Calculating Work Done Using Force Vector & Distance

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SUMMARY

The discussion centers on calculating work done by a constant force vector F = 4i + 2j + 2k while moving an object from point (1, -5, 1) to point (1, -1, 0). The correct method to calculate work involves using the dot product of the force vector and the displacement vector, which is derived from the difference between the initial and final position vectors. The work done is expressed as W = F · (r2 - r1), where r1 and r2 are the initial and final position vectors, respectively.

PREREQUISITES
  • Understanding of vector notation and operations
  • Familiarity with the concept of work in physics
  • Knowledge of the dot product of vectors
  • Basic understanding of calculus, specifically integrals
NEXT STEPS
  • Study the properties and applications of vector dot products
  • Learn about the integral definition of work in physics
  • Explore examples of work done by varying force vectors
  • Review the distance formula and its applications in three-dimensional space
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Students in physics or engineering courses, educators teaching mechanics, and anyone interested in understanding the application of vector mathematics in calculating work done by forces.

somebodyelse5
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Homework Statement



A constant force F=4i+2j+2k moves an object along a straight line from the point (1, -5, 1) to the point (1, -1, 0).

Find the work done if the force is measured in Newtons and the position is measured in meters.

Homework Equations



W=Fd

The Attempt at a Solution



I used the distance formula to find the distance between the two points which is sqrt(17)

But what is the correct way to multiply the Force vector by this distance?
 
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1. Add up the x y and z components of work or 2. use the dot product of F and D, where D is the displacement.
 
somebodyelse5 said:

Homework Equations



W=Fd

You should get in the habit of using the more general definition of work from [itex]\textbf{r}_1[/itex] to [itex]\textbf{r}_2[/itex]:

[tex]W=\int_{\textbf{r}_1}^{\textbf{r}_2} \textbf{F}\cdot d\textbf{r}[/tex]

In your case, [itex]\textbf{F}[/itex] is uniform so you can pull it out of the integral

[tex]W=\textbf{F}\cdot\int_{\textbf{r}_1}^{\textbf{r}_2}d\textbf{r}=\textbf{F}\cdot(\textbf{r}_2-\textbf{r}_1)[/tex]
 

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