Spring Problem?

Question

A vertical spring with k = 480 N/m is standing on the ground. You are holding a 5.1 kg block just above the spring, not quite touching it.

(a) How far does the spring compress if you let go of the block suddenly?

(b) How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it?

(c) Why do your answer differ?

Answer 1

i won't solve it for ya, but here's the approach:

a) use the potential energy of the block = mgh, choose a reference point for h, you can even have it h=0. as the block moves down its energy becomes mg(h-x)

then as you depress a spring, you give it potential energy = 1/2*kx^2

both potential energies must equal, so mg(h-x)=1/2*kx^2

b) F=kx
mg = kx
x=mg/k

c) difference lies in the fact that b) is in equilibrium and a) is not.

hope this helps

Answer 2

In the first case, the string absorbs all the potential energy in the block, which is m*g*∆h. That will equal the energy in the spring, 0.5*k*∆h^2

m*g*∆h = 0.5*k*∆h^2

∆h = 2*m*g / k

In the second case, the spring merely supports the block with a force = k*∆h, which = m*g, so

h = m * g / k

When you lower the block slowly using your hands, you are absorbing some of the potential energy, so the string does not deflect as far.