A block slides down an inclined plane that makes an angle `theta` with the horizontal. If the coefficient of kinetic friction is `mu`, find the acceleration of the block.

A block rests on an inclined plane that makes an angle  with the horizontal, if the coefficient of sliding friction is 0.50 and that of static friction is 0.75, the time required to slide the block 4 m along the inclined plane is -

A block of mass m slides down an inclined plane which makes an angle theta with the horizontal. The coefficient of friction between the block and the plane is mu . The force exerted by the block on the plane is

A block of 4 kg slides down an inclined plane which makes an angle theta = 60^(@) with the horizontal. The co-effcient of friction between the block and the plane is mu = 3//4 . The force exerted by the block on the plane is (g = 10 m//s^(2))

A small block starts sliding down an inclined plane subtending an angle theta with the horizontal . The coefficient of friction between the block and plane depends on the distance covered from rest along the plane as mu = mu_(0)x where mu_(0) is a constant . Find (a) the distance covered by the block down the plane till it stops sliding and (b) its maximum velocity during this journey .

A block is moving on an inclined plane making an angle 45^@ with the horizontal and the coefficient of friction is mu . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N=10mu , then N is