Respuesta :
Answer:
Work done = 1.69 lb-ft.
Step-by-step explanation:
Work required to stretch a spring 2 feet beyond its natural length W = 12 ft-lb
By Hook's law
[tex]W=\int\limits^2_0 {kx} \, dx[/tex]
[tex]12=k[\frac{x^{2} }{2}]^{2}_0[/tex]
12 = [tex]\frac{k}{2}(2)^{2}[/tex]
12 = 2k
k = 6 lb per feet
Now we have to find the work done to stretch the spring 9 inch or [tex]\frac{9}{12}[/tex] feet beyond its natural length.
We again use the Hook's law.
[tex]W=\int\limits^\frac{3}{4} _0 {kx} \, dx[/tex]
[tex]W=k\int\limits^\frac{3}{4} _0 {x} \, dx[/tex]
[tex]W=6\int\limits^\frac{3}{4} _0 {x} \, dx[/tex]
[tex]W=6[\frac{x^{2} }{2}]^{\frac{3}{4}}_{0}[/tex]
[tex]W=3(\frac{3}{4})^{2}[/tex]
[tex]W=\frac{27}{16}[/tex] lb-ft
Therefore, work done to stretch the spring 9 in. beyond the its natural length will be 1.69 lb-feet.
