Answer
(a) $13.06N/m$
(b) $0.30s$
(c) decrease
Work Step by Step
(a) We can find the spring constant as follows:
$\frac{1}{2}mv_i^2=\frac{1}{2}KA^2$
This simplifies to:
$K=m(\frac{v_i}{A})^2$
We plug in the known values to obtain:
$K=(0.505Kg)(\frac{1.18m/s}{0.232m})^2$
$K=13.06N/m$
(b) We can find the required time as follows:
$t=\frac{1}{4}T$
$\implies t=\frac{2\pi}{4}\sqrt{\frac{m}{K}}$
We plug in the known values to obtain:
$t=\frac{2\pi}{4}\sqrt{\frac{0.505Kg}{13.06}}$
$t=0.30s$
(c) We know that the time period is inversely proportional to the square root of the force constant. Thus, if the force constant of the spring increases then the time required to stop the block decreases.
