# Determine velocity of a mass suspended from a stretched spring

1. Oct 7, 2004

### physicsss

A mass m is attached to a spring which is held stretched a distance xmax by a force Fmax , and then released. The spring contracts, pulling the mass. For the following answers, assume that there is no friction. Use m for the mass, x for the distance xmax and F for the force Fmax.

(a) Determine the speed of the mass when the spring returns to its normal length (x = 0).

(b) Determine the speed of the mass when the spring returns to half its original extension (x = xmax/2).

2. Oct 7, 2004

### Diane_

a) You should have an equation for the energy stored in the spring as a function of the spring constant and the distance it's stretched. If the surface is frictionless, then all of the energy in the spring will appear as kinetic energy in the mass. Do an energy balance.

b) Ditto.

3. Oct 7, 2004

### Pyrrhus

No friction?? only Conservative Forces, great!

Conservation of Mechanical Energy

$$\Delta K + \Delta \Omega = 0$$

or

$$\Delta E = 0$$

4. Oct 7, 2004

### physicsss

what does Omega mean?

5. Oct 7, 2004

### Pyrrhus

Sorry, i was busy, it means Potential Energy.

6. Oct 8, 2004

### physicsss

For the first part is it 1/2mv^2=1/2kx^2, solve for v i got v=sqrt(kx^2)/m
, substituting kx for f i got sqrt(Fx/m). This is right.

But for part b, 1/2mv^2=1/2k*(x/2)^2, and when I carried this all out, I got v=1/2*sqrt(Fx/m)...but I got it wrong.

7. Oct 8, 2004

### Pyrrhus

Your second equal makes no sense...

You should use the point where it's at its max potential energy (max length) and equal it to its energy at half lenght.

$$\frac{1}{2}kx_{max}^{2} = \frac{1}{2}mv^2 + \frac{1}{2}k(\frac{x_{max}}{2})^{2}$$