Suppose a force of 30 N is required to stretch and hold a spring 0.2 m from its equilibrium position.
a. Assuming the spring obeys Hooke's law, find the spring constant k.
b. How much work is required to compress the spring 0.1 m from its equilibrium position?
c. How much work is required to stretch the spring 0.4 m from its equilibrium position?
d. How much additional work is required to stretch the spring 0.2 m if it has already been stretched 0.2 m from its equilibrium?

a. The Hooke's Law states[tex]F=kx[/tex], where [tex]F[/tex] is the force, [tex]k[/tex] the spring constant and [tex]x[/tex] the displacement, Knowing the force and the displacement you can find [tex]k[/tex]:

[tex]F=kx[/tex]

[tex]k=\frac{F}x=\frac{30}{0.2}=150N/m[/tex]

b. The work done by a force [tex]F(s)[/tex] that moves along a displacement [tex]s[/tex] is:

[tex]W=\int\limits^{x_2}_{x_1} {F(s)} \, ds[/tex]

Then[tex]F(s)=ks[/tex] (Hooke's Law)

[tex]W=\int\limits^{x_2}_{x_1} {F(s)} \, ds=\int\limits^{x_2}_{x_1} {ks} \, ds=\frac{1}{2} ks^2|\limits^{x_2}_{x_1}=\frac{1}{2} k(x_2^2-x_1^2)[/tex]

Work needed to go from [tex]x_1=0[/tex] to [tex]x_2=0.1[/tex]

[tex]W=\frac{1}{2} k(x_2^2-x_1^2)=\frac{1}{2} 150(0.1^2-0^2)=75(0.01)=0.75 J[/tex]

c. Work needed to go from [tex]x_1=0[/tex] to [tex]x_2=0.4[/tex]

[tex]W=\frac{1}{2} k(x_2^2-x_1^2)=\frac{1}{2} 150(0.4^2-0^2)=75(0.16)=12 J[/tex]

d. Work needed to go from [tex]x_1=0.2[/tex] to [tex]x_2=0.4[/tex]

[tex]W=\frac{1}{2} k(x_2^2-x_1^2)=\frac{1}{2} 150(0.4^2-0.2^2)=75(0.16-0.04)=75(0.12)=9 J[/tex]