# Spring questions

1. Jul 26, 2011

### Hypercubes

1. The problem statement, all variables and given/known data
1. An unstretched spring with spring constant 20 N/cm is suspended from the ceiling. A 4.72-kg mass is attached to the spring and let fall. To the nearest tenth of a centimeter, how far does it stretch the spring?

2. An unstretched spring with spring constant 5.9 N/cm is suspended from the ceiling. A 4-kg mass is attached to the spring and slowly let down. To the nearest tenth of a centimeter, how far does it stretch the spring?

2. Relevant equations

F=-kx
F=ma

3. The attempt at a solution
1.
ma=kx
(4.72 kg)(9.81 m/s^2)=(2000 N/m)(x)
x=2.3 cm
Wrong. The answer is 4.6 cm.

2.
ma=kx
(4 kg)(9.81 m/s^2)=(590 N/m)(x)
x=6.7 cm
Correct. Even though I used the same steps, this one is correct and the other is wrong. Am I missing a crucial difference between "let fall" and "slowly let down"?

2. Jul 26, 2011

### FireStorm000

yes you are. All bodies in motion have kinetic energy equal to mv2/2
Because one is let down (V=0), and the other is dropped (V=/=0)

In other words, you assumed both were static equilibrium problems, when in reality only the second one is.

Hope that helps :)

3. Jul 26, 2011

### Hypercubes

Thank you, that makes sense.

How does this fit into the equations though, since I am using forces and this relates to energy?

4. Jul 26, 2011

### FireStorm000

Simply remember the definition of Work, W=F*d or W=integrate(F,d,d1,d2)
If you're in calc based physics, you should be able to derive the equation, if not, then find the equation for PE of a spring on your equation sheet.

5. Jul 26, 2011

### Hypercubes

Oh, I see.

Just for future reference, this is the solution:

Potential energy due to gravity is mgh; h in this case is x.

mgx=1/2kx^2
mg=1/2kx
(4.72 kg)(9.81 m/s^2)=1/2(2000 N/m)(x)
x=4.6 cm

Thanks again for the help!