A 10.0-kg box is pulled by a horizontal wire in a circle on a rough horizontal surface for which the coefficient of kinetic friction is 0.250. Calculate the work done by friction during one complete circular trip if the radius is (a) 2.00 m and (b) 4.00 m. (c) On the basis of the results you just obtained, would you say that friction is a conservative or nonconservative force? Explain.
Solution 27E Step 1: Given data: Mass of a box m = 10 kg Coefficient of kinetic friction between box and rough horizontal surface k = 0.250 Normal force on a box on a rough horizontal surface N = mg N = (10 kg)(9.8) N = 98 N (downward) Kinetic friction force k = k * N k = 0.250 *8 N k = 24.5 N Step 2: Normal reaction of the ground Fn= N = 98 N(upward) (a). istance traveled in one complete circular trip against friction is = 2r = 2(2.00 m) = 12.56 m. Work done by friction W = f 2r k* W = (24.5 N) (12.56 m) * W = 307.72 J
