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 12:44:53 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Physics
this is and energy conservation problem
mgh=1/2mv squared
the mass factors out leaving
v=sqrt2gh
=sqrt2*9.8*.18
=sqrt3.5+1.87 or 1.9
2006-12-19 13:04:57 · answer #1 · answered by Keith E 1 · 0⤊ 0⤋
Hi,
the potential energy at this height = mgh = m*9.8*0.18 = 1.764*m Joule
when the pendulum returned to its equilibrium position, all its potential energy was converted into kinetic energy so,
(mv^2)/2 = 1.764*m
(now it is the time to eliminate the mass "m" from both sides)
v^2 = 2*1.764 = 3.528
so, by taking the square root
v = 1.878 m/s = 1.9 m/s
2006-12-19 21:12:24 · answer #2 · answered by AHMED A 1 · 0⤊ 0⤋
Here is a hint: the potential energy when the speed is zero and the position highest relative to the lower position is the same as the kinetic energy when the elevation is minimal.
Put the two equations together and equal them, you will that the mass component cancel out...
2006-12-19 20:50:39 · answer #3 · answered by Vincent G 7 · 0⤊ 0⤋
In this case, potential energy is equal to kinetic energy. The equation for potential energy is PE=mgh (with m=mass, g=acceleration due to gravity, h=height) and the equation for kinetic energy is KE=(1/2)mv^2 (m=mass, v^2=velocity squared). So you set PE=KE and solve for velocity. (The masses (m's) cancel out) I got 1.89m/s by using the acceleration due to gravity as 10m/s^2. Hope that helps! (Don't worry, I highly dislike physics too right now.)
2006-12-19 20:52:37 · answer #4 · answered by ahua 1 · 0⤊ 0⤋
umg= 1/2 m v^2- this is from setting potential energy equal to kinetic energy
m cancels and may be removed from both sides of the equation
ug=1/2 v^2
.18 x 9.81 x 2 = v^2
v= 1.9 m/s
2006-12-19 20:49:25 · answer #5 · answered by MrWiz 4 · 0⤊ 0⤋
mass is not important you need the length of the pendulum in order to find it out. By the way gravity is 9.8 meters per second not feet
2006-12-19 20:48:53 · answer #6 · answered by Markus W 2 · 0⤊ 0⤋
Gravity is 9.8 feet per second squared. (I don't know it in meters, sorry.)
You don't need to know the mass, because gravity works regardless of the object's mass. (Think of the experiment with the hammer and the feather on the moon.)
Hope this is a good starting point.
2006-12-19 20:48:27 · answer #7 · answered by Anonymous · 0⤊ 0⤋