Using parameterisation to calculate work done by force

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SUMMARY

The forum discussion centers on calculating the work done by a force F = -K(yi + xj) on a particle moving in the x-y plane. The user attempted to parameterize the path from the origin to the point (a, a) but incorrectly assumed a diagonal path instead of following the specified square path. The correct approach involves calculating the work done along two segments: from (0, 0) to (a, 0) and then from (a, 0) to (a, a). The total work done must account for the path taken, confirming that the work done is path-dependent for this force.

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feetnappy
A force F = -K(yi + xj) (K is a positive constant) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). What is the total work done by the force F on the particle.

My attempt:

As the final position of the particle is (a, a) so I get the following using parameterisation for 0 <= t <= 1

x = at
y = at

Thus the work done,

W = integration -k(ydx + xdy)
= integration -k(atdt + atdt)
= integration -k(2atdt)
= -2ak integration (tdt)
= -ak [t]^2
= -ak

What am I doing wrong here?
 
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What are dx and dy?
 
feetnappy said:
A force F = -K(yi + xj) (K is a positive constant) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). What is the total work done by the force F on the particle.

My attempt:

As the final position of the particle is (a, a) so I get the following using parameterisation for 0 <= t <= 1

x = at
y = at

Thus the work done,

W = integration -k(ydx + xdy)
= integration -k(atdt + atdt)
= integration -k(2atdt)
= -2ak integration (tdt)
= -ak [t]^2
= -ak

What am I doing wrong here?

Why do you go along a straight line from (0,0) to (a,a)? The question said to go along the sides of the square. Of course, for some types of forces the work done is independent of the path, but is that the case here?

Anyway, even for the diagonal path, have you double-checked your work?
 
Last edited:

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