Kinetic Energy - Force done by friction

AI Thread Summary
Kelli, weighing 435 N, swings from a height of 1.00 m to 0.44 m, achieving a speed of 3.31 m/s at the lowest point in a closed system. When moving at 2.0 m/s, the work done by friction is calculated to be 154.284 Joules. The conservation of energy principle is applied, where potential energy minus kinetic energy equals work done by friction. The correct height for potential energy calculations is determined to be 0.56 m, as it represents the change in height from the lowest point. The discussion concludes with a clear understanding that the difference in potential and kinetic energy indicates the work lost to friction.
RichardCory
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Homework Statement



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

Homework Equations



K = .5mv2 = J(oules)

W = Fd = J(oules)

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?
 
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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).
 
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.
 
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You should use .56 since the bottom of the swing is your refrence point
 
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?
 
RichardCory said:
I now know the answer to part (b) is 154. 824 J, obtained by using the equation :

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

Thanks a bunch. :D
 
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