## College Physics (4th Edition)

(a) $k = 0.0286~N/m$ (b) The elastic potential energy stored in the membrane is $2.80\times 10^{-18}~J$
(a) From the graph, we can see the force at $x = 14~nm$ is $0.4~nN$. We can find the effective spring constant: $F = kx$ $k = \frac{F}{x}$ $k = \frac{0.4~nN}{14~nm}$ $k = 0.0286~N/m$ (b) We can find the elastic potential energy stored in the membrane: $U_s = \frac{1}{2}kx^2$ $U_s = \frac{1}{2}(0.0286~N/m)(14\times 10^{-9}~m)^2$ $U_s = 2.80\times 10^{-18}~J$ The elastic potential energy stored in the membrane is $2.80\times 10^{-18}~J$