# Springs / elastic potential energy / grav pot energy

1. Oct 12, 2004

### vtech

You are given an 8.00 kg box resting on a spring. The spring is compressed 7.0 cm by the box.

(a) What is the spring constant? ____ N/cm

(b) The box is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
____ J

(c) What is the change in the gravitational potential energy of the box-Earth system when the box moves from the release point to its maximum height? ____ J

(d) What is that maximum height, measured from the release point?
_____ m

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I've got part A) which is 11.2 N/cm, I believe... but I'm having trouble getting past this....

help anyone????

2. Oct 13, 2004

### Integral

Staff Emeritus
You should have an expression for the potential energy of a spring, something like $\frac {k x^2} 2$ you need to check that.

The change in gravitational potential energy is related to the change in height.

for d, consider that energy is conserved, at what point will the stored spring potiential energy be the same as the gravitational potential energy?

3. Oct 13, 2004

### HallsofIvy

Staff Emeritus
You should know that "F= kx" where F is the force, x is the distance the spring is stretched or compressed and k is the spring constant: so k= F/x.
In this case, the force is the weight of the box. Since you were told that the answer must be in units of N/cm, you need to find the weight of the box in kg and then divide by 7 cm.

As Integral said, the potential energy of a spring compressed a distance x is (1/2)kx2. Note that the TOTAL compression here is 37 cm.

If the spring was compressed 30 cm from the point at which the box was at 'equilibrium' (where it was at rest originally), it should be easy to see that the boxes maximum height will be 30 cm above its equilibrium point. What is the change in gravitational potential energy of the box in moving up 60 cm?

Oops, I think I just handed you that one!