## Essential University Physics: Volume 1 (4th Edition)

We know that the work done is given by: $$W = Fd cos \theta$$ The distance is the same, so we compare $Fcos \theta$. In the first case, the angle is 0 degrees and the force is F, so we get a work of F times the distance. In the second case, we have a force of 2F and an angle of 45 degrees. When plugged into the equation, this simplifies to $\sqrt{2} F$ times the distance, which is greater than the work done in the first scenario.