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Calc 2 help

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Peppy
Posts: 346 - Karma: 341
I'm having a hard time with this question...

A spring that obeys Hooke's Law (F(x) = kx) has length 12" at rest. A force of 15 lbs. compresses the spring to 11"
a) What is the vale of the spring constant k?
b) How much work is done to stretch the spring from 12" to 15"?

I think a) is 15/11.
and on b) i got this far.

(15/11) Integral of 3x on the intervals [12,15]

anybody know what i am talking about?
Mar 6 2012 10:14PM
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JordanButtfart
Posts: 1013 - Karma: 3,027
42.



Mar 6 2012 10:15PM
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Peppy
Posts: 346 - Karma: 341
i got 165.68 but im not entirely sure
Mar 6 2012 10:18PM
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.AR6rider.
Posts: 15976 - Karma: 8,799
for a:

F=kx
15=k(1)
k=15

Pretty sure x is 1, since you're compressing from 12in to 11in. I haven't done this shit in forever.

W = Fd, so you have to find the force to displace the spring 3 inches. F = 15(3) F = 45. W=Fd = 45(3)=135

I could be totally wrong here, just trying to rely on my memory of basic physics.
Mar 6 2012 10:20PM
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Peppy
Posts: 346 - Karma: 341
i but x=11 because 15 is the force required to move it to 11 but i understand what you are saying
Mar 6 2012 10:23PM
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Ski_Alot
Posts: 5900 - Karma: 3,658
the answer is clearly 4, geometry ftw
Mar 6 2012 10:58PM
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angus_k_
Posts: 438 - Karma: 24
K = 15lbs/(12-11)" = 15lb/in

Aha you are close!
Since there is no integral sign on ns, just assume int(__) means integrate. You have to use integrals because the force of the spring is not constant.

W= int(Fdx)

F = kx

so

W = int(Kxdx)
W = K int(xdx)

Since the spring is relaxed at x=12, at 12" there will be no force, while at 15" the force will be (15-12)K. So integrate from (12-12) to (15-12)

W = K int (xdx) from 0 to 3
W = K [ (1/2)x^2 ] from 0 to 3
W = (15) [ (1/2)(3^2) - (1/2)(0^3) ]
W = 67.5 lb*in

I hope thats right
Mar 6 2012 11:10PM
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angus_k_
Posts: 438 - Karma: 24
Yup im right. The work put into the spring to stretch it must equal the increase of potential energy in the spring.

The potential energy of a spring happens to = (1/2)Kx^2, x being the amount it is displaced from rest.
So deltaE = (1/2)(15)(3^2) = 67.5
Mar 6 2012 11:14PM
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