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

  • Thread starter Thread starter chocolatelover
  • Start date Start date
  • Tags Tags
    Work
Click For Summary
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.
chocolatelover
Messages
238
Reaction score
0
"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
 
Physics news on Phys.org
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.
 
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?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
29
Views
3K
Replies
25
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K