Evaluate the work done by the two-dimensional force F = (2xy, -y^2] along the fo

Question

Evaluate the work done by the two-dimensional force F = (2xy, -y^2] along the following three paths, each going from the origin to point (1, 1). Path (1) first goes along the y-axis to (0, 1), and then parallel to the x-axis to (1, 1) Path (2) follows a parabola, y = x^2 Path (3) is given parametrically as x = tA3 and y = t^4 For this case, rewrite all of x, y, dx and dy in terms of t and dot and do the path integral as an integral over t. It might help you to sketch this path but It is not necessary.

Explanation / Answer

first path:

step 1: from (0,0) to (0,1)

displacement vector=(0,dy)

then work done=integration of dot product of force and displacement

=integration of -y^2*dy from y=0 to y=1

=(-y^3/3) from y=0 to y=1

=-(1/3)+(0)=-1/3

step 2:

from (0,1) to (1,1)

displacement vector=(dx,0)

work done=integration of 2*x*y*dx

here y=1

so work done=integration of 2*x*dx

=x^2 from x=0 to x=1

=1^2-0^2=1J

so net work done=(-1/3)+1=2/3 J

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