A block of mass `m` is placed at the bottom of a massless smooth wedge which if placed on a horizontal surface. When we push the wedge with a constant force, the block moves up the wedge. Find the work done by the external agent when the block has a speed v and is reaches the top of the wedge.

Explanation

`W-E` theorem: Applying work-energy theorem for wedge bead system,

`W_(ext)+W_(gravity)=DeltaK`
We have `W_(ext)-mgy=DeltaK`
This gives `W_(ext)=mgy+DeltaK`
`DeltaK:` The KE of the bead changes by
`DeltaK=1/2mv^2`
Using Eqs. (i) and (ii), we have `W_(ext)=1/2mv^2+mgy`