Question

In the figure below, a spring with k = 170 is at the top of a frictionless incline of angle theta = 37 degree. The lower end of the incline is distance D = 1.00 m from the end of the spring, which is at its relaxed length. A 2.00 kg canister is pushed against the spring until the spring is compressed 0.2 m and released from rest. What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? What is the speed of the canister when it reaches the lower end of the incline?

Answer 1

a) Applying consevation of energy

gain in kinetic energy = potential energy stored in spring + loss in gravitational potential energy

1/2mv² = 1/2kx² + mgsin37x

=> 1/2 * 2 * v² = 1/2 * 170 * 0.2² + 2 * 9.8 * sin37 * 0.2

=> v = 2.4 m/s             --------------------> speed of canister

b)   Here,    1/2mv² = mgsin37 * D + 1/2mv1²

=>    1/2 * 2 * v² =   2 * 9.8 * sin37 * 1 + 1/2 * 2 * 2.4²

=> v = 4.189 m/s          --------------------> speed of canister

Answer 2

a) Applying consevation of energy

gain in kinetic energy = potential energy stored in spring + loss in gravitational potential energy

1/2mv² = 1/2kx² + mgsin37x

=> 1/2 * 2 * v² = 1/2 * 170 * 0.2² + 2 * 9.8 * sin37 * 0.2

=> v = 2.4 m/s             --------------------> speed of canister

b)   Here,    1/2mv² = mgsin37 * D + 1/2mv1²

=>    1/2 * 2 * v² =   2 * 9.8 * sin37 * 1 + 1/2 * 2 * 2.4²

=> v = 4.189 m/s          --------------------> speed of canister