Vector Notation for Work: An Exploration of Force and Displacement

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SUMMARY

The discussion focuses on calculating work using vector notation, specifically the equation Work = F * Δx. The user successfully determines the displacement vector as 5i - 21j + 14k and computes the work done as 622 Joules. The confusion arises from the absence of a k component in the initial displacement, which is clarified by noting that the k component equals zero at the initial position. The calculation is confirmed as correct, demonstrating the application of vector multiplication in physics.

PREREQUISITES
  • Understanding of vector notation in physics
  • Familiarity with the concept of work and its calculation
  • Knowledge of vector components (i, j, k)
  • Basic algebra for vector operations
NEXT STEPS
  • Study vector operations in physics, focusing on vector addition and subtraction
  • Learn about the physical interpretation of force vectors in work calculations
  • Explore the implications of missing components in vector analysis
  • Review the principles of energy and work in a physics context
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators seeking to clarify concepts related to work and displacement in vector notation.

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Homework Statement


workVectors.png



Homework Equations


Work = F* delta x


The Attempt at a Solution



My thought process since the force is given in vector notations multiply the pair of i components together same applies with j and k components. However this process actually strangely does not work. And I am baffled by why a k component missing for the initial displacement.
 
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The k component happens to equal 0 at the initial position. So what?

Figure out the displacement: r2-r1
 
after substraction I get 5 i - 21 j + 14k.

5i(67i) + 23j(-21j) + 55k(14k) = 335 - 483 + 770 = 622 J <- correct! =]

thanks for the guidance, Doc Al.
 

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