## College Physics (4th Edition)

(a) $k = \frac{k_1~k_2}{k_1+k_2}$ (b) The elastic potential energy stored in the spring is $0.15~J$
(a) Let's consider the two springs attached in series. When a force of $F$ is exerted on the springs, both springs pull with a force of $F$. Suppose the total stretch distance of the system is $x$. $F = k_1~x_1 = k_2~x_2$ $k_1~x_1 = k_2~(x-x_1)$ $x = \frac{(k_1+k_2)~x_1}{k_2}$ We can find the effective spring constant $k$ of the combination: $k = \frac{F}{x} = \frac{k_1~x_1}{\frac{(k_1+k_2)~x_1}{k_2}} = \frac{k_1~k_2}{k_1+k_2}$ (b) We can find the value of $k$: $k = \frac{k_1~k_2}{k_1+k_2}$ $k = \frac{(500~N/m)(300~N/m)}{500~N/m+300~N/m}$ $k = 187.5~N/m$ We can find the elastic potential energy: $U_s = \frac{1}{2}kx^2$ $U_s = \frac{1}{2}(187.5~N/m)(0.040~m)^2$ $U_s = 0.15~J$ The elastic potential energy stored in the spring is $0.15~J$