Calculating Work on a Compressed Spring

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The discussion focuses on calculating the work done in compressing a spring using Hooke's Law. A force of 1100 lb compresses a spring from its natural length of 17 inches to 14 inches, resulting in a spring constant (k) of 366.7 lb/in. The work done in further compressing the spring from 14 inches to 9 inches is calculated using the integral W=∫366.7x dx from x=3 to x=8, yielding a result of approximately 10,084.25 in-lb. The closest multiple-choice answer provided is option c, 10000.

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kari82
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I'm not able to get any of the multiple choice answers. Please help.

a force of 1100 lb compresses a spring from its natural length of 17in to a length of 14in. How much work is done in compressing it from 14in to 9in?

a)4600
b)0.075
c)10000
d)20000

What i did was F(x)=kx => 1100=3k => k=366.7

W=∫366.7x dx from x=3 to x=8, which equals 10084.25

is this answer correct?
 
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I get 10,083.3 (in-lb), which is close to what you got. My answer differs from yours because I didn't round 1100/3.

The answer closest to this is c, for what it's worth.
 
thanks!
 

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