a pendulum is drawn back so that its bob is now .18 m higher than when it is at its normal equilibrium position. How fast will it go when it passes its normal equilibrium postiion once it is let go? (assume 100% energy is transferred)
I don't know how to approach this without a mass of anything
The answer is v=1.9 m/s (according to the bottom of the sheet)...how did he do this? HOW!?
2006-12-19 12:44:30 · 3 answers · asked by Anonymous in Education & Reference ➔ Homework Help
Set the equations up. You know that gravitational potential energy is mgh and kinetic energy is (1/2) m v^2. We will assume no air friction, so energy will be conserved - all of the initial potential being converted to kinetic. So:
PE0 = KEf
mgh = (1/2) m v^2
Note that the masses cancel out, and you have the rest of the information you need. This tells you, by the way, that the mass is irrelevant. In this case, any mass pendulum will behave the same way.
2006-12-19 12:48:55 · answer #1 · answered by Anonymous · 0⤊ 0⤋
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2006-12-19 20:46:23 · answer #2 · answered by hello. it's me. 4 · 1⤊ 2⤋
here is a link that explains it, and nope, mass doesn't come into it!
2006-12-19 20:49:04 · answer #3 · answered by firefly 6 · 0⤊ 0⤋