Work-Energy Theorem: Is Lifting a Rigid Body Really 0 Work?

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The discussion centers on the application of the Work-Energy Theorem in the context of lifting a rigid body vertically. When lifting an object slowly, the initial and final kinetic energy is zero, leading to the conclusion that net work done is zero. However, work is done against gravity, resulting in an increase in gravitational potential energy. The distinction between the work done on the object and the net work is emphasized, highlighting the importance of specifying the system in question. Ultimately, while the net work is zero, the work done against gravity contributes to potential energy, clarifying the apparent contradiction.
PeterPeter
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I read on the Internet that the work done by a (rigid) body = the change in Kinetic energy.

What if I lift a rigid body slowly and vertically by 1 meter above the Earth's surface so that the initial velocity = final velocity =0?

According to the Work Energy theorem as stated on many sites on the Internet (you can google these for yourself) the net work done = 0 because the change in KE =0.

Yet I have done work against gravity. Clearly something is wrong! It's all very confusing!
 
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PeterPeter said:
Yet I have done work against gravity.
You have done positive work, gravity has done negative work, the net work is zero. Alternatively, you have done positive work that went into gravitational potential energy.
 
PeterPeter said:
the net work done = 0 because the change in KE =0.

Yet I have done work against gravity.
Is the work you did equal to the net work?
 
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It is helpful (necessary) to specify the system on which you are doing work. You can't just say "I did 10 J of work," it has to be "I did 10 J of work on the object." Then like @scottdave said, you have to take into account all of the forces doing work on the object.

When you lifted the object which system were you doing work on? The object? How about the object-Earth system?
 
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