Is the conservation of energy being applied correctly in this scenario?

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SUMMARY

The discussion focuses on the application of energy conservation principles in a system involving two masses connected by a light string over a frictionless pulley. The calculations confirm that the speed of the 3.00 kg object (m2) just as the 5.00 kg object (m1) hits the table is 4.43 m/s. The energy conservation equation used is Ei = Ef, where the initial potential energy of m1 is converted into the kinetic energy of both masses and the potential energy of m2 at its maximum height.

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  • Understanding of classical mechanics principles, specifically energy conservation.
  • Familiarity with kinetic energy (KE) and potential energy (U) formulas.
  • Knowledge of isolated systems and their behavior in physics.
  • Ability to solve equations involving multiple variables and units.
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Students of physics, educators teaching mechanics, and engineers interested in energy systems will benefit from this discussion.

Kuklinski
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I don't know if I'm setting up the energies correctly.

1. Two objects are connected by a light string passing over a light, frictionless pulley. The object of mass m1=5.00kg is released from rest at a height h=4m above the table. Using the isolated system model, (a) determine the speed of the object of mass m2=3.00kg just as the 5.00-kg object hits the table and (b) find the maximum height above the table which the 3.00-kg object rises.
2. E1=Ef
KE=.5mv2
U=mgh
3. Ei=Ef
m1gh=m2gh + .5(m1+m2)v2
5(9.8)(4) = 3(9.8)(4) + .5(8)(v2
v=4.43m/s
 
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Welcome to PF!

Hi Kuklinski! Welcome to PF! :smile:
Kuklinski said:
Two objects are connected by a light string passing over a light, frictionless pulley. The object of mass m1=5.00kg is released from rest at a height h=4m above the table. Using the isolated system model, (a) determine the speed of the object of mass m2=3.00kg just as the 5.00-kg object hits the table and (b) find the maximum height above the table which the 3.00-kg object rises.

Ei=Ef
m1gh=m2gh + .5(m1+m2)v2
5(9.8)(4) = 3(9.8)(4) + .5(8)(v2
v=4.43m/s

Yes, that looks fine. :smile:

What's worrying you about it? :confused:​
 

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