Home
Class 11
PHYSICS
A marble block of mass 2 kg lying on ice...

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

A

0.06

B

0.01

C

0.02

D

0.03

Text Solution

Verified by Experts

The correct Answer is:
A
v=ut+at
`:." "0=6+(-a)xx10`
`:." "10a=6" ":.a=0.6m//s^2`
a is the retardation.
But F`=muR=mumg` and `F=mxxa`
`:." "ma=mumg`
`:." "mu=a/g=6/(10)xx1/(10)=6/(100)`
`:." "mu=0.06`
Promotional Banner

Topper's Solved these Questions

  • FRICTIONAL IN SOLIDS AND LIQUIDS

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos

  • FORCE, WORK AND TORQUE

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos

  • MAGNETIC EFFECT OF ELECTRIC CURRENT

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos

Similar Questions

Explore conceptually related problems

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

A ball rolling on ice with a velocity of 4 .9 m//s stops after travelling 4 m . If g = 9.8 m//s^(2) what is the coefficient of friction ?

A ball rolling on ice with a velocity of 4 .9 m//s stops after travelling 4 m. If g = 9.8 m//s^(2) what is the coefficient of friction ?

A cricket ball is rolled on ice with a velocity of 5.6 m//s and comes to rest after travelling 8 m . Find the coefficient of friction Given g = 9.8 m//s^(2) .

If a block moving up an inclined plane at 30^(@) with a velocity of 5 m/s, stops after 0.5 s, then coefficient of friction will be nearly

If a block moving up an inclined plane at 30^(@) with a velocity of 5 m/s , stops after 0*5 s , then coefficient of friction will be nearly

A block 'A' of mass 1kg is kept on a block 'B' of mass 2kg .Block A is imparted a velocity of 16m/s at t=0.If the coefficient of friction between A and B is 0.2, find the time (in sec) after which momentum of A and B relative to ground will be equal.(Take g=10m/s^(2) ).