Question

An engine can pull $4$ coaches at a maximum speed of $20 m/s$. Mass of the engine is twice the mass of every coach. Assuming resistive forces proportional to the weight, approximate maximum speeds of the engine when it pulls 12 and 6 coaches are
(A) $8.5 m/s$ and $15 m/s$ respectively
(B) $6.5 m/s$ and $8 m/s$ respectively
(c) $8.5 m/s$ and $13 m/s$ respectively
(d) $10.5 m/s$ and $15 m/s$ respectively

Answer 1

Hint: The power of the engine will remain constant, so first, we will obtain the equation of power in the first given condition when the maximum speed of the engine is given. After this, obtain the equation of power for the condition when the engine pulls 12 coaches and then obtain the equation of power for the condition when the engine pulls 6 coaches. Now use these three equations for the determination of required maximum speed.

Complete step by step answer:
It is given to us that the engine can pull 4 coaches at a maximum speed of 20 m/s., the mass of the engine is twice the mass of every coach. So, we will use this information for the determination of power.
Write the equation of power for the above condition.
$p = FV$ …… (1)
Here, $F$ is the force and $V$ is the speed.
We know that the resistive force is proportional to the weight, so the expression of force becomes
$
F = W\\
F = \left( {{M_E} + 4{M_C}} \right)g
$
Here, ${M_E}$ is the mass of the engine and ${M_C}$ is the mass of each coach, we know that the mass of the engine is twice the mass of each coach.
Therefore, we get
$
F = \left( {2m + 4m} \right)g\\
F = 6mg
$ …… (2)
From equation (1) and (2), we get
\[p = 6mgV\] …… (3)
Now write the equation of power for the condition when the engine pulls 12 coaches.
$p = F{v_1}$ …… (4)
Here, ${v_1}$ is the maximum speed of the engine when it pulls 12 coaches.
From the above condition, calculate the magnitude of the force, so
$
F = \left( {{M_E} + 12{M_c}} \right)\\
F = \left( {2m + 12m} \right)g\\
F = 14mg
$
We will put $F = 14mg$ in the equation (4), so we get
$P = 14mg{v_1}$ …… (5)
Now write the equation of power for the condition when the engine pulls 6 coaches.
$p = F{v_2}$ …… (6)
Here, ${v_2}$ is the maximum speed of the engine when it pulls 6 coaches.
From the above condition, calculate the magnitude of the force, so
$
F = \left( {{M_E} + 6{M_c}} \right)\\
F = \left( {2m + 6m} \right)g\\
F = 8mg
$
We will put $F = 8mg$ in the equation (6), so we get
$P = 8mg{v_2}$ …… (7)
We know that the power of the engines remains constant, so we will equate equation (3) and (5) for the calculation of maximum speed when the engine pulls 12 coaches.
Therefore, we get
$
6mgV = 14mg{v_1}\\
{v_1} = \dfrac{{6mgV}}{{14mg}}\\
{v_1} = 0.428V
$
Substitute $V = 20\;{\rm{m/s}}$ in the above equation, so
$
{v_1} = 0.428 \times 20\;{\rm{m/s}}\\
{{\rm{v}}_1} = 8.56\;{\rm{m/s}}
$
Similarly, we will equate equation (3) and (7) for the calculation of maximum speed when the engine pulls 6 coaches.
Therefore, we get
$
6mgV = 8mg{v_2}\\
{v_2} = \dfrac{{6mgV}}{{8mg}}\\
{v_2} = 0.75V
$
Substitute $V = 20\;{\rm{m/s}}$ in the above equation, so
$
{v_2} = 0.75 \times 20\;{\rm{m/s}}\\
{{\rm{v}}_2} = 15\;{\rm{m/s}}
$
Therefore, approximate maximum speeds of the engine are $8.5\;{\rm{m/s}}$ and \[15\;{\rm{m/s}}\] , and option (A) is correct.

Note: Here the unit of the speed given in the question and options are the same, so the conversion of the unit is not required. But remember that if the unit of the speed of the engine in question and given option are not same then convert the unit of given speed same as the unit of speed given in the option.