A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.50 m/s. The coefficient of kinetic friction between the box and the surface is 0.20. (a) What horizontal force must the worker apply to maintain the motion? (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?

Solution 27E Mass of the box is 11.2 kg. The weight or the normal reaction it will get is, Normal force = N = mg = 11.2 × 9.8 = 109.76 N The kinetic friction coefficient is,K = 0.20. The frictional force is, F = NK= 0.20 × 109.76 = 21.952 N . a)The horizontal force is 21.952 N. b)The box was left after some time. So the initial velocity was 3.50 m/s and the acceleration will be 1.96 m/s . The kinetic energy of the box when it was moving with the velocity of 3.50 m/s was, 2 2 K.E = 1/2 mv = 1/2 × 11.2 × 3.50 = 68.6 jules. The work done by the frictional force to stop the block is, W = force × distance = 21.952 × x But as we know that , the kinetic energy at the end would be zero, so, W = K.E = 68.6 jules 21.952 × x = 68.6 jules x = 68.6 / 21.952 = 3.125 meters. So, before stop, it would travel 3.125 meters of distance.