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 $35 \,m / s$ in $15$ minutes. The power of the engine is

• A. $1.36\, hp$
• B. $1.36\, MW$
• C. $1.36 \,W$
• D. None

Answer 1

Answer: (B)

$P=\frac{w}{t}=\frac{\Delta\, K}{t}=\frac{1}{2} \times \frac{m\left(v^{2}-u^{2}\right)}{t}$
$=\frac{2.05 \times 10^{6}\left((35)^{2}-(5)^{2}\right)}{2 \times 15 \times 60}$
$P=\frac{2.05 \times 10^{6}(1225-25)}{1800}$
$=1.36 \times 10^{6}=1.36 \,M$ Watt