Question

he figure, a spring with spring constant k = 170 N/m s at the top of a frictionless incline of angle θ 36". The lower end of the incline is distance D = 1.0 m from the end of the spring, which is at its relax length. A 2.1 kg canister is pushed against the spring until the spring is compressed 0.24 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? (a) Number Units (b) Number Units

Answer 1

answer) part a) here we have initial kinetic energy=potential kinetic energy of the canister

so we have

1/2mv²=1/2kx²+mghsin$\Theta$

putting the values we have

2.1v²/2=0.5*170*0.24²+2.1*9.8*sin36*0.24

1.05v²=4.896+2.9032

v²=7.7992/1.05=7.4278

v=2.73 m/s

so the answer is 2.73m/s or 2.7 m/s.

b) applying conservation of energy we have

Emehcnical initial=Emehcnical final

1/2mvi²+mghsin$\Theta$=1/2mvf²+0

1/2*2.1*2.73²+2.1*9.8*sin36*1=1/2*2.1*vf²

vf²=19.8958/2.1*0.5

vf²=18.9484

vf=4.35m/s

so the answer is 4.35m/s or 4.4 m/s