Finding the speed of an object on a pulley system

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The discussion focuses on calculating the speed of an object in a pulley system using potential and kinetic energy. The user attempts to analyze the system by considering only one mass (B) instead of the entire system, leading to an incorrect calculation of kinetic energy. They initially derive the tension in the string and the work done against it, but ultimately arrive at the wrong speed due to using the wrong mass in their kinetic energy equation. The correct approach involves recognizing the total mass in the system for accurate calculations. The user acknowledges the mistake and expresses gratitude for the clarification.
TiernanW
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Homework Statement


tsdVoFy.jpg


Homework Equations


PE = mgh
KE = 0.5mv^2
WD = F * s

The Attempt at a Solution


Its part i. I understand the solution when you look at the whole system. You do not need to consider the tension in the string because they cancel out, but I want to be able to do it just considering B's KE and PE.

So I said that for B, the PE at the start is mgl and at the end it is considered 0. For its kinetic energy, at the start it is 0 and at the end it is 0.5mv^2. I also realize that it is subject to resistance from the tension in the string, so I resolved the forces to 3mg - T = 3ma (For B) and T - mgSin30 = ma (for A). From this I solved for T to get T = 9/8mg. Then I calculated the work done against tension as WD = -9/8mgl.

So then using the conservation of energy I said that 3mgl = 0.5mv^2 + 9/8mgl and for v I got (√15gl)/2. (g and l also in root)

The correct answer however is v = (√5gl)/2. (g and l also in root).

I know its easier to do it considering the whole system, but I want to do it this way, so what have I done wrong?
 
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TiernanW said:

Homework Statement


[ IMG]http://i.imgur.com/tsdVoFy.jpg[/PLAIN]

Homework Equations


PE = mgh
KE = 0.5mv^2
WD = F * s

The Attempt at a Solution


Its part i. I understand the solution when you look at the whole system. You do not need to consider the tension in the string because they cancel out, but I want to be able to do it just considering B's KE and PE.

So I said that for B, the PE at the start is mgl and at the end it is considered 0. For its kinetic energy, at the start it is 0 and at the end it is 0.5mv^2. I also realize that it is subject to resistance from the tension in the string, so I resolved the forces to 3mg - T = 3ma (For B) and T - mgSin30 = ma (for A). From this I solved for T to get T = 9/8mg. Then I calculated the work done against tension as WD = -9/8mgl.

So then using the conservation of energy I said that 3mgl = 0.5mv^2 + 9/8mgl and for v I got (√15gl)/2. (g and l also in root)

The correct answer however is v = (√5gl)/2. (g and l also in root).

I know its easier to do it considering the whole system, but I want to do it this way, so what have I done wrong?
You used a mass of m, not 3m, in the K.E.
 

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SammyS said:
You used a mass of m, not 3m, in the K.E.
Shoot. Awkward haha. Thanks! :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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