Archived Potential Energy and Conservation Of Energy Atwood's Machine

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In the discussion about Atwood's machine, participants analyze the conservation of energy as two masses, m1 and m2, are released from rest. The initial potential energy is calculated as the sum of the heights of both masses, while the final potential energy considers m1 rising to a height of 2h. The velocity of the masses just before m2 lands is derived, resulting in a speed of 3.52 m/s for the given values of m1 and m2. Confusion arises regarding the final potential energy and the algebra used to arrive at the expressions, particularly the inclusion of the change in potential energy for m1. Clarifications emphasize the importance of considering both masses' energy changes in the calculations.
godkills
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The two masses in the Atwood's machine shown in Figure 8-23 areinitially at rest at the same height. After they are released, thelarge mass, m2, falls through a heighth and hits the floor, and the small mass,m1, rises through a height h.


(a) Find the speed of the masses just beforem2 lands. Assume the ropes and pulley have negligible massand that friction can be ignored. (Use m1 form1, m2 for m2, andh and g as necessary.)

(b) Evaluate your answer to part (a) for the case where h= 1.2 m, m1 =3.7 kg, and m2 =4.1 kg.

08-20.gif



So what I basically did was state the fact that Ei= Ef

Initial Potential Energy Ui = m1gh + m2gh
Initial Kinetic Energy would be 0 for both masses due to the masses at rest.

Final Potential Energy Uf = m1gh + 0 (due to the object being at rest on floor)
Final Kinetic Energy = .5v2(m1 + m2)

Ei= Ef

m1gh cancels out for both sides.
Leaving this (4.1)(9.81)(1.2) = 5v2(3.7+4.1)

Velocity is 3.52 m/s

Anyone have any idea if answer is correct? I am currently unsure if i did it correctly but i compared it to this procedure.

http://tinyurl.com/4xcrvea"

If the link above shows correct procedure can somebody explain why the potential energy of final is 2h? Also somehow the algebra doesn't make sense where the person managed to get (m2-m1) someone care to explain this? This kind of makes me furious on how I am not able to understand how the person did this!
 

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This isn't correct. I haven't analyzed your process but it follows the basic form that the energy is conserved. This special case, however, in the final stage, the height of the first mass is doubled, so potential energy has increased instead of going to zero as it did for the second mass. KEf = 1/2m(1)v^2+1/2m(2)v^2+m(1)g(2)h
 
godkills said:
If the link above shows correct procedure can somebody explain why the potential energy of final is 2h?
m1 rises from the initial height of h to a final height of 2h, so the final PE (using the given reference level) is 2m1gh. (You forgot the change in PE of m1.)
 
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