# Homework Help: Work and Energy: Springs

1. Jul 23, 2013

### RandomGuy1

1. The problem statement, all variables and given/known data

A block of mass m is suspended through a spring of spring constant k[I/] and is in equilibrium. A sharp blow gives the block an initial downward velocity v. How far below the equilibrum position does the block come to an instantaneous rest?

2. Relevant equations

Potential energy of a compressed or extended spring: 1/2 kx2
Gravitational Potential energy: mgh

3. The attempt at a solution

When the block is at equilibrium, kx = mg.
This implies x = mg/k
Potential energy of the spring in this position = 1/2 kx2 = (mg)2/2k
Taking the gravitational potential energy to be zero in this position,
total mechanical energy = 1/2 mv2 + (mg)2/2k.

I can't figure out where to go from there.

2. Jul 23, 2013

### voko

What is the energy immediately after the blow? What is it at the instantaneous rest?

3. Jul 23, 2013

### RandomGuy1

Energy immediately after blow is 1/2 mv2 + (mg)2/2k. It is at energy at instantaneous rest that I'm not sure. As the kinetic energy is zero, only elastic and gravitational potential energies have values. So, I get

1/2mv2 + m2g2/2k = 1/2 k[(mg/k) + h2] - mgh

Solving that, I get h = v√(m/k).

Just one thing, how do I know if the spring potential energy is negative or positive? I mean, should the 1/2 k[(mg/k) + h2] term be negative as the spring is being brought from its natural state to an elongated state?

4. Jul 23, 2013

### voko

First of all, it is not 1/2 k[(mg/k) + h2] but 1/2 k[(mg/k) + h]2. Since you got the correct answer above, I assume that was a typo.

From the equation it follows immediately that the potential energy, as a function of elongation, is always positive. This is logical, too. If the elongation is positive, that the spring is stretched, so it has some energy stored. If the elongation is negative, then the spring is compressed, again with energy in it. The lowest energy state of a spring is zero, when it is left alone.