Answer
$0\;\rm J$
Work Step by Step
We know, from the work-kinetic energy theorem, that
$$\Delta K=W$$
Thus,
$$W=K_f-K_i $$
$$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$
We can assume that the initial velocity of the object was in the positive direction, so the opposite direction of the velocity is in the negative direction. We need the final velocity to be at the same magnitude as the initial velocity, so
$$v_f=-v_i$$
$$W=\frac{1}{2}m(-v_i)^2-\frac{1}{2}mv_i^2$$
Therefore,
$$W=\frac{1}{2}mv_i^2-\frac{1}{2}mv_i^2=\color{red}{\bf 0}\;\rm J$$
This means that the negative work done on the object to bring it to rest from the initial velocity is equal to the positive work done on the object to give it the same initial speed. Hence, the net work is zero.
