Why do I not get the same result when I use change in PE?

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Faris Tulbah
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I found the answer to this problem using the change in KE, but when I try to relate the work done on the 12.0-N block in terms of potential energy i don't get the same result. Is the change potential energy not equal to the work done? I would also like to know what situations is the change PE equal to work. I know that if an object height isn't changing then i wouldn't need PE, but is this also true in free fall? Thank you!

1. Homework Statement
Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure (Figure 1)). The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward.

Find the total work done on 12.0-N block if μs=0.500 and μk=0.325 between the table and the 20.0-N block.

Homework Equations


PEi + KEi = PEf + KEf + Elost
w=KEf-KEi
w=PEf-PEi ?
Elost=MGUxD

The Attempt at a Solution


I found velocity of for this system which was 1.59 m/s. Found that the work done was simply mv^2/2.
 
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Faris Tulbah said:
I found the answer to this problem using the change in KE, but when I try to relate the work done on the 12.0-N block in terms of potential energy i don't get the same result. Is the change potential energy not equal to the work done? I would also like to know what situations is the change PE equal to work. I know that if an object height isn't changing then i wouldn't need PE, but is this also true in free fall? Thank you!

1. Homework Statement
Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure (Figure 1)). The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward.

Find the total work done on 12.0-N block if μs=0.500 and μk=0.325 between the table and the 20.0-N block.

Homework Equations


PEi + KEi = PEf + KEf + Elost
w=KEf-KEi
w=PEf-PEi ?
Elost=MGUxD

The Attempt at a Solution


I found velocity of for this system which was 1.59 m/s. Found that the work done was simply mv^2/2.
Hello Faris Tulbah. Welcome to PF.

Friction is a non-conservative force. Therefore, mechanical energy is not conserved, and the gain in KE is not equal to the decrease in PE.

You seem to have stumbled upon the "Work - Energy Theorem", which states that the change in Kinetic Energy is equal to the work done on the system by the net external Force.
 
So in other words the final work done on the object would be what was initially in the system minus the other variables that took away from the system. Since there is no PE in the "final" state of the equation we don't need to worry about it because the energy that would have been used up by it is being taken up by other variables in the equation, such as the other blocks KE and the energy used to counteract friction. To find the work on the object I would just need to isolate its components, KEf and PEf, right? Just want to make sure this concept is solid in my mind. Thanks for the response I really appreciate it!
 
Faris Tulbah said:
I would also like to know what situations is the change PE equal to work.

One case that this would hold true is if there's no work done on the system and the system is not moving.