Answer:
[tex]E=0.0452 \ J[/tex]
Explanation:
Work in a Spring
Suppose we have a spring of constant k and natural length xo. When it's unstretched, the force exerted by the spring is 0. When the spring is stretched to a length x1, the spring exerts a force given by the Hooke's law:
[tex]\displaystyle F=k.\Delta x[/tex]
Where
[tex]\Delta x=x_1-x_o[/tex]
We have the following data
[tex]\displaystyle x_o= 8\ in=8.0.0254=0.2032\ m\\\displaystyle F=16\ lb =16.4.45=71.2\ N\\\displaystyle X_1= 6\ in\ =6.0.0254=0.1524\ m[/tex]
Solving the first equation for k
[tex]\displaystyle k=\frac{F}{\Delta x}[/tex]
Plugging in the given values
[tex]\displaystyle k=\frac{71.2}{0.2032-0.1524}[/tex]
[tex]\displaystyle k=140.16\ Nw/ m[/tex]
When the sping is compressed to 7 inches:
[tex]\displaystyle \Delta x=8-7=1\ in=0.0254\ m[/tex]
The work required to compress the spring is
[tex]\displaystyle W=\frac{1}{2}\ k\ (\Delta x)^2=\frac{1}{2}\cdot 140.16\cdot 0.0254^2[/tex]
[tex]\displaystyle \boxed{E=0.0452 \ J}[/tex]
