Solving 2 Weights on a Pulley Problem

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The discussion focuses on solving a physics problem involving two weights on a pulley system: a 3kg weight and a 2kg weight. The 2kg weight is initially positioned 4 meters lower than the 3kg weight. The net force was calculated as 9.81N, leading to an acceleration of 3.27 m/s². However, the initial calculations for speed using the kinematic equation were incorrect, prompting suggestions to utilize free body diagrams and Newton's second law for accurate analysis.

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1. A 3kg weight is attached to a 2kg weight by a rope. This rope is placed over a pully so the weights are hanging. The 2kg weight starts 4 meters lower than the 3kg weight. If the system starts at rest what is the speed of the objects when they are at the same height.



2. Fnet = ma
v^2=2*a*(x2-x1)




3. I took the net force to be the normal force of the 3kg weight minus the normal force of the 2 kg weight. Fnet=9.81N I then divided by 3kg to get a=3.27 m/s^2 I then plugged knowns into v^2=2*a*(x2-x1) v^2=2*3.27*2 and got v=3.617 This answer does not work. What am i doing wrong?
 
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The normal force usually refers to a force provided by contact between the system and the environment. Try drawing a free body diagram for both objects, then write down Newtons second law for both objects. Think about the acceleration of each object, then solve for the acceleration of either. Plug into the kinematic equations.
 

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