How to solve for the work done by a force on an object in the x direction?

  • Thread starter chocolatelover
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In summary, the conversation discusses finding the work done by a force as an object moves in the x direction from the origin to a specific point. The correct calculation for the integral of 2x is 2x^2/2, resulting in a total work of 41.6J. The conversation also highlights the importance of fully understanding the concept before attempting calculations.
  • #1
chocolatelover
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"work" problem

Homework Statement


A force F(2xi+5yi)N acts on an object as it moves in the x direction from the origin to x=4.96m. Find the work W=integral vector F(dr) done by the force on the object


Homework Equations





The Attempt at a Solution



integral 0 to 4.96 (2xi+5yi)N=2x^2/2|0 to 4.96=24.6

integral 5yi=5y^2/2=5(4.96)/2=61.5

61.5+24.6=
86.1

Could someone please tell me if this is correct?

Thank you very much
 
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  • #2
Incorrect. There's no displacement in the y, that is, j direction.

Judging by your other posts on related topics, I would recommend another thorough reading up on this chapter.
 
  • #3
Would it be like this?

Integral (2xi+5yj)(idx)|0 to 4.96=
integral 0 to 4.96(2x)dx=
2xx|0 to 4.56=
2x^2|0 to 4.56=
41.6

Work done in the y direction=0

Total work=41.6J

Thank you
 
  • #4
Integral of 2x is not 2x^2. Might want to recheck that.
 
  • #5
chocolatelover said:
Would it be like this?

Integral (2xi+5yj)(idx)|0 to 4.96=

It's a dot product: (2xi+5yj).(idx)

integral 0 to 4.96(2x)dx=
2xx|0 to 4.56=
2x^2|0 to 4.56=
41.6

(Why did it become 4.56?)

Antiderivative of 2x is 2x^2/2.

Correct your calculation.
 

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