Finding Work along a displacement

[c42]

Hi

The problem given is that there is a Force vector:

F=3i(x^2 - y^2) + 2j(x+y)

Then we are given coordinates, where all values are in meters and the magnitude of force is in Newtons.

The question asks to calculate the work done in Joules by this force in a displacement along line y =0 from point 1 (0,0) to point 2 (2,0).




Now I followed the equation W = Fd.
So F= (3x^2 - 3y^2)i + (2x + 2y)j
d= 2i

So W = Fd = 6x^2 - 6y^2

But now I am not quite sure what to plug in for x or y, unless I can do W(final) - W(initial), with point 1 = initial and point 2 = final, which would be 24 Joules.

 

[HallsofIvy]

You are on the straight line from (0, 0) to (2, 0). y= 0 at every point on that line.

Now, W is NOT Fd because F is not a constant. Work is, instead, \int\vec{F}\cdot d\vec{x}. x is the variable of integration.

 

[nasu]

As the force is not constant but dependent on position you'll have to integrate along the given path.
W=∫F(x,y)ds where ds is the displacement along the given direction. (F and ds are both vectors, of course)
In the given problem the things are a simplified by the path being along the x-axis (y=0). In this case ds=i dx and the force depends only on x (put y=0 in the expression for force, before doing the integral).

 

[c42]

Thank you for responding.

So I would take F=3i(x^2 - y^2) + 2j(x+y) plug in y=0, which becomes

F = 3i(x^2) + 2j(x) or F = (3x^2)i + (2x)j

Then take F dot product ds, with ds = 2i dx.

So i would take the integral from 0 to 2 for 6x^2

Which would be 2x^3 from 0 to 2, with an answer of 16 J.

 

[nasu]

Why do you think that ds=2idx? I mean, why the factor of 2?