A block of mass 11 kg is pulled along a rough horizontal floor at a constant velocity by a force applied at an angle of 37^ above the horizontal. If the work done by this pulling force over a displacement of 10 m is 400 J , the coefficient of kinetic friction between the block and the floor is: (Take g = 10 m/s ^2 , 37^ = 3 5 , 37^ = 4 5 )

Step-by-step solution

Let the pulling force be F . The work done by this force is given by: W = F d 400 = F 10 37^ 400 = 10 F 4 5 F = 50 N The normal force N on the block is reduced by the vertical component of the pulling force: N = mg - F 37^ N = 11 10 - 50 3 5 = 110 - 30 = 80 N Since the block moves at a constant velocity, the net horizontal force is zero. Thus, the horizontal component of the pulling force equals the kinetic friction: F 37^ = _k N 50 4 5 = _k 80 40 = 80 _k _k = 1 2 Answer: 1 2