A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m (Fig. P7.39). When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The force constant k is 100 N/m. What is the coefficient of kinetic friction µk between the block and the tabletop?

Solution 43P Step 1: Elongation of the spring, x = 0.20 m Force constant of the spring, k = 100 N/m 2 2 2 The potential energy of the spring, PE = ½ kx = ½ ×100 N/m × 0.20 m = 2 J The equation for frictional force, F = mg k Where, - cokficient of static friction m - Mass of the body g - Acceleration due to gravity The work done by the spring, W = FX Where,X - Distance of travel Mass, m = 0.5 kg Distance, X = 1 m Therefore, W = mg X = × 0.5 kg × 9.8 m/s × 1 m = 4.9 J 2 k k k