Simple Work Done by Force Problem

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SUMMARY

The discussion focuses on calculating the work done by the force vector \(\vec F = (-4.0\hat\imath-6.0\hat\jmath)\;{\rm N}\) on a particle moving through the displacement vector \(\Delta\vec{r}=(-3.0\hat\imath+2.0\hat\jmath)\;{\rm m}\). The correct method involves using the dot product formula \(A_{x}B_{x} + A_{y}B_{y}\) to find the work, which results in a scalar value. The participant initially miscalculated by creating an unnecessary vector C and incorrectly applying the square root, leading to an erroneous answer of 17. The correct calculation yields a work done of zero, confirming that the dot product is indeed a scalar quantity.

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  • Understanding of vector notation and components
  • Familiarity with the dot product operation
  • Basic knowledge of work-energy principles in physics
  • Ability to perform vector arithmetic
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  • Study the properties of dot products in vector mathematics
  • Learn about work done by forces in physics
  • Explore scalar versus vector quantities in physics
  • Practice problems involving force and displacement vectors
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Students in physics, particularly those studying mechanics, as well as educators looking for examples of vector operations and work calculations.

tangibleLime
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Homework Statement


How much work is done by the force [tex]\vec F = (-4.0\hat\imath-6.0\hat\jmath)\;{\rm N}[/tex] on a particle that moves through displacement [tex]\Delta\vec{r}=(-3.0\hat\imath+2.0\hat\jmath)\;{\rm m}[/tex]?

Homework Equations


[tex]A_{x}B_{x} + A_{y}B_{y}[/tex]

The Attempt at a Solution


I decided to use the component form of the dot product method for this problem. (Or is the component form an alternative to the dot product?) Anyways, I set F as A and R as B, resulting in the following:

[tex]A_{x}B_{x} + A_{y}B_{y}[/tex]
[tex]F_{x}r_{x} + F_{y}r_{y}[/tex]

[tex]\vec F = <-4, -6>[/tex]
[tex]\vec r = <-3, 2>[/tex]

[tex]-4(-3) + (-6)(2)[/tex]
[tex]\vec C = <12,-12>[/tex]

Then I took the square root of the sum of the sqaures:

[tex]\sqrt{12^{2}+(-12)^{2}}[/tex]
[tex]\sqrt{288} \approx 17[/tex]

My answer of 17 was marked incorrect. Where did I go wrong? Any input is greatly appreciated.

I just realized that I must have made some major mistake. I just kind of "created" vector C out of nowhere. Any direction on this problem would be great.

EDIT: Nevermind, I figured out my error. Instead of creating vector C, I should have just stopped there, added 12 and -12, and got the correct answer of zero. Huzzah!
 
Physics news on Phys.org
The dot product is not a vector! It is scalar.

ehild
 

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