A question on work, power and energy

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The discussion revolves around calculating the work done by gravity in an Atwood machine with masses of 2 kg and 3 kg. The initial calculations incorrectly used the total distance traveled over four seconds instead of focusing on the distance during the specific fourth second. The correct approach involves determining the distance traveled between the third and fourth seconds to find the work done. The user realizes the mistake after clarification, leading to an understanding of the need for precise time intervals in such calculations. This highlights the importance of careful analysis in physics problems.
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Homework Statement


The two blocks in an Atwood machine have masses 2 kg and 3 kg. Find the work done by gravity during the fourth second after the system is released from rest.


Homework Equations


Work done by gravity= mgh



The Attempt at a Solution


Both the masses will have the same acceleration 'a'. Let T be the tension in the string.
m1= 2 kg, m2= 3 kg, g= 10 m/s^2

T-m1g=m1a
therefore, T- 20= 2a

m2g- T= m2a
therefore, 30 - T= 3a
on solving both the equations, a= 2 m/s^2

distance traveled by the blocks would be 1/2(at^2). So, the heavier block would travel 16 m downwards and the lighter block 16 m upwards.
Net mass= 3-2= 1 kg
hence, the work done should be
W= mgh= 1 x 10 x 16 = 160J

but the answer is 67J.
where am I going wrong?
 
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ritik.dutta3 said:
distance traveled by the blocks would be 1/2(at^2). So, the heavier block would travel 16 m downwards and the lighter block 16 m upwards.

You have used distance traveled in 4 seconds .Whereas you need distance traveled during the 4th second i.e between 3rd and 4th sec.
 
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Got it! Thanks! :D
 
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