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

(a) $\frac{1}{2}kx^2 = 3.20~J$ $x^2 = \frac{6.40~J}{k}$ $x = \sqrt{\frac{6.40~J}{1600~N/m}}$ $x = 0.0632~m = 6.32~cm$ The spring must be compressed a distance of 6.32 cm. (b) The initial potential energy of the book will be equal to the potential energy stored in the spring when it is compressed a distance of x. Note that the initial potential energy of the book depends on the total change of height 0.800 m + x. $\frac{1}{2}kx^2 = mg(0.800~m+x)$ $\frac{1}{2}kx^2 - mgx - mg(0.800~m) = 0$ $\frac{1}{2}(1600~N/m)~x^2 - (1.20~kg)(9.80~m/s^2)~x - (1.20~kg)(9.80~m/s^2)(0.800~m) = 0$ $(800~N/m)~x^2 - (11.76~N)~x - (9.408~J) = 0$ We can use the quadratic formula to find $x$. $x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{(11.76)\pm \sqrt{(-11.76)^2-(4)(800)(-9.408)}}{(2)(800)}$ $x = 0.116~m, -0.101~m$ Since the negative value is unphysical, the solution is $x = 0.116~m$ which is 11.6 cm The spring will be compressed a maximum distance of 11.6 cm.