Hooke's law is written as `F = kx`
where:
`F` = force
`k ` = proportionality constant or spring constant
`x` = length displacement from its natural length
Apply Hooke's Law to the integral application for work: `W = int_a^b F dx` , we get:
`W = int_a^b kx dx`
`W = k * int_a^b x dx`
Apply Power rule for integration: `int x^n(dx) = x^(n+1)/(n+1).`
`W = k * x^(1+1)/(1+1)|_a^b`
`W = k *...
Hooke's law is written as `F = kx`
where:
`F` = force
`k ` = proportionality constant or spring constant
`x` = length displacement from its natural length
Apply Hooke's Law to the integral application for work: `W = int_a^b F dx` , we get:
`W = int_a^b kx dx`
`W = k * int_a^b x dx`
Apply Power rule for integration: `int x^n(dx) = x^(n+1)/(n+1).`
`W = k * x^(1+1)/(1+1)|_a^b`
`W = k * x^2/2|_a^b`
From the required work 18 ft-lbs, note that the units has "ft" instead of inches. To be consistent, apply the conversion factor: 12 inches = 1 foot then:
`4` inches = `1/3` ft
`7` inches = `7/12` ft
To solve for k, we consider the initial condition: W =18 ft-lbs to stretch a spring `4` inches or `1/3` ft from its natural length. Stretching `1/3` ft of it natural length implies the boundary values: `a=0` to `b=1/3` ft.
Applying `W = k * x^2/2|_a^b` , we get:
`18= k * x^2/2|_0^(1/3)`
Apply definite integral formula: `F(x)|_a^b = F(b)-F(a)` .
`18 =k [(1/3)^2/2-(0)^2/2]`
`18 = k * [(1/9)/2 -0]`
`18 = k *[1/18]`
`18 = k/18`
`k =18*18`
`k= 324`
To solve for the work need to stretch the spring with additional 3 inches, we plug-in: `k =324` , `a=1/3,` and `b = 7/12` on `W = k * x^2/2|_a^b` .
Note that stretching "additional 3 inches" from its initial stretch of 4 inches is the same as stretching 7 inches from its natural length.
`W= 324 * x^2/2|_((1/3))^((7/12))`
`W =324 [ (7/12)^2/2 -(1/3)^2/2]`
`W = 324 [ 49/288 -1/18]`
`W = 324[11/96]`
`W=297/8` or `37.125` ft-lbs
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