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

(a) The gravitational potential energy at the top will be equal to the potential energy in the spring at the bottom. $\frac{1}{2}ky^2 = mgy$ $y = \frac{2mg}{k}$ $y = \frac{(2)(3.0~kg)(9.80~m/s^2)}{1500~N/m}$ $y = 0.0392~m = 3.92 ~cm$ The spring will stretch a maximum distance of 3.92 cm. (b) The gravitational potential energy at the top will be equal to the potential energy in the spring at the bottom. $\frac{1}{2}ky^2 = mg(y+1.0)$ $\frac{1}{2}ky^2 - mgy -mg(1.0) = 0$ $\frac{1}{2}(1500~N/m)y^2 - (3.0~kg)(9.80~m/s^2)y -(3.0~kg)(9.80~m/s^2)(1.0~m) = 0$ $(750~N/m)y^2 - (29.4~N)y -(29.4~J) = 0$ We can use the quadratic formula to find $y$. $y = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $y = \frac{-(-29.4)\pm \sqrt{(-29.4)^2-(4)(750)(-29.4)}}{(2)(750)}$ $y = 0.219~m, -0.179~m$ Since the negative value means that the spring is compressed, the solution is $y = 0.219~m$, which is 21.9 cm.