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Hooke's law and Muscles

  1. Jan 13, 2015 #1
    1. The problem statement, all variables and given/known data
    A muscle of 1 cm^2 cross section and 10 cm length is stretched to 11 cm by hanging a mass on it. The muscle behaves like a spring that follows Hooke's law. The Hooke's law constant for the muscle was determined by finding that the muscle exerts a force of 5 N when it is stretched from 10 cm to 10.5 cm.

    2. Relevant equations

    The formula I used was: w = 1/2k (x2 - x0)^2 - (x1-x0)^2
    3. The attempt at a solution
    I came up with -20 Joules for the answer but I'm not sure if this is right. I don't know why they included the cross section.

    First I solved for k using the first set of numbers. So... 5 N = 1/2 k (0.105m - 0.1m)^2 - (0.1m - 0.1m)^2

    But I'm not sure if that is right because (w) should be in Newton meters not just Newtons, right?
     
  2. jcsd
  3. Jan 13, 2015 #2

    Stephen Tashi

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    What formula have you studied in connection with Hooke's law ?

    I don't know what w and the x's represent in the formula that you gave.
     
  4. Jan 13, 2015 #3
    w is work, k is a constant, x2 is the final length, x0 is the spring at equilibrium, x1 is the initial spring length. I assumed x1 and x0 to be the same values when solving for k.
     
  5. Jan 13, 2015 #4

    Stephen Tashi

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    The problem you quoted doesn't ask a specific question. Did you quote the entire text of it?
    Does the problem concern finding an amount of work done?

    One formula called "Hooke's Law" is [itex] F = k L [/itex] where [itex] F [/itex] is the (constant) applied force needed to hold the spring stretched by a distance [itex] L [/itex] above the length it had when zero force was applied, and [itex] k [/itex] is Hooke's constant.

    Your formula can be interpreted as the work [itex] w [/itex] done by a spring when a varying force is applied that is just sufficient to stretch it from length [itex] x_1 [/itex] to [itex] x_2 [/itex]. The equilibrium length of the spring is [itex] x_0 [/itex].

    The formula for [itex] w [/itex] requires that you know enough information to express the varying force. The problem (as I interpret it) only gives the value of the force when the extension of the spring is eactly 10.5 cm.

    You could use [itex] F = k L [/itex] to find [itex] k [/itex]. Then find [itex] w [/itex] using your formula.

    The constant [itex] k [/itex] has units of measure associated with it. If you give it the correct units, the units in the problem will work out correctly.

    There is a formula for elongation involving the length and cross sectional area. The formula is associated with "Young's modulus" or "the modulus of elasticity" ( http://www.engineersedge.com/material_science/hookes_law.htm ). I don't know whether you are studying Young's modulus in your course.
     
  6. Jan 13, 2015 #5
    Thank you for your response. It is a physical chemistry question. You are correct in your interpretation. I don't think the Young's modulus equation can be applicable since it is a muscle in the question not a metal. Is it safe to assume the cross sectional number was just put in to throw me off? Also, since the system is the muscle, and it is doing work, it would be negative, right?
     
  7. Jan 13, 2015 #6

    Stephen Tashi

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    It's possible that it was, but I don't know how devious your course materials tend to be.

    I don't know much about physical chemistry, so I don't know if there is emphasis on the sign of work vis-a-vis whether it is done "by" the system or "on" the system. From my point of view, the experimenter (by means of the weight) did work "on" the muscle.
     
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