mass on vertical spring

Question

A spring with spring constant k = 35 N/m and unstretched length of L₀ = 2.4 is attached to the ceiling. A block of mass m = 2 kg is hung gently on the end of the spring.

1) How far does the spring stretch?

2) Now the block is pulled down until the total amount the spring is stretched is twice the amount found in the above question. The block is then pushed upward with an initial speed v_i = 3.8 m/sec. What is the maximum speed of the block?

Answer 1

1. We assume that the spring is ideal, i.e. it obeys Hooke's law (F = -kx). F is the restoring force exerted by the spring on the mass. It is equal to the weight of the mass, which is 19.6 N; however, it is up (positive), opposite the direction of the gravity force. x = F / (-k) = 19.6 N / (-35 N/m) = -0.56 m

This means the spring stretches 0.56 m down.

2, The energy stored in a spring is (1/2)kx². When it is in equilibrium, that would be (1/2)(35 N/m)(.56 m)² = 5.488 J.

If the stretch doubles, the stored energy quadruples to 21.952 J, an increase of 16.464 J. The push upward also adds energy equal to (1/2)mv² = (1/2)(2 kg)(3.8 m/s)²= 14.44 J.

A total of 30.904 J has been added. This then will be the kinetic energy of the mass when it passes the equilibrium point, which is where it is maximized.

The last step is to solve 30.904 J = (1/2)(2 kg)v² for v. This comes out to be 5.6 m/s