Question

A relaxed spring (k = 120 N/m) is set up on a ramp inclined at θ =28 degrees. A mass of    0.012 kg is placed on the spring, compressing it 0.025 m. It is then released.

a)   What is the velocity of the mass once it loses contact with the spring?

b)   How far up the ramp will the mass itself travel after it is released?

Answer 1

Given K = 120 N/m theta = 28 degree mass m = 0.012 kg compressing x = 0.025 m According to the conservation of energy P.E = K.E 1/2 kx^2 = 1/2 n theta^2 = > v^2 = kx^2/m = > v = squareroot kx^2/m = squareroot (120 N/m) (0.0025 m)^2/(0.012 kg) = > v = 2.5 m/s v^2 - u^2 = 2 as find v = 0 velocity initial velocity u = 2.5 m/s^2, a = -g sin theta = > (0)^2 - u^2 = 2(-g sin theta) delta x = > delta x = u^2/2g sin theta = (2.5 m/s)^2/2(9.8 m/s^2) sin (28 degree) = > delta x = 0.679 m