- #1
Seiya
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Tipler5 6.P.060.] A simple Atwood's machine uses two masses, m1 and m2 (Figure 6-38). Starting from rest, the speed of the two masses is 5.0 m/s at the end of 3.0 s. At that instant, the kinetic energy of the system is 90 J and each mass has moved a distance of 7.5 m. Determine the values of m1 and m2.
Okay this problem is giving me a hard time.. i don't know why i must be missing something...
Here is what I have done, i said that at the beginning KE is 0 and PE must be 90 J.
Once it falls down the 7.5m the KE is 90J therefore the PE is 0J.
There is the "3 seconds time" that is there and i know I am supposed to use it somehow but i could use some help here... i calculated the acceleration with that but fail to see whre i could use it right now...
here are my equations... for the beginning:
initial
total energy =90 = PEi + KEi = 7.5(9.8)(m1+m2) + 0
final
total eneryg = 90 = PEf + KEf = ((m1+m2)(v^2))/2
this homework problem is due in 3 hours, its the last problem i have to do , just a hint would help a lot, thanks guys
Okay this problem is giving me a hard time.. i don't know why i must be missing something...
Here is what I have done, i said that at the beginning KE is 0 and PE must be 90 J.
Once it falls down the 7.5m the KE is 90J therefore the PE is 0J.
There is the "3 seconds time" that is there and i know I am supposed to use it somehow but i could use some help here... i calculated the acceleration with that but fail to see whre i could use it right now...
here are my equations... for the beginning:
initial
total energy =90 = PEi + KEi = 7.5(9.8)(m1+m2) + 0
final
total eneryg = 90 = PEf + KEf = ((m1+m2)(v^2))/2
this homework problem is due in 3 hours, its the last problem i have to do , just a hint would help a lot, thanks guys