Question

- Find the power of an engine which can draw a train of 400 metric ton up the inclined plane of 1 in 98 at the rate of \[10\,m{s^{ - 1}}\]. The resistance due to friction acting on the train is 10 N per ton.
A. 350 kW
B. 400 kW
C. 440 kW
D. 500 kW

Answer 1

Hint: Draw the free body diagram of the forces acting on the train. Express the total force opposing the motion of the train. Use the relation between power, force and velocity to determine the power of the engine.

Formula used:
Power, \[P = Fv\]
Here, F is the force and v is the velocity

Complete step by step solution:
We have given the mass of the train \[m = 400 \times {10^3}\,kg\] and velocity \[v = 10\,m{s^{ - 1}}\]. The plane is inclined such that \[\sin \theta = \dfrac{1}{{98}}\], where, \[\theta \] is an angle of inclination.
Also, we have given the frictional force acting on the train per ton is, \[{f_k} = 10\,N/{\text{ton}} \times 400\,{\text{ton}} = 4000\,N\] .
We have to draw the free body diagram of the forces acting on the train as follows,
In the above figure, \[mg\sin \theta \] is the horizontal component of weight of the train and \[{f_k}\] is the frictional force.
Therefore, from the above figure, we see, the net force in the opposite direction of the train is,
\[F = mg\sin \theta + {f_k}\]
The power required to draw the train should be,
\[P = Fv\]
$\implies$ \[P = \left( {mg\sin \theta + {f_k}} \right)v\]
Substituting\[m = 400 \times {10^3}\,kg\], \[{f_k} = 4000\,N\], \[\sin \theta = \dfrac{1}{{98}}\] and \[g = 9.8\,m/{s^2}\] in the above equation.
\[P = \left( {\left( {400 \times {{10}^3}} \right)\left( {9.8} \right)\left( {\dfrac{1}{{98}}} \right) + 4000} \right)\left( {10} \right)\]
\[ \Rightarrow P = \left( {\left( {400 \times {{10}^3}} \right)\left( {9.8} \right)\left( {\dfrac{1}{{98}}} \right) + 4000} \right)\left( {10} \right)\]
\[ \Rightarrow P = \left( {44000} \right)\left( {10} \right)\]
\[ \Rightarrow P = 440\,{\text{kW}}\]
Therefore, the power of the engine is 440 kW.

So, the correct answer is “Option C”.

Note:
The direction of the kinetic friction \[{f_k}\] is always opposite to the direction of the motion of the body. The relation \[P = Fv\] is acquired from the relation, \[P = \dfrac{W}{t}\], by substituting \[Fd\] for work done W. Therefore, if you don’t remember \[P = Fv\], you can also \[P = \dfrac{W}{t}\] and then simplify it.