Question

A force applied by an engine on train of mass $2.05 \times 10^6 \text{ Kg}$ changes its velocity from $5 \text{ m/s}$ to $25 \text{ m/s}$ in $5 \text{ minutes}$. The power of the engine is

• A. $1.025 \text{ MW}$
• B. $2.05 \text{ MW}$
• C. $5 \text{ MW}$
• D. $6 \text{ MW}$

Hint:

Always convert non-SI units (like minutes) to SI units (seconds) before calculating power. $1 \text{ Watt} = 1 \text{ Joule/second}$.

Answer 1

The Correct Option is B

Step 1: Calculate Change in Kinetic Energy ($\Delta KE$): Given $m = 2.05 \times 10^6 \text{ kg}$, $u = 5 \text{ m/s}$, and $v = 25 \text{ m/s}$: $$\Delta KE = \frac{1}{2} m (v^2 - u^2)$$ $$\Delta KE = \frac{1}{2} (2.05 \times 10^6) (25^2 - 5^2)$$ $$\Delta KE = 1.025 \times 10^6 (625 - 25) = 1.025 \times 10^6 (600)$$ $$\Delta KE = 615 \times 10^6 \text{ J}\lt strong\gt Step 2: Convert Time to Seconds\lt /strong\gt t = 5 \text{ minutes} = 5 \times 60 = 300 \text{ s}\lt strong\gt Step 3: Calculate Power ($P$)\lt /strong\gt P = \frac{\Delta KE}{t} = \frac{615 \times 10^6}{300}$$ $$P = 2.05 \times 10^6 \text{ Watts} = 2.05 \text{ MW}$$