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

(a) $K_2+U_2 = K_1+U_1$ $\frac{1}{2}mv_2^2+0=\frac{1}{2}mv_1^2+mgh$ $v_2^2=v_1^2+2gh$ $v_2=\sqrt{v_1^2+2gh}$ $v_2=\sqrt{(10.0~m/s)^2+(2)(9.80~m/s^2)(20~m)}$ $v_2 = 22.2~m/s$ The speed of the stone when it reaches point B is 22.2 m/s. (b) $\frac{1}{2}kx^2 = \frac{1}{2}mv^2+W_f$ $kx^2 = mv^2-2mg~\mu_k~(100+x)$ $kx^2 +2mg~\mu_k~x+2mg~\mu_k(100)- mv^2 = 0$ $(2.00~N/m)x^2 +(2)(15.0~kg)(9.80~m/s^2)(0.20)~x+(2)(15.0~kg)(9.80~m/s^2)(0.20)(100~m)- (15.0~kg)(22.2~m/s)^2 = 0$ $(2.00~N/m)x^2 +(58.8~N)~x-1512.6~J = 0$ We can use the quadratic formula to find $x$. $x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{-(58.8)\pm \sqrt{(58.8)^2-(4)(2.00)(-1512.6)}}{(2)(2.00)}$ $x = 16.5~m, -45.9~m$ Since the negative value is unphysical, the solution is $x = 16.5~m$. The stone will compress the spring a distance of 16.5 meters. (c) We can find the force exerted on the rock by the spring when the spring is compressed a distance of 16.5 meters. $F = kx = (2.00~N/m)(16.5~m)$ $F = 33.0~N$ We can find the maximum possible force of static friction. $F_f = mg~\mu_s = (15.0~kg)(9.80~m/s^2)(0.80)$ $F_f = 118~N$ Since the maximum possible force of static friction is greater than the force from the spring, the rock will not start moving again.