A block of mass 5 kg is moving on an inclined plane which makes an angle of 30° with the horizontal. Friction coefficient between the block and inclined plane surface is √3/2. The force to be applied on the block so that the block will move down without acceleration i $s ______$ N.
Options

Question

A block of mass 5 kg is moving on an inclined plane which makes an angle of 30° with the horizontal. Friction coefficient between the block and inclined plane surface is √3/2. The force to be applied on the block so that the block will move down without acceleration i $s ______$ N.

Options

G1 Physics · Accepted Answer

The block is to move downward with constant velocity. Therefore the net force along the incline must be zero. Downward component of weight: mg sin30° = 5×10×1/2 = 25 N. Since the block moves downward, kinetic friction acts upward. Normal reaction is N = mg cos30° = 5×10×√3/2. Therefore, friction = μN = (√3/2) (5×10×√3/2) = 75/2 = 37.5 N. Friction exceeds the downward component of weight by 37.5-25 = 12.5 N. Hence an additional force of magnitude 12.5 N must be applied downward along the incline…

A block of mass 5 kg is moving on an inclined plane which makes an angle of 30° with the horizontal. Friction coefficient between the block and inclined plane surface is √3/2. The force to be applied on the block so that the block will move down without acceleration i s ______ N.