Answer
(a) $k = 0.0286~N/m$
(b) The elastic potential energy stored in the membrane is $2.80\times 10^{-18}~J$
Work Step by Step
(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 $2.80\times 10^{-18}~J$
