Calculus/ physics question HELP!!! thanks?

F=kx=>

x..........x

SFdx=Skxdx=>

0.........0

w=k(x^2)/2

where k=constant;

x=the net increment in length;

F=the force applied

w=the work done needed to

stretch the length by x.

When W=6 J,

x=36-24=12 cm=0.12 m

=>k=2500/3

x=29-24=5cm=0.05m

=>

w(0.05)=(2500/3)(0.05^2)/2

=>

w(0.05)=1.0417 J

x=34-24=10 cm=0.1 m

=>

w(0.1)=(2500/3)(0.1^2)/2

=>

w(0.1)=4.1666 J

Ans. it needs w(o.1)-w(0.05)=

3.12 J approximately to stretch

the spring from 29 cm to 34 cm.

How do you expect it to be more than 6 J when it starts further than 24 and stretches to less than 36? Just sayin.

W = ½kx²

6 = ½k(36-24)²

k= 12/12²

k = 1/12

W(29 to 34) = W(34) - W(29)

W(29 to 34) = ½(1/12)(34-24)² - ½(1/12)(29-24)²

W(29 to 34) = (1/24)[(10)²-(5)²]

W(29 to 34) = 3.125 J

No