Answer:
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]
k =7.5*72
k =540
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 =540[1/128]
W=135/32 or 4.21875 ft-lbs
Step-by-step explanation:
≈ 6.08
≈ 8.06
≈ 4.47
≈ 11.40