Answer:
Work done in stretching the spring = 7.56 lb-ft.
Step-by-step explanation:
Normal length of the spring = 8 in or [tex]\frac{8}{12}[/tex] ft
= [tex]\frac{2}{3}[/tex] ft
If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in
= [tex]\frac{11}{12}[/tex] ft
Force applied to stretch the spring = 12 lb
By Hook's law,
F = kx [where k is the spring constant and x = length by which the spring is stretched]
12 = k([tex]\frac{2}{3}[/tex])
k = 18
Work done (W) to stretch the spring by [tex]\frac{11}{12}[/tex] ft will be
W = [tex]\int\limits^\frac{11}{12} _0 {kx} \, dx[/tex]
= [tex]\int\limits^\frac{11}{12} _0 {(18x)} \, dx[/tex]
= [tex]18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0[/tex]
= 9([tex]\frac{11}{12}[/tex])²
= 7.56 lb-ft
