Calculus...?

Question

For a certain type of nonlinear spring, the force required to keep the spring stretched a distance s is given by the formula

F = ks^(4/3).

If the force required to keep it stretched 8 inches is 2 pounds, how much work is done in stretching this spring 27 inches?

Amount of work done:________ inch pound(s).

Answer 1

F = ks^(4/3)
W = ∫F∙ds = k∫s^(4/3)ds
Integrating from 0 to s,
W = (3/7)ks^(7/3)
F(8) = 2
k(8^(4/3)) = 2
16k = 2
k = 1/8
W = (3/7)(1/8)(27^(7/3))
W = (3^8)/56
W ≈ 117.16 in lb.

Answer 2

F=k s^4/3
s=8'',F=2pounds
2=k 8^4/3=16k
k=1/8
work=integral of ks^4/3 ds[] from zero to 27''] =3/7k s^7/3[ from 0 to 27''=1/827^7/3=3/7x1/8 x 3^7 pounds=3^8/56
=117.16pounds ANS

Answer 3

2 = k 8^(4/3) = k (2^3)^(4/3) = k (2^4)
k = 2/ (2^4) = 1/8

Work = integrate F ds over s=0 to 27
= int( 0.125 s^(4/3) ds )
= 0.125 (3/7) s^7/3 over s=0 to 27
= (0.0536) (27)^7/3
= 0.0536 (3^3)^(7/3)
= 0.0536 (3^7)
= 117.16 inch lb.

(EDITED: Had math error)