Answer:
5.5 lb-ft
Explanation:
k = Spring constant
x = Stretched length = 3 in = 3/12 = 1/4 feet
F = Force = 11 lb
[tex]F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{11}{\frac{1}{4}}\\\Rightarrow k=44[/tex]
So,
[tex]F=44x[/tex]
Work done when x = 6 in = 6/12 = 0.5 feet
[tex]W=\int_0^6Fdx\\\Rightarrow W=\int_0^{0.5} 44xdx\\\Rightarrow W=\left[\frac{44x^2}{2}\right]^{0.5}_0\\\Rightarrow W=\frac{44\times 0.5^2}{2}=5.5[/tex]
Work done is 5.5 lb-ft
