# A simple Atwood's machine (conservation of energy)

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 im 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

why does the PE have to be 0 at the end of three seconds? it doesn't say that the masses have come to a halt doest. one of the masses has to have some PE because it is moving in th positive y direction right.

the total kinetic energy is 90 joules, there is also some PE being made here, from the one mass moving up, the system is transferring some KE to PE

you are right, i assumed all of the PE converted to KE... damn..

illl write my new equations now, thank you!

alright is this better?

KEf+PEf=PEi+KEi

KEi = 0
KEf= 90J

KEf=1/2((m1+m2)v^2)

90= PEi - PEf

90=(m1+m2)(g)(h) - ((m1(g)(-7.5))+(m2(g)(7.5)))

a=v/t
a=1.667

d=(at^2)/2

update:
im going to put the lowest height as height 0 and PE 0, where they start from ast 7.5 and the highest height one as 15.... is this right? what should i do next?

Last edited:
wee i tried it like i stated above and it worked out :) awesome, thanks for the hint mathmike

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