Answer to Question #269991 in Classical Mechanics for Ghadeer

Question

In the figure below, a spring with k = 170 N/m is at the top of a frictionless incline of angle θ = 370. 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 0.200 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?

Answer 1

a)

$K_1+U_1=K_2+U_2,$

$mg(D+x)\sinθ+ \frac 12 kx ^2= \frac 12 mv_2^2+mgD\sinθ,$

$v _2= \sqrt{2gx\sinθ+\frac{kx ^ 2}m}=2.4~\frac ms.$

b)

$\\ K_1+U_1=K_3+U_3,$

$mg(D+x)\sinθ+ \frac 12 kx ^2= \frac 12 mv_3^2,$

$v _3= \sqrt{2g(D+x)\sinθ+\frac{kx ^ 2}m}=4.2~\frac ms.$