A small block is projected up a rough inclined plane with an initial velocity of 8 m s ⁻¹ . The angle of inclination of the plane is 37^ and the coefficient of kinetic friction between the block and the plane is 0.25 . The distance travelled by the block along the incline before coming to instantaneous rest is : [Take g = 10 m s ⁻² , 37^ = 3 5 and 37^ = 4 5 ]

Step-by-step solution

Let the mass of the block be m . The forces acting on the block along the incline are the component of gravity downwards and the kinetic friction downwards. The normal reaction is N = mg 37^ . The frictional force is f_k = N = mg 37^ . The net retarding force is F = mg 37^ + mg 37^ . The deceleration of the block is: a = g 37^ + g 37^ a = 10 3 5 + 0.25 10 4 5 a = 6 + 2 = 8 m s ⁻² Using the third equation of motion, v^2 = u^2 - 2as . Since the block comes to rest, v = 0 . 0 = (8)^2 - 2(8)s 16s = 64 s = 4 m . Answer: