A spring with spring constant k=170 N/m is at the top of a 37 degree frictionles

Question

A spring with spring constant k=170 N/m is at the top of a 37 degree frictionless incline. The lower end of the incline is 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 0.200 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?

Explanation / Answer

total vertical height of canister when spring is compressed 0.2 m = (1 + 0.2) sin 37 = 0.722 m

vertical height gained due to compression of spring sin 37 = 0.12036 m

energy stored in spring = 0.5kx2 = 0.5*170*0.2^2 = 3.4 J

energy gained due to additional height gained by compression = mgh = 2 * 9.81 * 0.12036 = 2.36146 J

So velocity when spring reaches relaxed state, by conservation of energy

3.4 + 2.36146 = 0.5 * 2 * v^2

=> v = 2.4 m/s

So  the speed of the canister at the instant the spring returns to its relaxed length = 2.4 m/s

Velocity when low lower end is reached

3.4 + 2*9.81*0.722 = 0.5 * 2 * v^2

=> v = 4.191 m/s

So the speed of the canister when it reaches the lower end of the incline = 4.191 m/s

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