Work done by force on moving particle

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To calculate the work done by the force F=-(5yx^2)i + (4y^3)j as a particle moves from point (-2,4) to point (5,10), one must evaluate a path integral. The work is defined as W = ∫ F · dr, which involves integrating the force components separately for each coordinate. The integral for F_x only requires consideration of the x variable, while F_y involves the y variable. Properly setting the limits of integration for each coordinate is essential for accurate calculation. Understanding these integrals is crucial for determining the total work done on the particle.
PinkDaisy
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My problem is:

A particle is acted on by a force, F=-(5yx^2)i + (4y^3)j
Calculate the work done by F as the particle moves from point (-2,4) to point (5,10)? F is in Newtons and all x's and y's are in meters.

I think that I need to integrate each piece using the points as limits, but I'm not sure what I do with -(5yx^2) Do I only need to integrate with respect to x since it is with the "i" portion of the F?
Thanks!
 
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The integral you evaluate when calculating work is called a "Path Integral". From the definition of work,

W = \int_a^b \vec F \cdot d\vec r = \int_a^b F_x dx + F_y dy + F_z dz

Thus you have three integrals: one over each coordinate.

So you are right...the integral over F_x only affects the x variable.
 
Thanks so much!
 

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