## University Physics with Modern Physics (14th Edition)

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.