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

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To calculate the work done by the force F(2xi + 5yi)N on an object moving in the x direction from the origin to x=4.96m, the integral of the force must be evaluated. The correct approach involves recognizing that there is no displacement in the y direction, leading to a focus solely on the x component. The integral simplifies to W = integral from 0 to 4.96 of (2x)dx, resulting in a total work of 41.6J. Several participants noted errors in the initial calculations and emphasized the importance of correctly applying the dot product in this context. Ultimately, the discussion highlights the need for careful evaluation of integrals in physics problems.
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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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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.
 
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
 
Integral of 2x is not 2x^2. Might want to recheck that.
 
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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