## College Physics (4th Edition)

The maximum compression distance of the spring is $8.12~m$
We can use conservation of momentum to find the speed of the block just after the collision with the dart: $m_f~v_f= m_0~v_0$ $v_f= \frac{m_0~v_0}{m_f}$ $v_f= \frac{(0.122~kg)~(132~m/s)}{5.00~kg+0.122~kg}$ $v_f = 3.144~m/s$ We can use work and energy to find the maximum compression distance of the spring: $U_s+Work = KE_0$ $\frac{1}{2}kx^2 -(m_fg~\mu_k~x) = \frac{1}{2}m_f~v_f^2$ $kx^2 -(2m_fg~\mu_k~x) - m_f~v_f^2 = 0$ $(8.56~N/m)x^2 -(2)(5.122~kg)(9.80~m/s^2)(0.630)~x - (5.122~kg)(3.144~m/s)^2 = 0$ $(8.56~N/m)x^2 -(63.25~N)~x - (50.63~J) = 0$ We can use the quadratic formula to find $x$: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{63.25 \pm\sqrt{(-63.25)^2 - 4(8.56)(-50.63)}}{(2)(8.56)}$ $x = -0.729~m, 8.12~m$ Since $x$ must be positive, $x = 8.12~m$. The maximum compression distance of the spring is $8.12~m$