A lift moving with a uniform velocity v in which a block of mass `m` is lying .The friction force offered by the when coefficient of friction is `mu = 0.5` will be

A lift is moving upwards with a uniform velocity v in which a block of mass m is lying: The frictional force offered by the block when the coefficient of friction is mu will be

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 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 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 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 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 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 -