# Conservation of mechanical energy ?

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?

## Answers and Replies

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!