A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 40.0o above the horizontal. The glider has mass 0.0900 kg. The spring has k = 640 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.80 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Solution 45E Step 1: Provided that, the glider and spring system makes an angle of 40° with the horizontal. The distance moved, S = 1.80 m According to the work-energy theorem, work done = Change in kinetic energy At maximum point, the KE will be completely PE. Step 2: According to work-energy theorem, kinetic energy+Potential energy = total energy ½ mv + ½ kx = Total energy 2 Therefore, at maximum point, ½ kx = total energy (v = 0 at maximum height) 2 ½ kx = mg (h sin40) Provided, k = 640 N/m ½ ×640 N/m × x = 0.0900 kg × 9.8 m/s × 1.8 m × 0.643 (where, sin 40 = 0.643) Therefore, we can write, 2 x = 1.02 J × 2 / k Therefore, x = 2.04 J / 640 N/m = 0.0032 m 2 Taking square root on both sides, x = 0.0564 m = 5.64 cm The spring was originally compressed to 5.64 cm. Step 4: b) At 0.8 m, the spring will not have any contact with the glider because, it will come back to its initial position at 5.64 cm. The potential energy at this point will be, PE = mg hsin 40 2 That is, PE = 0.0900 kg × 9.8 m/s × 0.8 m × 0.643 = 0.454 J Therefore, according to the work-energy principle, W = PE + KE Or, KE = W - PE W = mgh = 0.0900 kg × 9.8 m/s × 1.8 m × 0.643 = 1.02 J KE = 1.02 J - 0.454 J = 0.566 J At 0.8 m, the KE of the glider will be, 0.566 J