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 is _ _ _ _ N. (g=10 ~m / s ² ) .

Step-by-step solution

For motion without acceleration, net force = 0. Component of weight along incline: mg (30°) = 5 10 0.5 = 25 N (down). Normal force: N = mg (30°) = 50 3 2 = 25 3 N. Friction force opposing downward motion: f = N = 3 2 25 3 = 37.5 N (up). For equilibrium along the incline, applying force F down the incline: mg + F = f , so 25 + F = 37.5 , giving F = 12.5 N.