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The two masses in the Atwood's machine shown in Figure 823 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.
So what I basically did was state the fact that E_{i}= E_{f}
Initial Potential Energy U_{i} = m_{1}gh + m_{2}gh
Initial Kinetic Energy would be 0 for both masses due to the masses at rest.
Final Potential Energy U_{f} = m_{1}gh + 0 (due to the object being at rest on floor)
Final Kinetic Energy = .5v^{2}(m_{1} + m_{2})
E_{i}= E_{f}
m_{1}gh cancels out for both sides.
Leaving this (4.1)(9.81)(1.2) = 5v^{2}(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 doesnt make sense where the person managed to get (m2m1) someone care to explain this? This kind of makes me furious on how I am not able to understand how the person did this!
(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.
So what I basically did was state the fact that E_{i}= E_{f}
Initial Potential Energy U_{i} = m_{1}gh + m_{2}gh
Initial Kinetic Energy would be 0 for both masses due to the masses at rest.
Final Potential Energy U_{f} = m_{1}gh + 0 (due to the object being at rest on floor)
Final Kinetic Energy = .5v^{2}(m_{1} + m_{2})
E_{i}= E_{f}
m_{1}gh cancels out for both sides.
Leaving this (4.1)(9.81)(1.2) = 5v^{2}(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 doesnt make sense where the person managed to get (m2m1) 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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