Power input of a Force acting on a particle moving with Velocity

Click For Summary

Homework Help Overview

The discussion revolves around calculating the power input of a force acting on a particle moving with a given velocity. The problem involves vector components of force and velocity in different scenarios, requiring the application of the dot product to find power.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate power using both magnitudes and angles, expressing confusion about the dot product method. They question the correctness of their assumptions regarding the direction of vectors affecting the sign of the power.
  • Some participants suggest a more straightforward approach by directly using the components of the vectors for the dot product calculation.
  • Others raise concerns about the potential pitfalls of using arctan for angle calculations, particularly in determining the correct quadrant for the angles involved.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the calculations. There is an acknowledgment of the original poster's confusion and a suggestion to clarify the method for the third scenario. Multiple interpretations of the problem are being explored, particularly regarding the use of angles and the implications of vector directions.

Contextual Notes

The original poster expresses difficulty with the concept of dot products and the implications of vector direction on the results. There is an indication that the homework may impose specific methods or approaches that are being questioned.

_h2tm
Messages
10
Reaction score
0

Homework Statement



Find the power input of a force \vec{F} acting on a particle that moves with a velocity \vec{V} for each of the following situations.

  • \vec{F} = 4\hat{i} N  + 3\hat{j} N , \vec{V} = 7\hat{i} m/s 
  • \vec{F} = 7\hat{i} N  - 5\hat{j} N , \vec{V} = -5\hat{i} m/s + 4\hat{j} m/s 
  • \vec{F} = 2\hat{i} N + 10\hat{j} N , \vec{V} = 2\hat{i} m/s  + 3\hat{j} m/s 

Homework Equations



P = \frac{dW}{dt} = \vec{F} • \vec{v}

The Attempt at a Solution



  • \sqrt{4^2 + 3^2} \times cos(arctan(3/4)) \times 7
    For this problem, I got the correct answer: 28 W.
  • \sqrt{7^2 + (-5)^2} \times \sqrt{(-5)^2 + 4^2}
    I also got the correct answer for this problem: 55 W, except that the answer is negative, I figure because the velocity vector's \hat{i} and \hat{j} components are in the opposite direction of the force's components (individually). Is this supposition correct?
  • \sqrt{2^2 + 10^2} \times sin(arctan(10/2)) \times \sqrt{2^2 + 3^2} \times sin(arctan(3/2))
    I cannot figure out what I am doing wrong on this one. (The correct answer is 34.)

I am having a hard time wrapping by head around dot products. I'm guess that is my problem here. Any and all help is greatly appreciated.
 
Physics news on Phys.org
When you do dot products, you take the sum of the products of each component.
For the first one:
P = 4N*7m/s (the i components) + 3N*0m/s (the j components) = 28 W.
For the second one:
P=7N*-5m/s (i components) + -5N*4m/s (j components) = -55W.
Try the third one out this way.
P= (product of i components) + (product of j components) = ?
 
What you are doing is finding it using magnitudes and angles, which you must first calculate. When calculating angles, you also have to think about the fact that it may be more than 90 degrees, something that arctan will not provide.
 
OK, thanks. I completely get how to do this in the future.

Regarding your second comment though -- I understand about arctan. I drew a diagram so I knew where my vectors were. Was this comment just FYI or was my attempted method something that would work, I was just simply doing it incorrectly?
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
Replies
12
Views
2K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
2
Views
884
Replies
2
Views
1K
Replies
3
Views
1K