In Example 8-10, mass m1 = 2.40 kg is connected to mass m2 = 1.80 kg as shown. Suppose the two masses start from rest and are moving with a speed of 2.10 m/s just before m2 hits the floor.
http://i214.photobucket.com/albums/cc30/bkoz319/08-10ex.gif
(a) How much conservative work was done on this system?
(b) How much nonconservative work was done on this system?
Thanks sooooo much for your time!!!!
2007-11-02 08:48:46 · 5 answers · asked by SurferGirl9 1 in Science & Mathematics ➔ Physics
Sorry, a genius wouldn't do your homework for you.
2007-11-02 17:20:04 · answer #1 · answered by Frank N 7 · 0⤊ 14⤋
The system starts out with purely potential energy (no motion). As the potiential energy decreases (i.e. as the weight falls), part of the initial PE is turned into motion--that's the "conservative" part--and the rest is turned into heat due to friction--that's the "nonconservative" part.
The system's change in potential energy is determined by how much the heights of the masses change:
ΔPE = (m2)(g)(d)
That's the amount of energy we have to work with.
> (a) How much conservative work was done on this system?
That equals the change in KE of the system. This is:
ΔKE = (m1)v²/2 + (m2)v²/2
They give you m1, m2 and v; so just plug in the values.
(b) How much nonconservative work was done on this system?
That equals the portion of the PE that did not get converted into KE. This is:
ΔPE − ΔKE
= (m2)(g)(d) − ((m1)v²/2 + (m2)v²/2)
2007-11-02 09:22:18 · answer #2 · answered by RickB 7 · 6⤊ 1⤋
i like puppies
2007-11-02 08:58:26 · answer #3 · answered by zorro1701e 5 · 0⤊ 2⤋
YEP.............
GOING TO TAKE A GENIUSES
2007-11-02 08:53:04 · answer #4 · answered by LADY_FIRE_1958 2 · 0⤊ 1⤋
a-some
b-not so much
2007-11-02 08:52:15 · answer #5 · answered by Anonymous · 1⤊ 2⤋