Finding the speed of an object on a pulley system

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SUMMARY

The discussion focuses on calculating the speed of an object in a pulley system using energy conservation principles. The user initially calculated the kinetic energy (KE) and potential energy (PE) for mass B incorrectly by using a mass of m instead of 3m, leading to an incorrect final velocity of (√15gl)/2 instead of the correct (√5gl)/2. The user also attempted to resolve forces and calculate work done against tension but misapplied the mass in the kinetic energy equation. The correct approach involves consistently using the total mass in energy calculations.

PREREQUISITES
  • Understanding of potential energy (PE) and kinetic energy (KE) equations
  • Knowledge of Newton's laws of motion and force resolution
  • Familiarity with conservation of energy principles
  • Basic understanding of pulley systems and tension forces
NEXT STEPS
  • Review the principles of conservation of energy in mechanical systems
  • Study the effects of tension in pulley systems
  • Learn about force resolution and its application in dynamics
  • Explore advanced problems involving multiple masses in pulley systems
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of common mistakes in pulley system problems.

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! :)
 

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