A spring-loaded launcher can be securely attached to a tabletop. When the spring is at its uncompressed length, the plunger is in the position shown above in Figure 1. A small wood block can be pressed against the end of the plunger, compressing the spring a distance $L$, as shown in Figure 2. When the block is released, the plunger pushes the block along the tabletop. Two pins are anached to the edge of the table to prevent the plunger from extending beyond the table. When the plumger hits the pins, the block then leaves the table. There is nonnegligible friction between the block and the tabletop. The launcher can be moved closer to or farther from the edge of the tabletop. Students set up tables 1 and 2, as shown above, with tabletops that are made of the same material. On table 1, the launcher is positioned so that the distance $s=A$ between the launcher and the edge of the tabletop is slightly less than $L$. On table 2, the launcher is positioned so that the distance $s=B$ between the launcher and the edge of the tabletop is significantly less than $L$. On both tables, the block is released from rest and loses contact with the plunger at the moment the plunger reaches the pins. Tables 1 and 2 are the same beight.

$k$ is the spring constant of the spring. $m$ is the mass of the block. $L$ is the distunce the spring is compressed. $\mu(\mathrm{mu})$ is the coefficient of kinetic friction between the block and the tabletop. $s$ is the distance between the launcher and the edge of the tabletop. Assume the spring is ideal, the plunger hus negligible mass, the block loses contact with the plunger immediately after it reaches the end of the table, and there is not-negligible friction between the tabletops and the blocks. (a) Without manipulating equations, expluin why the block launched from table 1 cuuld land farther from the table than the block launched from table 2 does. Please respond on separate paper, following directions from your teacher. (b) Does the block launched from table 1 spend more, less, or the same amount of time in the air than the block launched from table 2 does? Explain your reasoning. Please respond on separate puper, following directions from your teacher. (c) Consider the Earth-block system. Determine the change, if any, in the total mechanical energy of the system from the instant the block leaves the table to the instant immediately before it reaches the ground. Please respond on separate paper, following directions from your teacher. The students correctly derive an equation for the horizontal distance $D$ from the edge of the tabletop that the block lands, in terns of $s$ (the distance between the launcher and the edge of the tabletop): \[ D=t \sqrt{\frac{k}{m}\left(2 L s-s^{2}\right)-2 \mu g s} \] where $t$ is the time interval between the instant the block leaves the table and the instant immediately before it reaches the ground. (d) Does this equation for $D$ support your argument from part (a) that the block launched from table 1 could land farther from the table than the block launched from table 2 ? Briefly explain why or why not.