# Height with EPE question.

1. Jan 28, 2009

### kuhawkfan25

1. The problem statement, all variables and given/known data
A .250 kg block on a vertical spring with a spring constant of 5.00 x 10(3) N/m is pushed downward, compressing the spring .100 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above above the point of release?

Mass = .250 kg
Distance = .100m
Spring Constant/K = 5.00 x 10(3) N/m or 5000 N/m

2. Relevant equations
Have to find EPE I think, EPE = 1/2 K(^x)2....the ^ is a triangle which in this equation is change in length.
Maybe use mgh

3. The attempt at a solution

I got EPE which was 25. I think that to find h you can use formula mgh. I forgot how to do this but I recall doing something like (.250)(9.81)(h), 2.45(h), and you take the square root of 2.45 and that is how you get h. Then in this problem h would = 1.60m. I really think this is wrong but it was all I could think of.

Can someone please help me find the height? Or if I have the correct method please reply and tell me I did it right(don't think I did though)

Thank You
~Patrick

Last edited: Jan 28, 2009
2. Jan 28, 2009

### Kurdt

Staff Emeritus
All you have to do is equate the spring potential energy with the gravitational potential energy to find the height.

3. Jan 28, 2009

### kuhawkfan25

So PE = the GPE and divide like...2.45(h) = 25..25/2.45...h = 10.2m

Is this correct?

Thank You
~Patrick

Last edited: Jan 28, 2009
4. Jan 28, 2009

### Kurdt

Staff Emeritus
Equate just means setting one thing equal to the other. In your question you have set:

$$\frac{1}{2}kx^2 = mgh$$

With some algebraic manipulation you can make h the subject and plug in the numbers.

5. Jan 28, 2009

### kuhawkfan25

Okay, so I had done that, I'm pretty sure 10.2m is the answer than, can you confirm that though?

6. Jan 28, 2009

### Kurdt

Staff Emeritus
Yes, that looks good to me.

7. Jan 28, 2009

### kuhawkfan25

Okay, thank you very much.