Explanation:
Hook's law describes the force F a spring exerts on an object that is stretched or compressed by a distance x from its natural length and has spring constant k:
(1) [tex]F_S = -kx[/tex]
The force of gravity of an object with mass m and gravitational acceleration g is given by:
(2) [tex]F_G=mg[/tex]
(a) To find the natural length l of the spring you can use the example 1 and 3 of the spring. Use equations 1 and 2 to get the following system of equations with two unknowns k and l:
[tex]0.02g=(0.14-l)k\\0.03g=(0.16-l)k[/tex]
To solve for k you can subtract the equations:
[tex]0.01g=0.02k\\ k=\frac{g}{2}\frac{N}{m}[/tex]
Now plug in k to solve for l:
[tex]l=0.1m[/tex]
(b) Now use Hook's law to find x:
[tex]x=\frac{F}{k}+l=\frac{mg}{k}+l=\frac{2mg}{g}+l=2m+l=0.19m[/tex]
(c) Do the same to find the mass m:
[tex]m=\frac{k}{g} (0.17-l)=\frac{g}{2g}(0.17-l)=\frac{1}{2}(0.07)=0.035kg=35g[/tex]
(d) Work is given by:
[tex]W=\int\limits^a_b {F}\cdot \, dx=\int\limits^{0.17}_{0.14} {kx} \, dx=\frac{1}{2}kx^2|^{0.17}_{0.14}=\frac{1}{2}k(0.17^2-0.14^2)=2.2*10^{-2}J[/tex]
