Problem 3: In the figure below a spring with k- 170 N/m is at the top of a frict

Question

Problem 3: In the figure below a spring with k- 170 N/m is at the top of a frictionless incline of angle 370, The lower end of the incline is distance D = 1 m from the end of the spring, which is at its relaxed length. A 2 kg canister is pushed against the spring until the spring is compressed 0.2 m, and then released from rest. a) What is the speed of the canister at the instant the spring returns to its relaxed length? b) What is the speed of the canister when it reaches the lower end of the incline?

Explanation / Answer

a) Here the restoring Force of the spring = m*g*sin(theta) = 2*9.8*sin(37) = 11.8 N

spring constant is k = F/x = 11.8/0.2 = 59 N/m

Using law of conservation of energy

0.5*k*x^2 = 0.5*m*v^2

0.5*59*0.2^2 = 0.5*2*v^2

v =1.08 m/s

b) 0.5*m*v^2 + (m*g*D*sin(37)) = 0.5*m*Vf^2)

(0.5*2*1.08^2)+(2*9.8*1*sin(37)) = (0.5*2*Vf^2)

Vf = 3.6 m/s

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