Hooke's law (confirmation if I did it right)

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The discussion confirms the correct application of Hooke's Law to calculate the work required to compress a spring. The initial work of 7.5 foot-pounds is used to determine the spring constant (k), which is found to be 540. The integration of the force function, represented as ∫(540x)dx, is set up correctly to find the additional work needed for a further compression of one-half inch from the existing compression of 2 inches.

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Homework Statement


Seven and one half foot pounds of work is required to compress a spring 2 inches from its natural length. Find the work required to compress the spring an additional one-half inch


Homework Equations


W=kd
W=∫f(x)dx


The Attempt at a Solution


7.5=W
∫kxdx from (0,1/6)
=(kx^(2))/2 from (0,1/6)
=k/72=7.5
K=540

∫(540x)dx from (4/24, 5/24)

The rest is just algebra that's irrelevant; did i set up the problem right?
 
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Yes, that is correct.
 

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