Block M slides down on frictionless incline as shown. The minimum friction coefficient so that m does not slide with respect to M would be :

Block M slide donw on frictionless inclined plane as shown. Find frictional force acting on block m while it is at rest with respect to the block M. (m=2kg, M=5kg,g=10m//sec^(2))

Two blocks m_(1) and m_(2) are allowed to move without friction. Block m_(1) is on block m_(2) and m_(2) slides on smooth fixed incline as shown. The angle of inclination of inclined plane is theta The acceleration of m_(1) with respect to ground is:

Two blocks m_(1) and m_(2) are allowed to move without friction Block m_(1) is on block m_(2) and m_(2) slides on smooth fixed incline as shown The angle of inclination of inclined plane is theta . The acceleration of m_(2) with respect to ground is .

The time taken for an ice block to slide down on inclined surface of inclination 60^(@) is 1-2 times the time taken by the same ice block to slide down a frictionless inclined plane of the same inclination. Calculate the coefficient of friction between ice and the inclined plane ?

Two blocks of mass 2m and m slide down the slope AB having same friction coefficient Then:

A plank of mass 3m is placed on a rough inclined plane and a man of mass m, the coefficient of friction between the board and inclined plane is mu = 0.5 , the minimum acceleration of man so that plank does not slide is :

Two blocks m_(1) and m_(2) are allowed to move without friction Block m_(1) is on block m_(2) and m_(2) slides on smooth fixed incline as shown The angle of inclination of inclined plane is theta . Normal reaction on m_(1) is .

In the figure shown, the minimum force F to be applied perpendicular to the incline so that the block does not slide is