Question

When a car of mass $ 1200kg $ is moving with the velocity of $ 15m{s^{ - 1}} $ on a rough horizontal its engine is switched off. How far the car travel before it comes to rest if the coefficient of kinetic friction between the road and tyre of the car is 0.5? ( $ g = 10{m \mathord{\left/
 {\vphantom {m {{s^2}}}} \right.} {{s^2}}} $ )
(A) 21.6m
(B) 25m
(C) 23.5m
(D) 22.5m

Answer 1

Hint : The force of friction always opposes the motion and wants to put the object at rest. When the engine is switched off, the only force acting on the object is the force of friction.

Formula used: The formulae used in the solution are given here.
 $ Friction = {\mu _k}N $ where $ {\mu _k} $ is the coefficient of kinetic friction between the road and tyre of the car and $ N $ is the normal force.
 $ {v^2} = {u^2} + 2aS $ where $ v $ is the final velocity, $ u $ is the initial velocity, $ a $ is the acceleration and $ S $ is the distance travelled.
Stopping distance $ S $ is given by, $ S = \dfrac{{{u^{^2}}}}{{2{\mu _k}g}} $ where $ u $ is the initial velocity, $ {\mu _k} $ is the coefficient of kinetic friction and $ g $ is the acceleration due to gravity.

Complete step by step answer:
It is given that the coefficient of kinetic friction between the road and tyre of the car is 0.5.
The maximum amount of friction force that a surface can apply upon an object can be easily calculated with the use of the given formula:
 $ Friction = {\mu _k}N $ where $ {\mu _k} $ is the coefficient of kinetic friction between the road and tyre of the car and $ N $ is the normal force.
We know that the normal force acting on a body is given by $ N = mg $ where $ m $ is the mass of the body and $ g $ is the acceleration due to gravity.
Given that, the mass of the car is $ 1200kg $ and $ g = 10{m \mathord{\left/
 {\vphantom {m {{s^2}}}} \right.} {{s^2}}} $ .
Thus, $ N = 1200 \times 10 = 12000kgm{s^{ - 2}} $ .
Since, the coefficient of kinetic friction between the road and tyre of the car, $ {\mu _k} = 0.5 $ and $ N = 12000 $ ,
 $ \therefore Friction = {\mu _k}N = 0.5 \times 12000 = 6000N $
When the engine is switched off, the only force acting on the object is the force of friction.
We know that, $ F = ma $ where $ m $ is the mass and $ a $ is the acceleration. Assigning the values of force and mass,
 $ \therefore a = \dfrac{F}{m} = \dfrac{{6000}}{{1200}} = 5m{s^{ - 2}} $
From the laws of motion, we know, $ {v^2} = {u^2} + 2aS $ where $ v $ is the final velocity, $ u $ is the initial velocity, $ a $ is the acceleration and $ S $ is the distance travelled.
Given that, the final velocity is zero and the initial velocity is $ 15m{s^{ - 1}} $ .The acceleration is $ 5m{s^{ - 2}} $ .
Substituting the values,
 $ 0 = {15^2} - 2 \cdot 5g \cdot S $ , acceleration is negative, since the vehicle undergoes retardation.
 $ \Rightarrow 225 = 100S $
So, the distance travelled by the car before it comes to rest is $ S = 22.5m $ .
The correct answer is Option D.

Note:
Alternatively, stopping distance $ S $ is given by,
 $ S = \dfrac{{{u^{^2}}}}{{2{\mu _k}g}} $
Substituting the values,
 $ S = \dfrac{{{{15}^{^2}}}}{{2 \cdot 0.5 \cdot 10}} $
 $ \Rightarrow S = 22.5m $
So, the distance travelled by the car before it comes to rest is $ S = 22.5m $ .
The correct answer is Option D.