# Potential Energy

1. Jan 18, 2010

1. The problem statement, all variables and given/known data
A spring is compressed a distance of 1m and a 2 kg rock is shot straight up with the spring. The rock attains a maximum height of 500m. What is the force constant k of the spring?

Use g=10m/s^2

2. Relevant equations
P.E.= mgh
P.E. elastic= (1/2)(k)(x^2)

3. The attempt at a solution

P.E.= mgh
P.E.=(2000g)(10m/s^2)(500m)
P.E.= 10000000 newtons

P.E. elastic= (1/2)(k)(x^2)
k= P.E elastic/ (1/2)(x^2)
k= 10000000/ (1/2)(1)
k=2000000 newtons ?

2. Jan 18, 2010

### Fightfish

Use SI units ie kilogram, not gram. Furthermore, energy is in joules, not newtons! Also, the spring constant k is in newtons per meter, not newtons. Please read up on units, okay?

3. Jan 18, 2010

Thanks,

So, if I correct my units I would have:

P.E.= mgh
P.E.=(2kg)(10m/s^2)(500m)
P.E.= 10000 joules

P.E. elastic= (1/2)(k)(x^2)
k= P.E elastic/ (1/2)(x^2)
k= 10000/ (1/2)(1)
k=20000 N/m ?

Is this correct?

4. Jan 18, 2010

### Fightfish

Looks good; that is one very strong spring indeed.