Marble block of mass 2kg lying on ice when given velocity 6m/s is stopped by friction in 10 sec then coefficient of friction is
Answers
Answered by
83
Hi.
Mass of the marble block, m=2kg. Initial velocity,uo=6m/s. time for velocity to become zero,t=10s
The frictional force=uN=umg. Here u is coefficient of friction. N is normal force of reaction. N=mg.
The retardation( a ) produced by this force=umg/m=ug……………(1). We can find retardation from the equation of motion:
0=uo-at or a=uo/t=6/10=0.6m/s^2…………………………..(2)
From (1) and (2),
ug=0.6. Then u=0.6/10=0.06 . This is coefficient of friction...
Hope this helps u!!
Mass of the marble block, m=2kg. Initial velocity,uo=6m/s. time for velocity to become zero,t=10s
The frictional force=uN=umg. Here u is coefficient of friction. N is normal force of reaction. N=mg.
The retardation( a ) produced by this force=umg/m=ug……………(1). We can find retardation from the equation of motion:
0=uo-at or a=uo/t=6/10=0.6m/s^2…………………………..(2)
From (1) and (2),
ug=0.6. Then u=0.6/10=0.06 . This is coefficient of friction...
Hope this helps u!!
Answered by
9
Mass of the marble block, m=2kg. Initial velocity,uo=6m/s. time for velocity to become zero,t=10s
The frictional force=uN=umg. Here u is coefficient of friction. N is normal force of reaction. N=mg.
The retardation( a ) produced by this force=umg/m=ug……………(1). We can find retardation from the equation of motion:
0=uo-at or a=uo/t=6/10=0.6m/s^2…………………………..(2)
From (1) and (2),
ug=0.6. Then u=0.6/10=0.06 .
Similar questions