To compress a spring 2 inches from its natural length we need 7.5 ft/pounds, or 7.5*12 = 90 inch/pounds.
Therefore from Hooke's Law:
`int_0^2 kx dx = 90`
`(k2^2)/2 - (k0^2)/2 = 90`
k = 45
Therefore f(x) = 45*x
Now to compress the spring an additional .5 inches, or...
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To compress a spring 2 inches from its natural length we need 7.5 ft/pounds, or 7.5*12 = 90 inch/pounds.
Therefore from Hooke's Law:
`int_0^2 kx dx = 90`
`(k2^2)/2 - (k0^2)/2 = 90`
k = 45
Therefore f(x) = 45*x
Now to compress the spring an additional .5 inches, or from 2 to 2.5 inches we plug in
`int_2^2.5 45x dx`
`45x^2/2` , evaluate the end points:
`(45/2)(6.25-4) = 50.625`
To move the spring an additional .5 inches, we will need 50.625 inch/pounds of work.