Big Test Prep- Questions (2 of 3)

  • #1
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Big Test Prep-- Questions (2 of 3)

Alright we are having a review test and I needed to go back to the stuff we did earlier this year and I am having some trouble remembering what to do....
:shy: how can i forget this stuff you ask... I don't know


The Force exerted on an object is [tex] F=F_0(X/X_o-1) [/tex] Find the work done in movieng the object from [tex]x=0[/tex] to [tex] x=3X_o[/tex]

a. by plotting F(x) and finding the area under the curve and

b. by evaluting the intergral analytically.

Answer: [tex] \frac{3}{2}F_oX_o [/tex]
 

Answers and Replies

  • #2
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W = int (F(x)dx)

Remember that F0 and X0 are both just constants.
 
  • #3
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If you were to integrate that assuming xo and fo are constants would you not get

FoX(X-Xo)
----------
2Xo
 
  • #4
HallsofIvy
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Tom McCurdy said:
If you were to integrate that assuming xo and fo are constants would you not get

FoX(X-Xo)
----------
2Xo

Your notation is ambiguous. Is the original force function
[tex]F_0\frac{X}{X_0-1}[/tex]
or
[tex]F_0(\frac{X}{X_0}-1)[/tex]
??
If it is the first, then when X= 3X0, the force is 3F0X0/(X0-1) and when X= 0, it is 0. The area under is the area of a triangle: (1/2)(3X0)(3F0X0/(X0-1)
= 9F0X0/2(X0-1) which is exactly what you would get by integrating.

If it is the second, then when X= 3X0, the Force is 2F0 but when X=0, it is -F0 and the "force graph" has x-intercept at X= X0. Now the "area under the curve" is the area of the triangle with vertices at (X0,0), (3X0,0) and (3X0, 2F0) (which is 2F0X0) minus the area of the triangle with vertices at (X0,0), (0,0) and (0,-F0) which is (1/2)F0X0: work done (2- 1/2)F0X0=(3/2)F0X0. I guess that's the one you meant.

In terms of integration, The work would be the integral, from X= 0 to 3X0, of F0((X/X0- 1) dx. The anti-derivative of that is F0(X2/(2X0)- X). Evaluating that at 3X0 and 0 gives F0((9/2)X0-3X0)= F0(3/2)X0 as before.
 

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