Calculating the Angle Between a Force and Displacement

Click For Summary

Homework Help Overview

The problem involves calculating the angle between a force and a displacement vector in a physics context. The force is given as 13 N, and the displacement vector is specified as (3i - 5j + 3k) m, with the change in kinetic energy presented in two scenarios: +25.6 J and -25.6 J.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to apply the dot product to find the angle between the force and displacement. They express uncertainty about the relationship between the force, displacement, and the resulting work done.
  • Some participants clarify the concept of the dot product, mentioning its relation to the magnitudes of the vectors and the cosine of the angle between them.
  • Others suggest alternative expressions for the dot product that may simplify the calculations.

Discussion Status

The discussion is ongoing, with participants providing clarifications on the dot product and its application to the problem. There is no explicit consensus yet, but helpful insights have been shared that may guide the original poster's understanding.

Contextual Notes

The original poster appears to be grappling with the interpretation of the problem and the application of relevant equations, indicating a need for further exploration of the concepts involved.

TomFoolery
Messages
14
Reaction score
0

Homework Statement



A 13 N force with a fixed orientation does work on a particle as the particle moves through displacement = (3i - 5j + 3k) m. What is the angle between the force and the displacement if the change in the particle's kinetic energy is (a) +25.6 J and (b) -25.6 J?


Homework Equations



Angle = √(x^2 + y^2 + z^2)
Work = mad or F (dot) d

The Attempt at a Solution



I'm not sure how to approach this. I know that the dot product really just means to multiply the x parts by x parts, y parts by y parts, and z parts by z parts (or \hat{i}, \hat{j}, and \hat{k} parts, if you prefer).

If the particle starts at (0, 0, 0), then the force would have caused it to go to its current position (3, -5, 3). I'm still not sure that is what is being said though.
 
Physics news on Phys.org
There is another way to express the dot product, \vec{F}\cdot\vec{d}\,.

It involves the magnitude of each of the vectors as well as the cosine of the angle between their directions.
 
Yes, work is equal to Fx*dx+Fy*dy+Fz*dz where where Fx,Fy,Fz are the force components and dx,dy,dz are the displacement components...but there is another common way to write the the dot product of two vectors that should make quick work of this problem.
 
Got it, thanks for your help.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
Replies
2
Views
1K
Replies
11
Views
2K
Replies
1
Views
1K
Replies
12
Views
2K
Replies
2
Views
2K
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
1K
Replies
2
Views
2K