Calculate Work of Vector Force in XY Plane

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SUMMARY

The discussion focuses on calculating the work done by a vector force in the XY plane, specifically using the formula W = F·Dcos(θ). Participants detail the process of determining the displacement vector, with components sx = 1.57 m and sy = 3.21 m, and the force vector with components Fx = 5.81 N and Fy = 3.93 N. The correct approach involves using the dot product of the force and displacement vectors, which was initially overlooked by one participant. Ultimately, the importance of unit vector dot products in calculating work is emphasized.

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1. A particle moving in the xy plane undergoesa displacement ~s = (sxˆı + sy ˆ|), with sx =
1.57 m, sy = 3.21 m, while a constant force F=(Fxˆı+Fyˆ|),withFx=5.81N,Fy=3.93 N, acts on the particle.
Calculate the the work done by vector F.
.




2.W=FDcos(theta)
inverse tangent



3. first found the magnitude of the resultant vector for the displacement via the theorum of pathagoreas. Using the same method i found the magnitude of the force resultant. Used inverse tangent times y over x for the angle. with those values i pluged into W=F*Dcos(theta). answer was incorrect. Okay, thought about it some more, the only work done through the applied force is on the X axis using just the x vectors calculated the work, that is also incorrect.
 
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never mind i had a brain fart forgot unit vector dot products...lmao
 

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