Calculating work done using vectors

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SUMMARY

The discussion focuses on calculating work done by a constant force vector F = (2.84i + 4.60j) N on a 4.12 kg particle moving to the position vector r = (1.52i - 2.55j) m. The correct calculation of work is given by the formula W = F · d, resulting in W = -7.41 J. The conversation clarifies that the Pythagorean theorem cannot be used for this calculation because work is determined by the dot product of force and displacement vectors, not by their magnitudes alone.

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  • Understanding of vector notation and operations
  • Familiarity with the concept of work in physics
  • Knowledge of dot products in vector mathematics
  • Basic principles of force and motion
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  • Learn about the relationship between force, displacement, and work
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts of work and force in vector analysis.

ClassicRock
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Homework Statement


A single constant force F = (2.84i + 4.60j) N acts on a 4.12 kg particle. Calculate the work done by this force if the particle moves from the origin to the point having the vector position r = (1.52i - 2.55j) m.



Homework Equations


W=FD


The Attempt at a Solution



W=(2.84*1.52)+(4.60*-2.55)
W=-7.41 J

This is the correct answer. I want to know why I can't use the pythagorean identity in order to find Force and Distance. Why can't I do that instead of the what I have shown here (could someone help me out with the name?).

Thanks,
ClassicRock
 
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Your equation should be ##W=\vec{F}\cdot\vec{d}## ... which is a definition.
Notice that it is a vector dot-product, not a multiplication.
The force perpendicular to the displacement does not contribute to the work.

Consider - if you push a trunk along the floor by pressing at an angle downwards, some of your force goes into the ground doesn't it? Not all you effort moves anything.

But you may find it more intuitive to think another way - the component of the path perpendicular to the force requires no work.

It is fairly easy to walk around the contour of a hill - keeping the same height all the time.
It is only when you move up or down the hill that you have to work at it.
So only the component of your path that goes up or down the hill contributes to the overall work.
 
Last edited:

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