Calculating Height Using Elastic Potential Energy

[kuhawkfan25]

Homework Statement

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

Homework 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

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

 

[Kurdt]

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

 

[kuhawkfan25]

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

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

Is this correct?

Thank You
~Patrick

 

[Kurdt]

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.

 

[kuhawkfan25]

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

 

[Kurdt]

Yes, that looks good to me.

 

[kuhawkfan25]

Okay, thank you very much.