Answer:
0.037 N/m
Explanation:
The web acts as a spring, so it obeys Hook's law:
[tex]F=kx[/tex] (1)
where
F is the force exerted on the web
k is the spring constant
x is the stretching/compression of the web
In this problem, we have:
- The mass of the fly is [tex]m=15 mg=15\cdot 10^{-6} kg[/tex]
- The force exerted on the web is the weight of the fly, so:
[tex]F=mg=(15\cdot 10^{-6}kg)(9.81 m/s^2)=1.47\cdot 10^{-4}N[/tex]
- The stretching of the web is
[tex]x=4.0 mm=0.004 m[/tex]
So if we solve eq.(1) for k, we find the spring constant:
[tex]k=\frac{F}{x}=\frac{1.47\cdot 10^{-4} N}{0.004 m}=0.037 N/m[/tex]