Conservation of mechanical energy ?

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Homework Help Overview

The discussion revolves around the conservation of mechanical energy in the context of a coconut falling from a height. The original poster presents a scenario involving a coconut at different heights and seeks to determine its maximum speed upon falling.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between potential energy (Ep) and kinetic energy (Ek), with specific calculations for both scenarios of the coconut falling. There are questions about the sign of potential energy and how it relates to kinetic energy.

Discussion Status

Some participants have provided calculations for the maximum speed of the coconut in both scenarios, and there is acknowledgment of a correction regarding the height used in the calculations. The discussion appears to be progressing with participants confirming each other's work.

Contextual Notes

There is a mention of a potential misunderstanding regarding the height measurement in one of the calculations, which could affect the results. The original poster's initial assumption about the potential energy's sign is also under discussion.

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Conservation of mechanical energy ?

A 0.80kg coconut is growing 10m above the ground in its palm tree. The tree is just at the edge of a cliff that is 15 meters tall.

a>What would the maximum speed of the coconut be if it fell to theground beneath the tree?
b>What would the maximum speed be if it fell from the tree to the bottom of the cliff?

Equations:
Ep=mgh
Ek= -Ep
Ek= (1/2)mv^2


The Attempt at a Solution



So I've figured out Ep= 78.48J
However,Im not sure how to relate that to Ek so I could later solve for v?
 
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The second relation you wrote, relates Ek to Ep. Also notice that Ep would be -78.48 J (why?)
 


so for a>
Ep= -78.48J so Ek= 78.48J
Ek= (1/2)(0.80kg)v^2
v= 14.0 m/s

is that right??
 


Yes. For second part, the height changes (to?)
 


for b>
Ep= (.80)(9.81)(25cm)
Ek= 196.2J
196.2/ 0.4 = v^2
v= 22.14m/s
 


That looks good.

Edit : I assumed you meant 25m and not 25 cm.
 


Oops yes, I meant 25m.
Thank you! :)
 


You are welcome!
 

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