Answer
(a) $k = k_1+k_2$
(b) The elastic potential energy stored in the spring is $0.16~J$
Work Step by Step
(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 $0.16~J$.
