The bob of a pendulum has a mass of 2.0 kg. The bob is pulled sideways so that it is 0.25 m above the reference level.
What is the speed of the pendulum at the middle of the swing (resting position where there's only kinetic energy?)?
2007-12-17 09:44:22 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Hi:
This is simply Potential Energy of Gravity is equal to Kinetic Energy at the bottom.
KE = 1/2 * m v^2
PE=mgh
therefore
KE = PE
1/2 * m v^2 = mgh
v^2 = mgh / (1/2 *m)
v^2 = 2*g*h
v= sqr (2*g*h)
v = sqr (2 *9.8 m/sec^2 * .25 m)
v = 2.21 m/sec
2007-12-17 09:51:51 · answer #1 · answered by Frst Grade Rocks! Ω 7 · 0⤊ 0⤋
The initial energy of the mass is mgh. The energy at the bottom of the swing will be 1/2 m v²
mgh = 1/2 m v² or v = sqrt ( 2gh )
2007-12-17 17:48:00 · answer #2 · answered by jgoulden 7 · 0⤊ 0⤋