Question

Can someone help?
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.
(a) Calculate its maximum speed at that point by applying the law of conservation of energy and assuming that the efficiency is 100%.
(b) Repeat (a) if the efficiency is 94%.
I found the eg=mgh
eg=(0.200kg)(9.8)(0.22)
eg=0.4312 j

Answer 1

What do I do now?

Answer 2

m=2.0*10^-2g, h=2.2*10^-2m (a)K.E=100%P.E (1/2)mv^2=mgh, thus, v^2=2gh v^2=2*9.8*2.2*10^-2 v^2=0.4312, thus, v=0.657m/s=65.7cm/s (b)K.E=94%P.E K.E=0.94P.E (1/2)mv^2=0.94mgh v^2=1.88gh v^2=1.88*9.8*2.2*10^-2 v^2=0.4053 v=0.637m/s=63.7cm/s

Answer 3

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.

Answer 4

Thanks I get it now

Answer 5

I think

Answer 6

do u want me to explain it more?

Answer 7

No it's okay, I get it. Thank you

Answer 8

your welcome :)