In a truck-loading station at a post office, a small 0.200-kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m (Fig. P7.57). The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest. (a) What is the coefficient of kinetic friction on the horizontal surface? (b) How much work is done on the package by friction as it slides down the circular arc from A to B?

Solution 65P a) When the object reached point B, which is at the horizontal surface, it’s velocity was 4.80 m/s. So, it’s kinetic energy at that point would be, K.E = 1/2 mv = 0.5 × 0.200 kg × 4.80 2 = 2.304 J . The potential energy at that point would be zero as the height is zero. So, the total mechanical energy would be 2.304 J. Then it slided for next 3 meters to point C, where it stopped. The force which stopped it was the frictional force. So, the work done by the frictional force is, W = F .displacement fric fric = F × 3 m × cos 0 = 3 × F . fric fric According...