# Spring energy

1. Oct 5, 2008

### myk127

1. The problem statement, all variables and given/known data
A 2.0 kg block is placed on top of a spring which is compressed 0.5 m. The spring constant of the spring is 2.50 N/cm. when the block is released, find (a) the velocity at which the block leaves the spring (b) the velocity of the block when it has risen 1.00 m (c) the maximum height that the block can reach.

2. Relevant equations

3. The attempt at a solution
a) I tried using the formula W = ΔKE + ΔPE + ΔPEe
since there's no other work done W = 0. I canceled ΔKE because there's no motion in the object at the beggining and at the highest point. then I canceled PE1 because there's no height yet then I canceled PEe2 because it not in contact with the spring anymore. then my formula looked like this:
PEe2 = PE1 -----> ½kx² = mgh

how can i find the initial Velocity with the formula I got? pls tell me I did something wrong or if I forgot any thing.

2. Oct 5, 2008

### Rake-MC

What do your algebraic values stand for?

For this question $$E_{total} = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 + mg\Delta h$$

(a) velocity of block leaving spring. This implies all spring potential (.5kx^2) has become gravitational and kinetic. You have all values except for v, shouldn't be too hard to solve for.

(b)block after 1m. This means that once again all energy is gravitational and kinetic, exactly same form as (a) just a different h value.

(c) Once again you use only gravitational and kinetic energy, but let v = 0 and solve for h.

The reason that $$\frac{1}{2}kx^2$$ is present even though it was not mentioned in any of the guidelines I gave, is that before it is released, all energy is spring potential, therefore you can calculate the total energy of the system.

3. Oct 5, 2008

### myk127

the values I used are:
PEe - spring potential energy
k - spring constant
x - length of elongation or compression
mgh - mass, gravity and height.

sorry if i didn't understand, I'm kind of not good with physics. how come there's still ½mv² in the equation?

4. Oct 5, 2008

### Rake-MC

1/2 mv^2 is in the equation because for parts a) and b) we're solving for v.
for part c) it's not what we're solving for, although it is a part of the total energy in the system so it has to be included. Energy is always conserved.

5. Oct 5, 2008

### myk127

what would be my Δh then?

6. Oct 5, 2008

### Rake-MC

Well have a little think about it. Part (a) is asking for velocity right as it leaves the spring. This implies that the spring is completely extended and there is no more force acting (on its part). However the particle has not moved any further than the end of the spring. Therefore: your change in height is equal to that of the change in spring compression = 0.5m

for part (b) it asks for when it has risen 1m. I am not sure if they mean 1m from when the spring was fully compressed, or 1m above the top of the uncompressed spring (most likely the latter). If it was the case of the latter then your change in height would be 1 + 0.5m because you have earlier defined your gravitational potential to be measured from the top of the compressed spring. Therefore to prevent any errors entering your calculations you must keep this point the same.

7. Oct 5, 2008

### myk127

wow, thanks I never thought of that. part (a) was really my problem.