In the figure, a spring with spring constant k = 150 N/m is at the top of a fric

Question

In the figure, a spring with spring constant k = 150 N/m is at the top of a frictionless incline of angle = 38°. The lower end of the incline is distance D = 1.1 m from the end of the spring, which is at its relaxed length. A 1.8 kg canister is pushed against the spring until the spring is compressed 0.16 m and released from rest. (a) 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)? (b) What is the speed of the canister when it reaches the lower end of the incline?

Explanation / Answer

(a) energy stored in the spring = ½ kx 2 = ½ X 150 X 0.16^2 = 1.92 J

Let the potential energy of the canister–earth system be zero at the initial position of the canister from where it is released.

Therefore, potential energy of the canister when it loses contact with the spring = mgh

= -1.8 X 10 X (-0.16sin38) = -1.77 J

Applying conservation of energy,

K i + U i = K f + U f

0 + ½ kx 2 = ½ mv 2 + mgh

0 + 1.92 = 0.5*1.8*v 2 – 1.77

or v = 2.02 m/s

(b)Again, applying conservation of energy,

0 + ½ kx 2 = ½ mv 2 + mgh

Now, h = -1.26*sin38o = -0.78 m

Therefore, 1.92 = 0.5*1.8*v 2 – 1.8*10*0.78

or v = 4.21 m/s

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