Question

A force $F$ stops a body of mass $m$ moving with a velocity $u$ in a distance $s$. The force required to stop a body of double the mass moving with double the velocity in the same distance is:
(A) $4F$
(B) $5F$
(C) $2F$
(D) $8F$

Answer 1

Hint The force required to stop a body of double the mass moving with double the velocity in the same distance can be determined by equating the work done with the kinetic energy and it is equation as the first equation and in the second equation the mass is doubled, then the force is determined.

Useful formula:
The work done is given by,
$W = F.d$
Where, $W$ is the work done by the body, $F$ is the force of the body and $d$ is the distance of the body.
The kinetic energy is given by,
$KE = \dfrac{1}{2}m{v^2}$
Where, $KE$ is the kinetic energy of the body, $m$ is the mass of the body and $v$ is the velocity of the body.

Complete step by step answer
Given that,
The force of the body is given as, $F$,
The mass of the body is given as, $m$,
The velocity of the body is given as, $u$,
The distance of the body is given as, $s$.
By applying the law of the conversation of the energy
In first case,
$W = \Delta KE$
By using the formula of the work done and the kinetic energy in the above equation, then the above equation is written as,
$F.d = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{v^2}$
Then the above equation is written as,
$F.d = 0 - \dfrac{1}{2}m{v^2}$
By subtracting the terms in the above equation, then the above equation is written as,
$F.d = - \dfrac{1}{2}m{v^2}$
By rearranging the terms in the above equation, then the above equation is written as,
$2F.d = - m{v^2}\,................\left( 1 \right)$
In the second case,
$W = \Delta KE$
By using the formula of the work done and the kinetic energy in the above equation, then the above equation is written as,
$F.s = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{v^2}$
Then the above equation is written as,
$F.s = - \dfrac{1}{2}m{v^2}$
When the mass is doubled in the above equation, then
$F.s = - \dfrac{1}{2}\left( {2m} \right){v^2}$
By cancelling the same terms in the above equation, then
$F.s = - m{v^2}\,................\left( 2 \right)$
By substituting the equation (1) in the equation (2), then
$F.s = 2F.d$
Assume that $s = d$, then the above equation is written as,
$F = 2F$

Hence, the option (C) is the correct answer.

Note The work done is directly proportional to the force and the distance. As the force and the distance increases, then the work done also increases. As the force and the distance decreases, then the work done also decreases. The kinetic energy is directly proportional to the mass and the velocity.