## College Physics (4th Edition)

(a) $k = k_1+k_2$ (b) The elastic potential energy stored in the spring is $0.16~J$
(a) We can consider two springs connected in parallel. When a force of $F$ is applied to the system, both springs stretch the same distance $x$. The sum of the forces exerted by the springs is equal in magnitude to $F$: $F = k_1~x+k_2~x$ We can find the effective spring constant $k$: $k = \frac{F}{x} = \frac{k_1~x+k_2~x}{x} = k_1+k_2$ (b) We can find the value of $k$: $k = k_1+k_2 = (500~N/m)+(300~N/m) = 800~N/m$ We can find the elastic potential energy: $U_s = \frac{1}{2}kx^2$ $U_s = \frac{1}{2}(800~N/m)(0.020~m)^2$ $U_s = 0.16~J$ The elastic potential energy stored in the spring is $0.16~J$.