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.
Hooke's law is written as `F = kx`
`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 given work: seven and one-half foot-pounds (7.5 ft-lbs) , note that the units has "ft" instead of inches. To be consistent, apply the conversion factor: 12 inches = 1 foot then:
`2` inches = `1/6` ft
`1/2` or `0.5` inches =`1/24` ft
To solve for k, we consider the initial condition of applying 7.5 ft-lbs to compress a spring `2` inches or `1/6` ft from its natural length. Compressing `1/6` ft of it natural length implies the boundary values:` a=0` to `b=1/6` ft.
Applying `W = k * x^2/2|_a^b` , we get:
`7.5= k * x^2/2|_0^(1/6)`
Apply definite integral formula: `F(x)|_a^b = F(b)-F(a)` .
`7.5 =k [(1/6)^2/2-(0)^2/2]`
`7.5 = k * [(1/36)/2 -0]`
`7.5= k *[1/72]`
To solve for the work needed to compress the spring with additional 1/24 ft, we plug-in:` k =540` , `a=1/6` , and `b = 5/24` on `W = k * x^2/2|_a^b` .
Note that compressing "additional one-half inches" from its `2` inches compression is the same as to compress a spring `2.5` inches or `5/24` ft from its natural length.
`W= 540 * x^2/2|_((1/6))^((5/24))`
`W = 540 [ (5/24)^2/2-(1/6)^2/2 ]`
`W =540 [25/1152- 1/72 ]`
`W=135/32` or `4.21875` ft-lbs