Question

A force applied by the engine of a train of mass $2.05 \times {10^6}{\text{kg}}$ changes its velocity from $5{\text{m}}{{\text{s}}^{ - 1}}$ to $25{\text{m}}{{\text{s}}^{ - 1}}$ in 5 minutes. Find the power of the engine.
A) 1.025 MW
B) 2.05 MW
C) 5 MW
D) 6 MW

Answer 1

Power is defined as the rate of work done, ie., it refers to the work done in a given amount of time. According to the work-energy theorem, the change in kinetic energy of the train will contribute to the work done by its engine.

Formula Used:

1. The change in kinetic energy of an object is given by, $\Delta K = \dfrac{1}{2}m\left( {v_f^2 - v_i^2} \right)$ where $m$ is the mass of the object, ${v_f}$ is the final velocity of the object and ${v_i}$ is the initial velocity of the object.
2. Power is given by, $P = \dfrac{W}{t}$ where $W$ is the work done in $t$ seconds.

Complete step by step answer:
Step 1: List the parameters known from the question.
The mass of the train is $m = 2.05 \times {10^6}{\text{kg}}$ .
When a force is applied by the engine, the velocity of the train changes. The initial velocity of the train is ${v_i} = 5{\text{m}}{{\text{s}}^{ - 1}}$ and its final velocity is ${v_f} = 25{\text{m}}{{\text{s}}^{ - 1}}$ .
Step 2: Find the change in the kinetic energy of the train.
The change in kinetic energy of the train is given by, $\Delta K = \dfrac{1}{2}m\left( {v_f^2 - v_i^2} \right)$ -------- (1)
where $m$ is the mass of the train, ${v_f}$ is its final velocity and ${v_i}$ is its initial velocity.
Substituting the values for $m = 2.05 \times {10^6}{\text{kg}}$ , ${v_i} = 5{\text{m}}{{\text{s}}^{ - 1}}$ and ${v_f} = 25{\text{m}}{{\text{s}}^{ - 1}}$ in equation (1) we get, $\Delta K = \dfrac{1}{2} \times 2.05 \times {10^6} \times \left( {{{25}^2} - {5^2}} \right)$
Simplifying the right-hand side of the above equation we get the change in kinetic energy of the train as $\Delta K = 300 \times 2.05 \times {10^6}{\text{J}}$
According to the work-energy theorem, the change in the kinetic energy of the train is equal to the work done by the force acting on it. So, the work done will be $W = \Delta K = 300 \times 2.05 \times {10^6}{\text{J}}$
Step 3: Find the power of the engine.
The power of the engine is given by, $P = \dfrac{W}{t}$ ---------- (2) where $W$ is the work done in $t$ seconds.
Here the given amount of time to do the work is 5 minutes i.e., $t = 300{\text{sec}}$
Substituting the values for $W = 300 \times 2.05 \times {10^6}{\text{J}}$ and $t = 300{\text{sec}}$ in the equation (2) we get, $P = \dfrac{{300 \times 2.05 \times {{10}^6}}}{{300}} = 2.05 \times {10^6}{\text{W}} = 2.05{\text{MW}}$
Thus the power of the engine is $P = 2.05{\text{MW}}$ .
Hence the correct option is B.

Note: Alternate method
The work done by the force is given by, $W = Fs$ -------- (a) where $F$ is the force applied by the engine and $s$ is the resultant displacement.
Power of the engine is given by, $P = \dfrac{W}{t}$ where $W$ is the work done in $t$ seconds.
Using equation (a), we can express the power of the engine as $P = \dfrac{{Fs}}{t}$ ------ (b)
The force applied to the train will be $F = ma$ ------- (c) where $m$ is the mass of the train and $a$ is its acceleration.
Acceleration of the train is obtained using the equation of motion given by,${v_f} = {v_i} + at$ or, $a = \dfrac{{{v_f} - {v_i}}}{t}$
Substituting for final velocity ${v_f} = 25{\text{m}}{{\text{s}}^{ - 1}}$ , initial velocity ${v_i} = 5{\text{m}}{{\text{s}}^{ - 1}}$ and time $t = 300{\text{s}}$ in the above expression we get, $a = \dfrac{{25 - 5}}{{300}} = \dfrac{{20}}{{300}}{\text{m}}{{\text{s}}^{ - 2}}$
Now, substituting for $a = \dfrac{{20}}{{300}}{\text{m}}{{\text{s}}^{ - 2}}$ and $m = 2.05 \times {10^6}{\text{kg}}$ in (c) we get the applied force as $F = \dfrac{{2.05 \times {{10}^6} \times 20}}{{300}} = \dfrac{{41 \times {{10}^4}}}{3}{\text{N}}$
Displacement of the train can also be obtained using the equation of motion given by, $2as = v_f^2 - v_i^2$ or, $s = \dfrac{{v_f^2 - v_i^2}}{{2a}}$
Substituting values for $a = \dfrac{{20}}{{300}}{\text{m}}{{\text{s}}^{ - 2}}$ , ${v_f} = 25{\text{m}}{{\text{s}}^{ - 1}}$ and ${v_i} = 5{\text{m}}{{\text{s}}^{ - 1}}$ in the above expression and solving for $s$ we get the displacement as $s = 4500{\text{m}}$
Substituting the values for $F = \dfrac{{41 \times {{10}^4}}}{3}{\text{N}}$ , $s = 4500{\text{m}}$ and $t = 300{\text{s}}$ in equation (b) we get, $P = \dfrac{{41 \times {{10}^4} \times 4500}}{{3 \times 300}} = 2.05 \times {10^6}{\text{W}} = 2.05{\text{MW}}$
Thus the power of the engine will be $P = 2.05{\text{MW}}$ .