Calculating work done on nonlinear spring

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The discussion centers on calculating the work done on a nonlinear spring described by the force law F(x) = -10x + 3x^2. A participant initially calculates the work as -16 Nm using the formula W = F*d, but the expected answer is -12 Nm. Others clarify that the problem requires calculus to accurately determine the work done, as the force exerted by the spring is not constant throughout the stretch. The correct approach involves integrating the force over the distance to find the area under the curve of the force function. Ultimately, the consensus is that calculus is necessary to solve the problem correctly.
kivarocket
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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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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.
 
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.
 
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
 
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.
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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