(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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(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 Laws of Motion 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
$\frac{v_0}{\mu}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Laws of Motion

Subtopic

Kinetic Friction and Motion on Rough Horizontal Surface

$A$ block of mass $5 \,kg$ moving on a rough surface with a velocity of $4 \,ms^{-1}$ is stopped by friction in $2 \,s$. The coefficient of friction between the contact surfaces is (Acceleration due to gravity $g = 10 \,ms^{-2}$)

EasyAP EAMCET 2023

View Solution

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

Medium

View Solution

The coefficient of friction between the block and the surface is $0.03$. Find the acceleration of the system in $m/s^{2}$. $(g = 10\,m/s^{2})$

Medium

View Solution

$A$ block of mass $2 \,kg$ is being pushed against a wall by a force $F=90 \,N$ as shown in the figure. If the coefficient of friction is $0.25$, then the magnitude of acceleration of the block is (Take $g=10 \,ms^{-2}$, $\sin 37^{\circ}=\frac{3}{5}$). (in $\,ms^{-2}$)

EasyAP EAMCET 2018

View Solution

$A$ block of mass $M$ is moving on a rough horizontal surface with momentum $P$. The coefficient of friction between the block and the surface is $\mu$. The distance covered by the block before it stops is $[g = \text{acceleration due to gravity}]$.

MediumMHT CET 2020

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ 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

$A$ block of mass $5 \,kg$ moving on a rough surface with a velocity of $4 \,ms^{-1}$ is stopped by friction in $2 \,s$. The coefficient of friction between the contact surfaces is (Acceleration due to gravity $g = 10 \,ms^{-2}$)

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

The coefficient of friction between the block and the surface is $0.03$. Find the acceleration of the system in $m/s^{2}$. $(g = 10\,m/s^{2})$

$A$ block of mass $2 \,kg$ is being pushed against a wall by a force $F=90 \,N$ as shown in the figure. If the coefficient of friction is $0.25$, then the magnitude of acceleration of the block is (Take $g=10 \,ms^{-2}$, $\sin 37^{\circ}=\frac{3}{5}$). (in $\,ms^{-2}$)

$A$ block of mass $M$ is moving on a rough horizontal surface with momentum $P$. The coefficient of friction between the block and the surface is $\mu$. The distance covered by the block before it stops is $[g = \text{acceleration due to gravity}]$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module