Work Energy And Power Question 192
Question: A force applied by an engine of a train of mass $ 2.05\times 10^{6}kg $ changes its velocity from $ 5m/s $ to 25 $ m/s $ in $ 5 $ minutes. The power of the engine is [EAMCET 2001]
Options:
A) $ 1.025MW $
B) $ 2.05MW $
C) $ 5MW $
D) $ 6MW $
Show Answer
Answer:
Correct Answer: B
Solution:
Power $ =\frac{\text{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 [{{(25)}^{2}}-(5^{2})]}{5\times 60} $
$ P=2.05\times 10^{6}W=2.05MW $
