# Spring height question

1. Sep 14, 2007

### myelevatorbeat

1. The problem statement, all variables and given/known data

A 0.300 kg block is placed on a light vertical spring (k = 5.20 103 N/m) and pushed downward, compressing the spring 0.100 m. After the block is released, it leaves the spring and continues to travel upward. What height above the point of release will the block reach if air resistance is negligible?
2. Relevant equations

PEspring=1/2kx^2
Fspring=-kx

3. The attempt at a solution

PEspring=1/2(5.20 x 10^3 N/m)(0.100 m)
PEspring=260 Joules

Fspring=-(5.20 x 10^3 N/m)(0.100 m)
Fspring= -520 N

I'm not really sure where to go from here. I feel like I should take the force of the spring and the mass of the object and put it in F=ma to figure out the acceleration and then perhaps put it in a kinematic equation to solve for (y-yo) but I'm not sure if this would be correct. Could anyone tell me if I'm on the right track?

2. Sep 14, 2007

### myelevatorbeat

If I solve for a in F=ma, I get:

-520 N = (5.20x10^3 kg)a
a= -0.100

I don't think this is right because I'd expect an object sitting on a spring to accelerate positively, until the end when it would become negative and begin slowing down?

3. Sep 14, 2007

### rootX

what is its energy when it reaches the top?

[don't bring forces in this question]

4. Sep 14, 2007

### myelevatorbeat

I would assume the energy at the top would be 260 J since that was the potential energy of the spring and the box had no potential energy since it was sitting at rest?

5. Sep 14, 2007

### myelevatorbeat

So now, would I just plug 260 J into:

mgh to figure out h or is it more complicated than that?

6. Sep 14, 2007

### myelevatorbeat

I know that PEo+KEo=PE+KE also.

I would assume I'd need two equations since since the KEo of the spring is 260 Joules and the KEo of the box is 0?

7. Sep 14, 2007

### myelevatorbeat

Wait, I meant PE and I guess the spring has a PEo=260 J and the box has a PEo=0 since it is at rest.

8. Sep 14, 2007

### rootX

yea mgh = 260

and you get the equation above the point from where the mass started acceleration (generally/mostly then answer)

you are analyzing the situation when KE is 0.

9. Sep 14, 2007

### myelevatorbeat

Ok so I did this:

260 J =(5.20 x 10^3 kg)(9.80 m/s/s)h
and solving for h I got:

h=0.005 m

However, when I put this answer into webassign.net it tells me I'm incorrect because of "orders of magnitude". Did I do something wrong?

10. Sep 15, 2007

### myelevatorbeat

Still working on figuring this problem out.

So far, I figured out the Potential Energy of the spring and it is 260 Joules.

Now, I used this equation:

PEo+KEo=PE+KE (with KEo=0 because the box isn't in motion)
So:
PEo=PE+KE
260 J = mgH+1/2mv^2
260 J = (0.300 kg)(9.80 m/s^2)H+1/2(0.300 kg)v^2

How do I figure out what v^2 is?

Do I figure out the force of the spring and then divide to find the acceleration...but then plug it into a kinematic equation such as V^2=Vo^2=2g(Y-Yo)?

11. Sep 15, 2007

### myelevatorbeat

V^2=Vo^2-2g(Y-Yo), sorry.