Conservation of engery of a pulley system

In summary, two objects of masses 4.50 kg and 3.00 kg, connected by a light string and pulley system, are released from rest with one object being 3.00 m above the ground. Using the isolated system model and the law of conservation of energy, the speed of the 3.00 kg object just as the 4.50 kg object hits the ground is found to be 4.43 m/s. The maximum height to which the 3.00 kg object rises is 5 m. By assuming that the final velocity of both masses at the same instant is the same, the final velocity of the 3.00 kg object can be found using the data for both masses and the fact
  • #1
jwxie
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Homework Statement



Two objects, m1 = 4.50 kg and m2 = 3.00 kg, are connected by a light string passing over a light frictionless pulley as shown in the figure below. The object of mass 4.50 kg is released from rest, h = 3.00 m above the ground.

p8-13alt.gif
(a) Using the isolated system model, determine the speed of the 3.00 kg object just as the 4.50 kg object hits the ground.
(ans: 4.43)

(b) Find the maximum height to which the 3.00 kg object rises.
(ans: 5)

Homework Equations



[tex]\Delta K[/tex] = - [tex]\Delta Ug[/text]
mgy - mgyf = - [tex]\Delta Ug[/text]
KEi + PEi = KEf + PEf

The Attempt at a Solution



I know I can solve problem a differently. Suppose I solve each mass separately using law of conservation of energy, I can find Vf for mass 2 separately.

Since the system is connected by the same uniform massless string, so the accerlation must the same. But I did not assume their Vf will be the same (the impact of m1 as it hits the ground, and at that instant the Vf of m2 raising to 4 meter).

What I want to know is, how do you prove that the final velocity of m1 and m2 at that same instant is the same using the isolated system model KEf + PEf = KEi + PEi ?

I have the following data on my hands

for m2, where m= m2, h = 4
_________________________
| | KE | PE |
| i | 0 | 0 |
| f | 1/2mv^2 | m2h |for m1 where m = m1, h = 4
| | KE | PE |
| i | 1/2mv^2 | mgh |
| f | 0 | 0 |
 
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  • #2
Since the string is unstretchable, if m1 moves a small distance Δx in time Δt, m2 also moves the same distance in same time interval. So vf of m1 is the same as vf of m2.
 
  • #3
Hi, thank you.
 

1. What is the conservation of energy in a pulley system?

The conservation of energy in a pulley system refers to the principle that energy cannot be created or destroyed, but can only be transferred or converted from one form to another. This means that the total amount of energy in a pulley system will remain constant, even as the energy is transferred between different objects or forms.

2. How does a pulley system conserve energy?

A pulley system conserves energy by using the principle of mechanical advantage. This means that the force required to lift an object is reduced by spreading it out over multiple pulleys, resulting in less energy needed to lift the object.

3. Can a pulley system ever create energy?

No, a pulley system cannot create energy. As stated in the first question, the conservation of energy principle states that energy cannot be created. A pulley system can only transfer or convert energy, not create it.

4. How does friction affect the conservation of energy in a pulley system?

Friction can affect the conservation of energy in a pulley system by causing some energy to be lost due to heat production. This means that the total amount of energy in the system may decrease due to friction, but the principle of conservation of energy still holds true.

5. Are there any external factors that can impact the conservation of energy in a pulley system?

Yes, there are external factors that can impact the conservation of energy in a pulley system. For example, external forces such as air resistance or the weight of the pulley itself can affect the energy transfer within the system. Additionally, human error or mechanical limitations can also impact the conservation of energy in a pulley system.

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