Work of a Particle in the XY Plane

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SUMMARY

The discussion centers on calculating the work done by a force F = (2yi + 1.2x² j)N on a particle moving in the XY plane from the origin to the coordinates (5m, 5m). The work is computed using the integral W = ∫(Fxdx + Fydy), which is broken into two segments: Woa and Wac. A key point of confusion arises regarding the work done along the path OA, where participants clarify that the work is zero due to the force's direction and the particle's movement along the y-axis, resulting in no displacement in the x-direction.

PREREQUISITES
  • Understanding of vector forces in physics
  • Knowledge of line integrals in calculus
  • Familiarity with the concepts of work and energy
  • Basic understanding of particle motion in two dimensions
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  • Study the principles of line integrals in vector calculus
  • Learn about conservative forces and their implications on work done
  • Explore the concept of path independence in work calculations
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Students studying physics, particularly those focusing on mechanics and vector calculus, as well as educators seeking to clarify concepts related to work and force in two-dimensional motion.

Redfire66
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Homework Statement


A force acting on a particle moving in the xy-plane is F = (2yi + 1.2x^2 j)N. where x and y are in meters. The particle moves from the origin to a final position having coordinates x = 5m and y = 5m. Calculate the work done by F along OAC

Homework Equations


Integral of force by distance
W = ∫(Fxdx + Fydy)

The Attempt at a Solution


I have the answer. And I know what they did. But what I don't understand is how it makes sense.
The solution involves breaking up the integral into two partitions, Woa and Wac.
This is what I don't really understand - how is the work on OA zero? There is a force acting in the x direction and it moves from O to A. So by definition, since there is a force acting over a distance there should be work right? (Unless I'm not reading this properly)
 
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Where are points A and C?
 

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