A body of 5 kg weight kept on a rough inclined plane of angle `30^(@)` starts sliding with a constant velocity. Then the coefficient of friction is (assume `g=10m//s^(2)`)

A block is projected upwards with a velocity of 5 m/s from the bottom of a rough inclined plane of inclination 30° with the horizontal. It stops after 0.5 s. What is the coefficient of friction ? (g =10m//s^2 )

A body slides down a rough inclined plane and then over a rough plane surface. It starts at A and stops at D. The coefficient of friction is :

A body weighing 20 kg just slides down a rough inclined plane that rises 5 m in every 13 m. What is the coefficient of friction?

A block of mass 1 kg slides down on a rough inclined plane of inclination 60^(@) starting from its top. If the coefficient of kinetic friction is 0.5 and length of the plane is 1 m , then work done against friction is (Take g=9.8m//s^(2) )

A body of mass 5xx10^(-3) kg is launched upon a rough inclined plane making an angle of 30^(@) with the horizontal. Obtain the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent.