Conservation of Energy within a pulley system

AI Thread Summary
In a pulley system with two connected objects, the conservation of energy principle applies, meaning potential energy (PE) and kinetic energy (KE) must be considered for both masses. The 5.87 kg object, released from a height of 2.55 m, will convert its PE into KE as it falls, affecting the speed of both objects. The speed can be calculated using the formula v = sqrt(2gh), but it is essential to also account for the energy of the 3.24 kg object. Since the system is not closed, both masses must be analyzed to determine their respective speeds and the height the lighter object travels after the heavier one hits the floor. Understanding these dynamics is crucial for solving the problem accurately.
miamirulz29
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Homework Statement


Two objects are connected by a light string passing over a light frictionless pulley as shown in the figure. the 5.87 kg object is released from rest at a point 2.55m above the floor. The acceleration of gravity is 9.8 m/s^2. What is the speed of each object at the instant before the 5.87 kg object hits the floor. How much higher does the 3.24 kg object travel after the 5.87 kg object hits the floor.

Homework Equations


Ei = Ef

The Attempt at a Solution



The speed of the both objects are the same when both objects are moving. Because that is the case, I thought the answer would be v = sqrt(2gh), where h = 2.55m. However, that is not the case. Do I need to take into account the PE and KE of of the 3.24 kg object too?
 
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hi miamirulz29! :smile:
miamirulz29 said:
Do I need to take into account the PE and KE of of the 3.24 kg object too?

yup! :biggrin:

one mass on its own is not a closed system

conservation of energy (PE + KE = constant) only works for a closed system :wink:
 

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