Finding Work from Force Equation

mintsnapple
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Homework Statement


b7f66x.png



Homework Equations


W = F*d


The Attempt at a Solution


a. W = Ce^(ax)*2a^-1
b. W = Ce^(ax)*2a^-1
c. w = Ce^(ax)*(-4a^-1)

I feel like this problem is more than just this simple...
 
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mintsnapple said:

Homework Statement


b7f66x.png


Homework Equations


W = F*d

The Attempt at a Solution


a. W = Ce^(ax)*2a^-1
b. W = Ce^(ax)*2a^-1
c. w = Ce^(ax)*(-4a^-1)

I feel like this problem is more than just this simple...
Those results don't look right.

What do you get for the indefinite integral ## \displaystyle \int C e^{\alpha x}\ dx \ \ ? ##
 
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SammyS said:
Those results don't look right.
... in particular, there should not be any x in the answers.
 
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Ah, I just re-read the question and saw my mistake.

How would I approach this problem with a changing force though? Do I just plug in the end position, and multiply that force by displacement?
 
mintsnapple said:
Ah, I just re-read the question and saw my mistake.

How would I approach this problem with a changing force though? Do I just plug in the end position, and multiply that force by displacement?
dW = F dx .

You need to integrate to integrate the force with respect to x .
 
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SammyS said:
dW = F dx .

You need to integrate to integrate the force with respect to x .

Ahh, I see. So integrate the force with respect to x, and use the positions as my limits.
 
mintsnapple said:
Ahh, I see. So integrate the force with respect to x, and use the positions as my limits.

Yes.
 

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