An object which is thrown up rises to a height $h$ and and is stopped at height $h/2$. What is the work done? So, I had this question in mind  for a few days now, suppose an object of mass $m$ is thrown up by a person.
Now, work done on the object would be stored in the form of potential energy, now it rises to a height $h$ and the person wants to stop the object at height $h/2$ what would be the work required to stop it, Would it be $mgh$ itself or something else and if so what would happen to our "extra" muscular energy?
 A: Your muscular energy used =mgh (this energy will used fully whether it goes to height h or h/2
When you throw want to stop it at h/2 some external force is required to stop it
$\int F_{ext}$+$\int F_{non conservative}$ = $\delta$ K.E + potential energy
On plunging values
$\int F_{ext}$+0= mgh
Note# (sum of potential and kinetic energy at every point in motion is constant)

A response to others
If you consider some other forces like drag then you surely have to add the work generally questions are solved by ignoring drag
For practical situation
$\int F_{ext}$+$\int F_{non conservative}$ = $\delta$ K.E + potential energy
$\int F_{ext}$ = $\delta$ K.E + potential energy -$\int F_{non conservative}$
since work done by non conservative force is negative net external force is greater than mgh

For the acceleration which @burarian is talking you can plunge in the values in $\delta$ K.E
A: I think that this can be fairly simply answered, assuming that you're only considering a gravitational force. Although, I think you could still solve this problem with either quadratic or linear drag included (even though it would complicate things a little bit).
We know that the work done on an object is $W = \Delta K = \frac{1}{2}m(v_f^2-v_i^2)$. Since we know the gravitational acceleration is constant, you can solve for the velocity at any height using a kinematic equation, which could then be plugged into the equation above. Now you just need to think about where the given values need to be plugged into the equation above and solve for the work.
A: I disagree with @Anusha's answer.
The amount of work you have to put in for accelerating the object once it's in your hand depends on 'how fast' you want it to be deaccelerated. If you made it slow down very slowly then it'd be a small amount of work but if you wanted it to quickly deaccelerate then it'd be a large amount.
Refer to this answer here
