Answered

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if 132 j of work are needed to stretch a spring from 9 cm to 12 cm and another 588 j are needed to stretch it from 12 cm to 19 cm, what is the natural length of the spring?

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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