A block of mass 10 kg is pushed from rest on a rough horizontal floor by a constant force of 100 N acting at an angle of 37^ downwards from the horizontal. If the block travels a distance of 8 m in 2 s, the coefficient of kinetic friction between the block and the floor is (Take g = 10 m/s ^2 , 37^ = 0.6 , 37^ = 0.8 ):

Step-by-step solution

From the second equation of motion, s = ut + 1 2 at^2 8 = 0 + 1 2 a(2)^2 a = 4 m/s ^2 Resolving the applied force into horizontal and vertical components: Horizontal component, F_x = 100 37^ = 100 0.8 = 80 N Vertical component (downwards), F_y = 100 37^ = 100 0.6 = 60 N Applying Newton's second law in the vertical direction to find the normal reaction N : N = mg + F_y = 10 10 + 60 = 160 N Applying Newton's second law in the horizontal direction: F_x - _k N = ma 80 - _k(160) = 10 4 160 _k = 40 _k = 0.25 Answer: 0.25