(A) The retarding force $F$ acting on the car is due to friction,given by $F = \mu R = \mu mg$.According to Newton's second law,$F = ma$,so $ma = \mu

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(A) The retarding force $F$ acting on the car is due to friction,given by $F = \mu R = \mu mg$.According to Newton's second law,$F = ma$,so $ma = \mu mg$,which gives the retardation $a = \mu g$.Using the third equation of motion,$v^2 = u^2 - 2as$,where $v$ is the final velocity $(0)$,$u$ is the initial velocity $(v_0)$,and $s$ is the stopping distance.Substituting the values: $0^2 = v_0^2 - 2(\mu g)s$.Rearranging for $s$,we get $2\mu gs = v_0^2$.Therefore,the shortest stopping distance is $s = \frac{v_0^2}{2\mu g}$.

Class 11 Physics Newton's Laws of Motion and Friction Kinetic Friction and Motion on Rough Horizontal Surface

Medium

$A$ car is moving along a straight horizontal road with a speed $v_0$. If the coefficient of friction between the tyres and the road is $\mu$,the shortest distance in which the car can be stopped is

A
$\frac{v_0^2}{2\mu g}$
B
$\frac{v_0}{\mu g}$
C
$\left( \frac{v_0}{\mu g} \right)^2$
D
$\frac{v_0}{\mu}$

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Class 11 Questions

Class 11

Physics Questions

Chapter

Newton's Laws of Motion and Friction

Subtopic

Kinetic Friction and Motion on Rough Horizontal Surface

Explain kinetic friction. Write the laws of kinetic friction. Define the coefficient of kinetic friction.

Medium

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$A$ child weighing $25 \,kg$ slides down a rope hanging from a branch of a tall tree. If the force of friction acting against him is $200 \,N$,the acceleration of the child is ........... $m/s^2$ $\left(g=10 \,m/s^2\right)$.

Easy

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What is the maximum force $F$ (in $N$) that can be applied such that the block does not move? (Given $\mu = 1/(2\sqrt{3})$ and $m = \sqrt{3} \ kg$)

Difficult

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The power required for an engine to maintain a constant speed of $50 \,m \,s^{-1}$ for a train of mass $3 \times 10^6 \,kg$ on rough rails is (The coefficient of kinetic friction between the rails and wheels of the train is $0.05$ and acceleration due to gravity $= 10 \,m \,s^{-2}$).

MediumAP EAMCET 2025

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$A$ body is pulled along a rough horizontal surface with a velocity $6\,m/s$. If the body comes to rest after travelling $9\,m$,then the coefficient of sliding friction is- (Take $g = 10\,m/s^2$)

Difficult

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$A$ car is moving along a straight horizontal road with a speed $v_0$. If the coefficient of friction between the tyres and the road is $\mu$,the shortest distance in which the car can be stopped is

Similar Questions

Explain kinetic friction. Write the laws of kinetic friction. Define the coefficient of kinetic friction.

$A$ child weighing $25 \,kg$ slides down a rope hanging from a branch of a tall tree. If the force of friction acting against him is $200 \,N$,the acceleration of the child is ........... $m/s^2$ $\left(g=10 \,m/s^2\right)$.

What is the maximum force $F$ (in $N$) that can be applied such that the block does not move? (Given $\mu = 1/(2\sqrt{3})$ and $m = \sqrt{3} \ kg$)

The power required for an engine to maintain a constant speed of $50 \,m \,s^{-1}$ for a train of mass $3 \times 10^6 \,kg$ on rough rails is (The coefficient of kinetic friction between the rails and wheels of the train is $0.05$ and acceleration due to gravity $= 10 \,m \,s^{-2}$).

$A$ body is pulled along a rough horizontal surface with a velocity $6\,m/s$. If the body comes to rest after travelling $9\,m$,then the coefficient of sliding friction is- (Take $g = 10\,m/s^2$)

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