A block of mass 0.50 kg on a smooth horizontal table is joined to a light spring of force constant…

Step 1. Speed at the wall (a). $\omega = \sqrt{k/m} = 10\ \text{rad/s}$. At $x = 5.0\ \text{cm}$:

Step 2. Time between hits (b). From release, $-A\cos\omega t = d$ first holds at $\omega t = \cos^{-1}(-d/A) = 2.0944\ \text{rad}$, so $t_1 = 0.2094\ \text{s}$. After the bounce the block retraces the motion back to $x = -A$ and returns: the time between hits is $2t_1 = 0.419\ \text{s}$. Without the wall the period would be $0.628\ \text{s}$.

Step 3. Average force (c). Each hit gives the wall $2mv = 0.866\ \text{kg m/s}$. So the average force is

Step 4. Time to the right (d). The block is to the right of its rest position from $x = 0$ to the wall and back. Going out takes $\omega t = \sin^{-1}(d/A) = 0.5236\ \text{rad}$, so the time on that side is $0.1047\ \text{s}$ per cycle, a fraction $0.1047\ \text{s} / 0.4189\ \text{s} = 0.250$.