Physics - W.E.P. Question with Solution | TestHub

Method 1: Even though the normal force from the floor does no physical work on the person (energy transfers occur internally from the muscles), one can write a mathematically correct expression as though the contact force with the floor does change the person's energy. Hence, we can treat the center of mass motion as $W_{net}=\Delta K_{cm}$. Here, we treat it as though only gravity and the floor act on the person, and since the person is in contact with the floor until the center of mass is at a position $h/2$ from the ground, then

$$\left(F_{floor}-mg\right)\left(\frac{h}{2}-\frac{h}{4}\right)=mg\left(\frac{3h}{4}-\frac{h}{2}\right)\Rightarrow F_{floor}-mg=mg\Rightarrow F_{floor}=2mg$$

as the student's center of mass rises from $h/4$ to $3h/4$ overall, starting and ending at rest. Therefore, $F_{floor}=2mg=2\times 0.1\times 10=2$ N.

Method 2: Since the feet come off the floor when the center of mass is at $h/2$, the person accelerates upward from $h/4$ to $h/2$ with the same magnitude of acceleration as he decelerates from $h/2$ to $3h/4$, by symmetry. The deceleration above $h/2$ is $g$ (free rise), so the upward acceleration while in contact is also $g$. Hence, while in contact with the floor, Newton's second law gives $F_{floor}-mg=mg$, i.e. $F_{floor}=2mg=2$ N.