Question

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. 1.025 MW
• B. 2.05 MW
• C. 5MW
• D. 5 MW

Answer 1

Answer: (B)

2.05 MW

Answer 2

$\text{Power} = \frac{\text{Work done}}{\text{time}} = \frac{\text{Increase in kinetic energy}}{\text{time}} = \frac{\frac{1}{2}m(v_{2}^{2} - v_{1}^{2})}{t} = \frac{\frac{1}{2} \times 2.05 \times 10^{6} \times \lbrack 25^{2} - 5^{2}\rbrack}{5 \times 60} = 2.05 \times 10^{6}watt = 2.05MW$