A marble block of mass 2, kg lying on ice when given a velocity of 6 ,m/s is stopped by friction in

English

Gujarati

Hindi

4-2.Friction

medium

A marble block of mass $2\, kg$ lying on ice when given a velocity of $6 \,m/s$ is stopped by friction in $10\,s$. Then the coefficient of friction is

A

$0.01$

B

$0.02$

C

$0.03$

D

$0.06$

(AIEEE-2003)

Solution

(d) $v = u – at \Rightarrow u – \mu gt = 0$

$\therefore \mu = \frac{u}{{gt}} = \frac{6}{{10 \times 10}} = 0.06$

Standard 11

Physics

Share

Similar Questions

A body of mass $2$ kg is moving on the ground comes to rest after some time. The coefficient of kinetic friction between the body and the ground is $0.2$. The retardation in the body is …… $m/s^2$

easy

Which one of the following statements is correct

easy

A block of mass $5\, kg$ is on a rough horizontal surface and is at rest. Now a force of $24\, N $is imparted to it with negligible impulse. If the coefficient of kinetic friction is $0.4$ and $g = 9.8\,m/{s^2}$, then the acceleration of the block is …….. $m/s^2$

medium

A stone weighing $1$ kg and sliding on ice with a velocity of $2$ m/s is stopped by friction in $10$ sec. The force of friction (assuming it to be constant) will be ……… $N$

medium

A block of mass $M = 5\,kg$ is resting on a rough horizontal surface for which the coefficient of friction is $0.2$. When a force $F = 40\,\,N$ is applied, the acceleration of the block will be …….. $m/\sec^2$ $(g = 10\,\,m/{\sec^2})$

hard