At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large compressed spring. The spring, with force constant k = 40.0 N/cm and negligible mass, rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 70.0 kg are pushed against the other end, compressing the spring 0.375 m. The sled is then released with zero initial velocity. What is the sled’s speed when the spring (a) returns to its uncompressed length and (b) is still compressed 0.200 m?

Solution 43E Step 1: Every ideal spring obeys Hooke’s law, which is written below. F = kx---------------(1) The minus sign indicates that, the force is restoring force and acts in the opposite direction to the displacement. In the problem, the spring got pressed by 0.375 m. The spring constant k k is 40 N/cm= 4000 N/m. So, by equation (1) the force generated is, F = 4000 × 0.375 = 1500 N . Step 2: The total mass on the spring is 70 kg. The work done when the spring reached to it’s undisplaced position after release is, W = F.d = 1500 × 0.375 = 562.5 jules. Work energy theorem tells that, W = K.E final K.E initial 2 562.5 = 1/2 mv 0 2 562.5 = 1/2 × 70 × v 2 v = (562.5 × 2) / 70 = 16.07 v = 16.07 = 4 m/s (approximately)