What is the Hooke's Law Constant for a Stretched Muscle?

In summary: Since you are solving for a number, you should be able to give the correct numerical value whether you call it positive or negative. In summary, the problem involves a muscle behaving as a spring according to Hooke's law. The Hooke's law constant was determined by measuring the force exerted when the muscle was stretched from 10 cm to 10.5 cm. The formula used to solve the problem involved finding the work done by the spring when stretched from one length to another. The constant k has associated units of measure and should be given in order for the units to work out correctly. There may be a formula for elongation involving the length and cross sectional area, but it is not applicable in this case since the material is
  • #1
8008jsmith
15
0

Homework Statement


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.

Homework Equations



The formula I used was: w = 1/2k (x2 - x0)^2 - (x1-x0)^2

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?
 
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  • #2
8008jsmith said:
The formula I used was: w = 1/2k (x2 - x0)^2 - (x1-x0)^2

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.
 
  • #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.
 
  • #4
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?

8008jsmith said:
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.

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.
 
  • #5
Stephen Tashi said:
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.

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 into throw me off? Also, since the system is the muscle, and it is doing work, it would be negative, right?
 
  • #6
8008jsmith said:
Is it safe to assume the cross sectional number was just put into throw me off?

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

Also, since the system is the muscle, and it is doing work, it would be negative, right?

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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1. What is Hooke's Law?

Hooke's Law is a principle in physics that states the force needed to extend or compress a spring by some distance is directly proportional to that distance. It is represented by the equation F = kx, where F is the force, k is the spring constant, and x is the displacement.

2. How does Hooke's Law relate to muscles?

Hooke's Law can be applied to understand the behavior of muscles. Muscles are made up of elastic fibers that can be stretched and compressed, similar to a spring. When a muscle is contracted or stretched, it exerts a force that is proportional to the amount of displacement. This allows us to understand how muscles move and perform work.

3. What is the role of the spring constant in Hooke's Law?

The spring constant, represented by the symbol k, is a measure of the stiffness of a spring. It is a constant value that relates the force applied to a spring to the amount of displacement. In terms of muscles, the spring constant is related to the stiffness of the muscle fibers and how much force is required to stretch or compress them.

4. How does Hooke's Law help us understand muscle strength?

Hooke's Law allows us to understand the relationship between force and displacement in muscles. By measuring the amount of force a muscle can exert when stretched or compressed, we can determine its strength. This can be useful in understanding muscle development and identifying potential weaknesses or imbalances in the body.

5. Can Hooke's Law be applied to other biological systems besides muscles?

Yes, Hooke's Law can be applied to other biological systems that exhibit elastic behavior, such as tendons, ligaments, and even some organs. It can also be applied in non-biological systems, such as in the study of materials and structures. However, it is important to note that Hooke's Law is an idealized model and may not fully capture the complex behavior of these systems.

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