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Math Help - Applications of integration: Work

  1. #1
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    Applications of integration: Work

    A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

    This is what I did...
    F=10
    x=4
    4 going to 6

    F=kx
    10=k(4)
    k=2.5

    f=2.5x

    I then took the integral of 2.5x from 4 to 6 and my answer was 25 ft-lbs. The correct answer is 15/4 ft-lbs.
    Can someone explain?
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  2. #2
    Member Last_Singularity's Avatar
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    Quote Originally Posted by abclarinetuvwxyz View Post
    A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

    This is what I did...
    F=10
    x=4
    4 going to 6

    F=kx
    10=k(4)
    k=2.5

    f=2.5x

    I then took the integral of 2.5x from 4 to 6 and my answer was 25 ft-lbs. The correct answer is 15/4 ft-lbs.
    Can someone explain?
    Check your units xD

    The Hooke's constant is actually 30 (with units of lbs/ft)

    Then, the work done going from natural length (x=0) to x=1/2 (6 inches is half a foot) is:
    \int_0^{1/2} 30x dx =15/4

    But who the heck uses lbs and ft as their units? Everyone else uses SI units, with kg and m...lol
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  3. #3
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    [QUOTE=abclarinetuvwxyz;244519]A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length.
    So F= kx and 10= 4k. k= 10/4= 5/2= 2.5

    How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

    This is what I did...
    F=10
    x=4
    4 going to 6

    F=kx
    10=k(4)
    k=2.5
    so k= 2.5 pounds/inch, not "pounds/foot"!

    f=2.5x

    I then took the integral of 2.5x from 4 to 6 and my answer was 25 ft-lbs. The correct answer is 15/4 ft-lbs.
    Can someone explain?
    \int_4^6 2.5 x dx= c\left| \frac{2.5}{2}x^2\right|_4^6
    = \frac{2.5}{2} (6^2- 4^2)= \frac{2.5}{2}(36-16)
    = \frac{2.5}{2}(20)= 25inch-lbs. That is the correct answer but since there are 12 inches/foot, that is 25/12 foot-pounds.
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  4. #4
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    How to actually do it.

    Quote Originally Posted by abclarinetuvwxyz View Post
    A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

    This is what I did...
    F=10
    x=4
    4 going to 6

    F=kx
    10=k(4)
    k=2.5

    f=2.5x

    I then took the integral of 2.5x from 4 to 6 and my answer was 25 ft-lbs. The correct answer is 15/4 ft-lbs.
    Can someone explain?
    - OK! You just screwed up one thing. You took it from 4 inches past its natural length to 6 inches past its natural length. You were supposed to take it from 0 to 6. Your K will be the same form 0 to 6 just like 4 to 6 inches.


    - This would give you 45 inches per pound. (12 inches per foot)!

    - THIS GIVES YOU (45/12) OR 15/4 foot per pounds!!!
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