# Conservation of energy and a spring

1. Oct 29, 2007

### Th3Proj3ct

1. The problem statement, all variables and given/known data
A 1.50 kg object is held 1.20 m above a relaxed, massless vertical spring with a force constant of 320 N/m. The object is dropped onto the spring.

2. Relevant equations
(a) How far does the object compress the spring?
(0.381m)

(b) How far does the object compress the spring if the same experiment is performed on the moon, where g = 1.63 m/s2?
(0.143m)

(c) Repeat part (a), but now assume that a constant air-resistance force of 0.700 N acts on the object during its motion.
(0.371m)

3. The attempt at a solution
I have all the answers already (in parenthesis), but I need help on how to find them. For A I tried
I found velocity with Vf^2=Vi^2+2ad; Vf=Sqrt(2*9.8*1.2); Vf=4.8497
- K+U=K+U
- 0 + .5kx^2 = .5mv^2 + 0
- x = Sqrt(m/k)*v
- x = .332, which isn't the right answer, but Im not sure where I went wrong.

2. Oct 29, 2007

### PhanthomJay

looks like you got the final energy on the left side and the initial on the right side, and that -K is shaky, but in any case, you forgot to include the gravitational potential energy term on one side of the equation (which side and with what sign?)

3. Oct 29, 2007

### hotcommodity

I don't think he meant that to be a -K, I think that was just a dash.

You shouldn't have any kinetic energy in the equation, the block falls from rest, and comes to a stop once it reaches the maximum compression.

4. Oct 29, 2007

### Th3Proj3ct

Maybe this would have been better:
Kef +Usf +Ugf = Kei + Usi +Ugi
0 + .5kx^2 + 0 = .5mv^2 + 0 + 17.64(mgh = 1.5(9.8)(1.2)=17.64)
But then you get 0.5kx^2=17.64+17.64
x = 4.695

Yea it was just a dash, but not using the kinetic, I get
Usf = Ugi
.5kx^2 = mgh
x = sqrt(2mgh/k)
x = 0.332 agian >_<

5. Oct 29, 2007

### hotcommodity

I think you may be defining "h" incorrectly. I'm working on the problem as well and I'm getting a different answer. The initial hight needs to be in terms of the compression distance "x," because that's when the blocks potential energy becomes zero. You should end up with a quadratic equation.

6. Oct 29, 2007

### PhanthomJay

There are always several ways to attack a problem. Stick with one of them. You started by calculating the velocity as the object hit the spring. Fine. If that's your start point, then your mgh term should be mgx, where x is the compressed length of the spring.

7. Oct 29, 2007

### Th3Proj3ct

Thank's all, I combined what you two said and got:
0.5kx^2=17.64+14.7x;
160x^2-14.7x-17.64 which I put into the quadratic formula and got 0.381 as one of the answers. The rest should come pretty easily now. Thanks you two for all the help, I use these forums all the time and you guys really are such a huge help.

8. Nov 21, 2008

### Mindstein

I know this is an old thread, but why is 14.7 negative in your final quadratic equation?