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Math Help - application to physics and engineering

  1. #1
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    application to physics and engineering

    Hi everyone,

    Could someone please help me on this problem?

    A spring has a natural length of 20 cm. If a 25 N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?

    I know that work=(force)(distance)

    Force=25 N
    distance=25-20=5
    work=(25N)(5cm)

    Is this correct so far? If it isn't, could you please show me what to do?

    Thank you very much
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  2. #2
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    Quote Originally Posted by chocolatelover View Post
    A spring has a natural length of 20 cm. If a 25 N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?

    I know that work=(force)(distance)

    Force=25 N
    distance=25-20=5
    work=(25N)(5cm)

    Is this correct so far? ...
    Hello,

    unfortunately no.

    1. If you stretch a spring the necessary force is not constant but is proportional to the distance you have stretched the spring:

    If F is the force, d is the distance of stretching then:

    \frac Fd=k . This quotient is constant (in certain boarders: You can't stretch a spring of 10 cm to a length of 1 km). In your case:

    \frac{25\ N}{10\ cm}=\boxed{k=2.5\ \frac N{cm}}

    2. Since the force is not constant your formula to calculate the work is not valid. The work done to stretch a spring can be calculated by:

    w=\frac12 \cdot k \cdot d^2 . Plug in the values for k and d.

    w=\frac12 \cdot 2.5\ \frac N{cm} \cdot 5^2\ cm^2=31.25\ Ncm = 0.3125 \ J
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  3. #3
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    Thank you very much

    I haven't learned that formula

    Could you also do it this way?

    20cm to 30 cm
    change=
    .10m

    f(.10)=25
    .10k=25
    k=250

    f(x)=kx=250dx
    w=int. .10 to .25(250xdx)
    =(250x^2/2)from .10 to .25 (Are these limits correct?)
    w=6.5625J

    Thank you very much
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Thank you very much

    I haven't learned that formula

    Could you also do it this way?

    20cm to 30 cm
    change=
    .10m

    f(.10)=25
    .10k=25
    k=250

    f(x)=kx=250dx
    w=int. .10 to .25(250xdx)
    =(250x^2/2)from .10 to .25 (Are these limits correct?)
    w=6.5625J

    Thank you very much
    Actually, the two of you are using the same method:
    W = \int \vec{F} \cdot \vec{ds}

    When you put Hooke's law in for the force, you get the formula earboth used.

    -Dan
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  5. #5
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    Do you see where I made my mistake?

    Thank you very much
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