Question

A body is moving with a velocity $1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$. A force $F$ is needed to stop it within a distance $x$. If the speed of the body is $3\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$, the force needed to stop is within the same distance ($x$) will be:
A. $9F$
B. $6F$
C. $3F$
D. $1.5F$

Answer 1

Use the formula for the work done and kinetic energy of an object. According to the law of conservation of energy, the work done by the force to stop the body within a distance is equal to the kinetic energy of that object. Rewrite this relation for law of conservation of energy for the first and second case and solve it for the force required.

Formulae used:
The work done $W$ is given by
$W = Fd$ …… (1)
Here, $F$ is the force acting on the object and $d$ is the displacement of the object.
The kinetic energy $K$ of an object is given by
$K = \dfrac{1}{2}m{v^2}$ …… (2)
Here, $m$ is the mass of the object and $v$ is the velocity of the object.

Complete step by step answer:
We have given that the initially the body is moving with a velocity $1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ and $F$ is the force required to stop the body within a distance $x$.
${v_1} = 1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
We have asked to determine the force required to stop the body moving with velocity $3\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ within the same distance $x$.
${v_2} = 3\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
Let $m$ be the mass of the body.According to the law of conservation of energy, the work done ${W_1}$ to stop the body within the distance $x$ must be equal to kinetic energy ${K_1}$ of the moving body.
${W_1} = {K_1}$
$\Rightarrow Fx = \dfrac{1}{2}mv_1^2$ …… (3)

Substitute $1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ for ${v_1}$ in the above equation.
$\Rightarrow Fx = \dfrac{1}{2}m{\left( {1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)^2}$
$\Rightarrow Fx = \dfrac{1}{2}m$
Rewrite the equation (3) for her second case.
$\Rightarrow F'x = \dfrac{1}{2}mv_2^2$
Here, $F'$ is the force required to stop the body moving with velocity $3\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$.
Substitute $3\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ for ${v_2}$ in the above equation.
$\Rightarrow F'x = \dfrac{1}{2}m{\left( {3\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)^2}$
$\Rightarrow F'x = \dfrac{9}{2}m$
Substitute $Fx$ for $\dfrac{1}{2}m$ in the above equation.
$\Rightarrow F'x = 9Fx$
$\therefore F' = 9F$
Therefore, the required force is $9F$.

Hence, the correct option is A.

Note: One can also solve the same question by another method. One can use the formula for force in terms of acceleration and kinematic equation for the final velocity of an object in terms of displacement of the object. Rewrite these equations for the first and second case and solve them for the required force. The only thing to keep in mind is that the acceleration of the body in the kinematic equation should be used with negative sign as the body retards and stops.