Law of conservation, and efficiency problem?

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raininggently
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Homework Statement


A 2.0x10^2 g pendulum bob is raised 22cm above its rest position.The bob is released, and it reaches its maximum speed as it passes the rest position.

Calculate its maximum speed at that point by applying the law of conservation of energy and assuming that the efficiency is 100%.


Homework Equations


-eg=mgh
-ek=(mv^2)/2
-et=eg+ek
-eff=eoutput/einput x 100%


The Attempt at a Solution


-m=2.0 x 10^2 = 0.200 kg
-h=22cm=0.22 m
I know the law of conservation states when energy is changed from one form to another no energy is lost.

I tried using ek=(mv^2)/2 but I didn't know what ek so I coulden't solve for V.

eg=mgh
eg=(0.200kg)(9.8)(0.22)
eg=0.4312 j

I don't know what to do from here :o
eg=(
 
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raininggently said:

Homework Statement


A 2.0x10^2 g pendulum bob is raised 22cm above its rest position.The bob is released, and it reaches its maximum speed as it passes the rest position.

Calculate its maximum speed at that point by applying the law of conservation of energy and assuming that the efficiency is 100%.


Homework Equations


-eg=mgh
-ek=(mv^2)/2
-et=eg+ek
-eff=eoutput/einput x 100%


The Attempt at a Solution


-m=2.0 x 10^2 = 0.200 kg
-h=22cm=0.22 m
I know the law of conservation states when energy is changed from one form to another no energy is lost.

I tried using ek=(mv^2)/2 but I didn't know what ek so I coulden't solve for V.

eg=mgh
eg=(0.200kg)(9.8)(0.22)
eg=0.4312 j

I don't know what to do from here :o
eg=(

When a pendulum swings there is a repeated interchange of Kinetic Energy and Gravitaional Potential Energy.
When it was to the side - 22 cm higher than rest position, it was all PE.
When it swings past the middle position, all that energy has been transformed into KE.