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Sagot :
The natural length of the spring is 5 cm.
As Given, work done =132 J
Displacement= 12-9 =3 cm or 0.03 m
Average force:
F1=W/d=132/0.03=4400N.
That is the force at midpoint, or d1 = 10.5 cm.
Similarly,
By the spring force and Hooke’s Law
work done =588 J
Displacement= 19-12 =7 cm or 0.07 m
Average force:
F2=W/d=588/0.07=8400N.
That is the force at midpoint, or d2 = 15.5 cm
We can now calculate spring rate:
[tex]k=\frac{F2-F1}{d2-d1}[/tex]
[tex]=\frac{8400-4400}{0.155-0.105} \\=80000 N/m[/tex]
So, to reduce the 4400 N force to zero, the spring shrinks by 4400/80,000 = 0.055 m = 5.5 cm
Therefore, its natural length is 10.5 cm – 5.5 cm = 5 cm
To learn more about spring force and Hooke’s Law, click here:
https://brainly.com/question/4291098
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