# Kinetic Energy - Force done by friction

1. Dec 11, 2008

### RichardCory

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

Kelli weighs 435 N, and she is sitting on a playground swing seat that hangs 0.44 m above the ground. Tom pulls the swing back and releases it when the seat is 1.00 m above the ground. Assume that air resistance is negligible.

(a) How fast is Kelli moving when the swing passes through its lowest position?
SOLVED: 3.31 m/s

(b) If Kelli moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction?
SOLVED: 154.284 Joules

V at bottom in closed system = 3.31 m/s
V at bottom in open system = 2.0 m/s
Kelli's mass = 44.388 kg
K in closed system = 243.158 Joules

2. Relevant equations

K = .5mv2 = J(oules)

W = Fd = J(oules)

3. The attempt at a solution

I attempted to use the difference of the two velocities to find the kinetic energy in Joules, and it did NOT work.

Any suggestions?

Last edited: Dec 11, 2008
2. Dec 11, 2008

### glennpagano44

Not 100% sure but did this class today
For b) you have to do conservation of energy .5mv^2+mgh+(work done by friction)= 0. From there you should have enough informationto solve for (work done by friction).

3. Dec 11, 2008

### RichardCory

Which height should I use for the "change in potential energy part" or the formula? 1.00 m? .44 m? or .56 m?

I tried all three, and none gave me the answer I need.

Last edited: Dec 11, 2008
4. Dec 11, 2008

### glennpagano44

You should use .56 since the bottom of the swing is your refrence point

5. Dec 11, 2008

### RichardCory

I now know the answer to part (b) is 154. 824 J, obtained by using the equation :

$$\left(F\times\left(1-d\right)\right)-\left(\frac{2F}{g}\right)$$

Can anyone explain how this give me the answer? I'm aware that $$\left(F\times\left(1-d\right)\right)$$ is the formula for Work. How does $$-\left(\frac{2F}{g}\right)$$ apply?

6. Dec 11, 2008

### LowlyPion

No need to do all that.

You know m*g*h = 435*(.56) = 243.6 J
In part a) that yields your v that I presume you calculated correctly.

In part b) they give you what the actual KE was so the Frictional losses must be the difference between the 2.

Your mass is 435/9.8 = 44.38 kg which yields 88.776 J of KE.

PE - KE = Work by friction.

7. Dec 11, 2008

### RichardCory

"PE - KE = work by friction" makes so much sense now that you said it.

Thanks a bunch. :D