Work energy theorem

[PeterPeter]

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!

 

[scottdave]

Can you specify which of the "many" sites you refer to? This one was at the top of my search reaults. https://courses.lumenlearning.com/boundless-physics/chapter/work-energy-theorem/

Notice it says the work done by the sum of the *net force* on the object...

If you are moving the object very slowly against gravity, think about all of the forces acting on the object.

 

[A.T.]

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.

 

[Dale]

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?

 

[jfmcghee]

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?