A block of mass `4kg` is released from the top of a rough inclines plane of height `10 m`. If velocity of the block is` 10 m//s` at the bottom of the inclined then work done by the friction force on the block will be

A block is released from the top of the smooth inclined plane of height h . Find the speed of the block as it reaches the bottom of the plane.

A block of mass m is released from the top of a mooth inclined plane of height h its speed at the bottom of the plane is proportion to

A body of mass m is released from the top of a rough inclined plane as shown in figure. If the frictional force be F, then body will reach the bottom with a velocity

A block of mass m is released from the top of fixed inclined smooth plane. if theta is the angle of inclination then vertical accelertion of block is

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

A block of mass 10kg is placed on a rough surface inclined plane of inclination 37^@ with horizontal. Now the block is released . The work done by the gravitational force and friction force, when the block moves from A to B(sin37^@=0.6 , cos37^@=0.8)

A small blockofmass 'm' is released from rest at the top of a rough inclined plane as shown The coefficient of friction between the block and inclined plane is 'mu' Using work-energy theorem, find the speed of block as it passes the lowest point.

A block of mass l O kg is kept at the top of a rough inclined plane ofi_nclination 60° with the horizontal. The length of the inclined plane is 5 m and the coefficient of sliding friction between the block and the surface is 0.3. What is the work done on the block against the force of friction , when it reaches the bottom? (g =10m//s^2 )

A block of mass 2 kg is released from rest on a rough inclined ground as shown in figure. Find the work done on the block by. . (Take g 10 m//s^(2) ).