Help ly need someone to help me check pulley problem

AI Thread Summary
A user is seeking assistance with a pulley problem involving two masses: 1.00 kg on the left and 6.50 kg on the right, with the latter positioned 2.00 m above the ground. The system is released from rest, and the user is trying to calculate the speed of each mass just before the right mass hits the ground. They presented an energy conservation equation but questioned whether their approach would yield the correct answer. Another participant confirmed that, despite some initial symbol errors, the user's substitutions were correct, indicating they are on the right track to solve the problem. The discussion highlights the importance of careful equation setup in physics problems.
squintyeyes
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Help urgently need someone to help me check pulley problem

Two masses are connected over a small pulley using a light string. The mass on the left is 1.00 kg and the mass on the right is 6.50 kg. The mass on the right is a distance 2.00 m above the ground. The system is released from rest as shown in the animation below.



What is the speed of each object just before the mass on the right strikes the ground?
________________m/s

Attempt
mgh + (mv^2)/2 + (Mv^2)/2 = mgh + (mV^2)/2 + (Mv^2)/2
(6.5)(9.8)(2) + 0 + 0 = 1(9.8)(2) + (6.5v^2)/2 + V^2/2
(6.5)(9.8)(2) = 1(9.8)(2) + (7.5v^2)/2

If i solve for this will i get the right answer?
 
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squintyeyes said:
Attempt
mgh + (mv^2)/2 + (Mv^2)/2 = mgh + (mV^2)/2 + (Mv^2)/2
(6.5)(9.8)(2) + 0 + 0 = 1(9.8)(2) + (6.5v^2)/2 + V^2/2
(6.5)(9.8)(2) = 1(9.8)(2) + (7.5v^2)/2

If i solve for this will i get the right answer?

Yes. You switched some symbols in the first line which should be:

mgh + (mv^2)/2 + (Mv^2)/2 = Mgh + (mV^2)/2 + (MV^2)/2

but you substituted all the right constants anyway.
 


thanks
 
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