A block is projected upwards along a rough inclined plane with an initial speed of 10 m s ⁻¹ . The angle of inclination of the plane is 37^ . If the block travels a distance of 5 m up the incline before momentarily coming to rest, the coefficient of kinetic friction between the block and the incline is: (Take g = 10 m s ⁻² , 37^ = 3 5 , 37^ = 4 5 )

Step-by-step solution

Using the kinematic equation for uniform acceleration: v^2 = u^2 - 2as Given u = 10 m s ⁻¹ , v = 0 , and s = 5 m : 0 = (10)^2 - 2 a 5 10a = 100 a = 10 m s ⁻² For a block sliding up a rough inclined plane, the net deceleration is provided by the component of gravity and the kinetic friction acting downwards along the incline: m a = m g + _k m g a = g ( + _k ) Substituting the given values: 10 = 10 ( 3 5 + _k 4 5 ) 1 = 0.6 + 0.8 _k 0.8 _k = 0.4 _k = 0.4 0.8 = 0.5 Answer: 0.5