# Homework Help: Maximum speed on conservative forces problem

1. Oct 12, 2008

### fanie1031

1. A girl swings on a playground swing in such a way that at her highest point she is 3 m from the ground, while at her lowest point she is 0.5m from the ground. The acceleration of gravity is 9.8m/s^2. What is her maximum speed? Answer in units of m/s.

2. 1/2mVf^2 + mgyf = 1/2mVi^2 + mgyi

I made initial velocity to equal zero.
3. [Vf= (mghi - mghf)/ (1/2m)]^(1/2)

When I do this... I endup square rooting a negative number. What am I doing wrong?

2. Oct 12, 2008

### Redbelly98

Staff Emeritus
What do you use for initial and final height?

Is the velocity actually zero at the initial height?

3. Oct 12, 2008

### fanie1031

I'm using .5 as her initial height and 3 as her final height. Is that my problem? And velocity... I just assumed... that may be wrong. (Thanks for the quick reply by the way!)

4. Oct 12, 2008

### fanie1031

OOh I got it! I did this:

v=[2(gVi-gVf)]^(1/2)

You sparked my thinking. Thank you.

5. Oct 12, 2008

### WHarmon

The problem mentions nothing about air resistance so we can assume that our forces our conservative. This meas that the sum of your forces at position 1 (her highest point) is equal to the sum of the forces at position 2 (her lowest point).

KE1 + PE1 = KE2 + PE2

You can elimitate your KE1 because at her highest point the girls velocity is 0.

This leaves us with PE1 = KE2 + PE2 We expand this formula to:

mgh1 = (1/2)m[v(squared)] + mgh2

this can be simplified by canceling out all of your masses.

gh1 = (1/2)[v(squared)] + gh2