Question

A train weighing $ {10^7}N $ is running on a level track with uniform speed of 36 km/h. The frictional force is 0.5kgf per quintal. What is the power of the engine?
(A) $ 0.5kW $
(B) 5 kW
(C) 50 kW
(D) 500 kW

Answer 1

Hint : One kilogram-force (kgf) is approximately equal to $ 9.8 $ Newton, and one quintal is about 980 N. Since the object is moving at a constant speed, the force of the engine is equal to the frictional force

Formula used: In this solution we will be using the following formula;
 $ \Rightarrow P = Fv $ where $ P $ is the power of a body, the $ F $ is the force exerted by the body on an object, and $ v $ is the velocity of the object.

Complete step by step answer
When a force is exerted by a body on another object and that object is moving with a particular velocity or speed, the force is said to have used a power proportional to that velocity.
The power is hence given by,
 $ \Rightarrow P = Fv $ where the $ F $ is the force exerted by the body on an object, and $ v $ is the velocity of the object.
In the question, we have that a train is moving at 36 km/h and a frictional force of $ 0.5 $ kgf per quintal.
Now, since the train is moving with a constant velocity, the force exerted on the train by the engine is the same as the frictional force against the train.
Hence $ F = 0.5kgf/{\text{quintal}} $ . This unit of force means that for every quintal of weight the frictional force is $ 0.5kgf $ which is $ \left( {{\text{0}}{\text{.5}} \times {\text{9}}{\text{.8}}} \right){\text{N}} $ .
Hence, frictional force is given by
 $ \Rightarrow F = \dfrac{{\left( {{\text{0}}{\text{.5}} \times {\text{9}}{\text{.8}}} \right)}}{{980}} \times {\text{1}}{{\text{0}}^7}{\text{ N}} $ (since $ {\text{1quintal = 980 N}} $ )
Hence, by computation
 $ \Rightarrow F = 5 \times {10^4}N $
Also, $ v = 36\dfrac{{km}}{h} \times 1000\dfrac{m}{{km}} \times \dfrac{1}{{3600}}\dfrac{h}{s} $
 $ \Rightarrow v = 10m/s $
Thus,
 $ \Rightarrow P = Fv = 5 \times {10^4} \times 10 = 500000{\text{W}} $
 $ \therefore P = 500kW $
Hence, the correct option is D.

Note
For clarity, the force by the engine is equal to the frictional force because according to newton’s second law of motion, when a body is moving at constant velocity (no acceleration) then the net force must be zero as in
 $ \Rightarrow {F_{NET}} = F - f_r = ma $ since $ a = 0 $ then
 $ \Rightarrow F = f_r $ where $ f_r $ is friction.