A force exerts an impulse `J` on a body changing its speed from `u` to `upsilon`. The force and object's motion are along the same line. Show that the work done by the force is `J(u+upsilon)//2`

Explanation

According to work energy theorem we have
`W = DeltaK=1/2mv^(2)-1/2m u^(2)`
i.e, `W=1/2m(v-u)(v+u)` …..(i)
But according to impulse momentum theorem
`J=m(v-u)` ….(i)
So, eliminating m between Eqns. (i) and (ii),
`W=1/2J/((v-u))(v-u)(v+u)`
i.e, `W=J(v+u)//2`
This is the required result.