144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length
Force F=10 lb
From the hookes Law

Therefore, calculate k for the spring for

Work done in stretching a spring through a length dx is

For calculating the work done for stretching the spring to x=6in=(6/12)feet beyond the natural length, Integrate over the limits of x=0 to x=1/2
Therefore,
![\int_{0}^{W}dW=\int_{0}^{0.5}kxdx = > W=\frac{1}{2}k[x^2]_{0}^{0.5}=\frac{1}{2}30 \times (0.5)^2=\frac{30}{8}](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7BW%7DdW%3D%5Cint_%7B0%7D%5E%7B0.5%7Dkxdx%20%3D%20%3E%20W%3D%5Cfrac%7B1%7D%7B2%7Dk%5Bx%5E2%5D_%7B0%7D%5E%7B0.5%7D%3D%5Cfrac%7B1%7D%7B2%7D30%20%5Ctimes%20%280.5%29%5E2%3D%5Cfrac%7B30%7D%7B8%7D)
That is,

The answer that is given is for stretching the spring to 4 inches.
Mass of 10 m of chain length is 80kg. This implies, Mass per unit meter length of the chain is

Consider a small length dx of the chain at the end point A. Work done in lifting the small length dx over the height of x meters is

Now, integrate over the the value of x from
x=0 when end of the chain A is on the ground
x=6 when end of the chain A reaches 6 m above the ground
That is,
![\int_{0}^{W}dW=\int_{0}^{6}8gxdx=8g [\frac{x^2}{2}]_{0}^{6}=4g(6^2-0^2)](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7BW%7DdW%3D%5Cint_%7B0%7D%5E%7B6%7D8gxdx%3D8g%20%5B%5Cfrac%7Bx%5E2%7D%7B2%7D%5D_%7B0%7D%5E%7B6%7D%3D4g%286%5E2-0%5E2%29)

Hence,144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length
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