How Much Work is Required to Compress a Spring in Specific Scenarios?

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SUMMARY

The discussion focuses on calculating the work done in compressing a spring using Hooke's Law, specifically for a spring with a constant of k = 5/4 pounds per inch. The first problem calculates the work done in compressing the spring from 0 to 7 inches, resulting in 30.625 in-lb. The second problem determines the work done in compressing the spring from 10 inches to 6 inches, yielding 40 in-lb. The spring constant remains unchanged between the two scenarios, confirming the validity of the calculations.

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  • Understanding of Hooke's Law (F = kx)
  • Basic calculus, specifically integration
  • Knowledge of units of work (in-lb)
  • Familiarity with potential energy concepts in springs
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  • Learn advanced integration techniques for calculating work
  • Explore potential energy calculations in mechanical systems
  • Investigate the effects of varying spring constants on work done
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Homework Statement



1. a force of 5 pounds compresses a 15-inch spring a total of 4 inches. how much work is done in compressing the spring 7 inches?

2. how much work is done in compressing the spring in exercise 1 from a length of 10 inches to a length of 6 inches?



Homework Equations



hooke's law: F=kd


The Attempt at a Solution



1.

F(k)=kx
5=k(4)
k=5/4

W = \int^7_0 (\frac{5}{4}x)dx

\frac{5}{4}[\frac{1}{2}x^2]^7_0

=\frac{245}{8} in-lb

note: i know that the above is correct, but it is needed for the second question, which i am unsure of.


2.

W = \int\begin{array}{c} 10 \\ 6 \end{array}<br /> (\frac{5}{4}x)dx

\frac{5}{4}[\frac{1}{2}x^2]\begin{array}{c} 10 \\ 6 \end{array}

\frac{5}{8}[x^2]\begin{array}{c} 10 \\ 6 \end{array}

\frac{5}{8}[(10^2)-(6^2)]\begin{array}{c} 10 \\ 6 \end{array}

=40 in-lb

note: in the second problem, i couldn't figure out how to fix the intervals, but it is [6,10]. that is actually the part that I'm a little confused about. is that the right interval? do i use the same k from the first problem? thanks in advance to any who can help.
 
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Your work looks fine. Since you are using the same spring in part 2 as you are in part 1, your spring constant, k, won't change. For part two, to find the work, you are essentially finding the change in potential energy stored in the spring between compressions of 10in. and 6in., and this is exactly what your equation in part 2 is equal to so your interval is correct. Good job!
 

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