#### Answer

(a) k = 7230 N/m
(b) The maximum distance the spring will be compressed is 1.22 meters.

#### Work Step by Step

We can set up a force equation. We can assume that the acceleration is 3.00 g.
$F = ma$
$kx = (3.00~g)~m$
$k = \frac{(3.00~g)~m}{x}$
We can set up an energy equation.
$U_s = K$
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
$\frac{1}{2}(\frac{(3.00~g)~m}{x})x^2 = \frac{1}{2}mv^2$
$(3.00~g)x = v^2$
$x = \frac{v^2}{3.00~g}$
$x = \frac{(6.00~m/s)^2}{(3.00)(9.80~m/s^2)}$
$x = 1.22~m$
(a) We can use the compression of the spring to find the force constant.
$k = \frac{(3.00~g)~m}{x}$
$k = \frac{(3.00)(9.80~m/s^2)(300~kg)}{1.22~m}$
$k = 7230~N/m$
(b) The maximum distance the spring will be compressed is 1.22 meters.