Kinetic Energy - Force done by friction

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Homework Help Overview

The problem involves Kelli on a swing, analyzing her kinetic energy and the work done by friction as she moves through different heights. It explores concepts of energy conservation, potential energy, and kinetic energy in a physics context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using conservation of energy to find the work done by friction, questioning which height to use for potential energy calculations. There are attempts to reconcile differences in kinetic energy between closed and open systems.

Discussion Status

Some participants have provided guidance on using specific heights for potential energy calculations and have clarified the relationship between potential energy, kinetic energy, and work done by friction. There is ongoing exploration of the equations involved and their application.

Contextual Notes

Participants are navigating through different interpretations of the problem, including the reference height for potential energy and the implications of friction on kinetic energy. There is mention of specific values and calculations that have been attempted but not universally agreed upon.

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?
 
Last edited:
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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.
 
Last edited:
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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