Calculating work done on nonlinear spring

In summary, the problem involves finding the work done by a nonlinear spring with a force law of F(x) = -10x + 3x^2. Despite initial attempts using a constant force formula, the correct answer of -12 Nm can only be obtained through integral calculus due to the varying force produced by the spring throughout the expansion process.
  • #1
kivarocket
5
0
Can anyone help me solve this problem? It seemed straightforward at first, but I am not getting the correct answer of -12 Nm. Thank you!

A nonlinear spring is modeled by a force law given by F(x) = -10x + 3x^2, where F is measured in Newtons and x in meters. How much work is done stretching the spring to x = +2.0 m from its equilibrium position at x = 0.0 m?


2. Attempt: W = F*d; F = -8N, so -8N * 2m = -16J
 
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  • #2
If a force 'F' moves its point of application a distance 'dx', then the incremental work done is dW=F dx, so over an interval x, the entire work done is

∫ dW = ∫ F dx or W = ∫F dx.
 
  • #3
This problem doesn't require calculus, so I don't need to take the integral. However, if I use the equation provided to solve for F, I get -8 N. Since W = F*d (where d = distance traveled), then -8N * 2m = -16 Nm. The correct answer is supposedly -12 Nm, so I'm not sure what I could be missing in this scenario.
 
  • #4
kivarocket said:
This problem doesn't require calculus, so I don't need to take the integral. However, if I use the equation provided to solve for F, I get -8 N. Since W = F*d (where d = distance traveled), then -8N * 2m = -16 Nm. The correct answer is supposedly -12 Nm, so I'm not sure what I could be missing in this scenario.
Contrary to what you said, this problem does require calculus. If you follow rock.freak667's advice, you will obtain -12 Nm using calculus.

Chet
 
  • #5
kivarocket said:
This problem doesn't require calculus, so I don't need to take the integral. However, if I use the equation provided to solve for F, I get -8 N. Since W = F*d (where d = distance traveled), then -8N * 2m = -16 Nm. The correct answer is supposedly -12 Nm, so I'm not sure what I could be missing in this scenario.

If you draw the graph of F vs x, you will get a parabolic shaped curve. The work done will be the area under that curve from x=0 to x=2 and in order to get the exact area, integral calculus is required otherwise, anything you do will be an approximation and may be inaccurate.
 
  • #6
kivarocket said:
This problem doesn't require calculus, so I don't need to take the integral. However, if I use the equation provided to solve for F, I get -8 N. Since W = F*d (where d = distance traveled), then -8N * 2m = -16 Nm. The correct answer is supposedly -12 Nm, so I'm not sure what I could be missing in this scenario.

You're using the formula for the work of a constant force. The force produced by that spring is not constant. You cannot simply use the force obtained at the very end after stretching the spring and use it as if that had been the force produced by the spring all along throughout the whole expansion process. This problem indeed does require calculus.
 

1. How is work done on a nonlinear spring different from a linear spring?

The main difference is that a nonlinear spring does not follow Hooke's law, which states that the force applied to a spring is directly proportional to its displacement. This means that the amount of work done on a nonlinear spring will vary depending on the displacement, while for a linear spring, it will always be the same.

2. How do you calculate work done on a nonlinear spring?

To calculate work done on a nonlinear spring, you need to integrate the force applied over the displacement. This means you need to have a function that describes the force as a function of displacement. Then, you can use the formula W = ∫Fdx to find the work done.

3. Can you give an example of a nonlinear spring?

One example of a nonlinear spring is a rubber band. As the rubber band is stretched, the force applied to it increases, but not proportionally to the displacement. This means that the work done on the rubber band will vary depending on how far it is stretched.

4. How does the stiffness of a nonlinear spring affect the work done?

In general, a stiffer nonlinear spring will require more force to be applied in order to stretch it, and therefore, more work will be done on it. However, the specific relationship between stiffness and work done will depend on the specific spring and the displacement being considered.

5. Is there a way to simplify the calculation of work done on a nonlinear spring?

Yes, in some cases, the force-displacement relationship of a nonlinear spring can be approximated by a linear function in a certain range of displacement. In these cases, you can use the formula W = 1/2 kx2 to calculate the work done, where k is the spring constant of the linear approximation and x is the displacement. However, this method should only be used if the linear approximation is accurate enough for the desired level of precision.

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