
A) $1.025MW$
B) $2.05MW$
C) $5MW$
D) $6MW$
Explanation
(b) Power $= \frac{Work\ done}{time} = \frac{\frac{1}{2}\ m(v^{2} - u^{2})}{t}$
$P = \frac{1}{2} \times \frac{2.05 \times 10^{6} \times \left\lbrack (25)^{2} - (5^{2}) \right\rbrack}{5 \times 60}$
$P = 2.05 \times 10^{6}W = 2.05\ MW$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.