Question

A truck and a car moving with the same kinetic energy are brought to rest by the application of brakes which provides equal retarding forces. Which of them will come to rest in a shorter distance?
A. The truck
B. The car
C. Both will travel the same distance before coming to rest
D. Cannot be predicted

Answer 1

Hint: Here we can use the relation between force and change in kinetic energy.

Formula used:
${\text{K}}{\text{.E = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{mv}}_{}^{\text{2}}$
${\text{F = ma}}$
${\text{v}}_{}^{\text{2}}{\text{ - u}}_{}^{\text{2}}{\text{ = 2as}}$

Complete step by step answer:
Given that truck and car are moving with the same kinetic energy.
Using the formula for kinetic energy ${\text{K}}{\text{.E = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{mv}}_{}^{\text{2}}$, where m is the mass and v is the velocity of the given body.

Also they are brought to rest by the application of brakes which provides equal retarding forces.

Using the formula ${\text{F = ma}}$, where F is the force, m is the mass and a is the acceleration.

In this case the applied force is retarding force hence ${\text{F = - ma}}$
$\Rightarrow {\text{a = }}\dfrac{{{\text{ - F}}}}{{\text{m}}}$

Given that truck and car are brought to rest by the application of brakes which provides equal retarding forces.

Using the kinematical equation of motion ${\text{v}}_{}^{\text{2}}{\text{ - u}}_{}^{\text{2}}{\text{ = 2as}}$, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement of the body.

In this case truck and car are brought to rest by the application of brakes which provides equal retarding forces hence the final velocity, v=0
$\Rightarrow {\text{ - u}}_{}^{\text{2}}{\text{ = 2as}}$
Substituting the value of a, ${\text{a = }}\dfrac{{{\text{ - F}}}}{{\text{m}}}$ in this equation we get,
$\Rightarrow {\text{ - u = }}\dfrac{{{\text{ - 2Fs}}}}{{\text{m}}}$

$\Rightarrow {\text{u = }}\dfrac{{{\text{2Fs}}}}{{\text{m}}}$

$\Rightarrow {\text{s = }}\dfrac{{{\text{m}}u_{}^2}}{{{\text{2F}}}}$

$\Rightarrow {\text{s = }}\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{mu}}_{}^{\text{2}}} \right)\left( {\dfrac{{\text{1}}}{{\text{F}}}} \right)$

$\Rightarrow {\text{s = }}\left( {\dfrac{{{\text{K}}{\text{.E}}}}{{\text{F}}}} \right)$

The value of displacement or distance travelled, s is same for both truck and car as they are moving with the same kinetic energy.

Also they are brought to rest by the application of brakes which provides equal retarding forces.

Hence both the truck and the car will travel the same distance before coming to rest.

Therefore (C) both will travel the same distance before coming to rest is the required answer.

Note: Negative sign in the formula for force indicates that the applied force is retarding.
It can also be done using work energy theorem, as on both the objects work is done by the force applied by brakes.