Calculating Work Done by Force: Vectors & Joules

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Homework Help Overview

The discussion revolves around calculating the work done by a force using vectors, specifically focusing on the dot product of force and displacement vectors. The original poster presents a specific example involving a force vector and a displacement vector, seeking validation of their calculation.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the work done by applying the dot product to the given vectors and questions the correctness of their result. Other participants confirm the method used and the result, while also providing context about negative work in certain scenarios.

Discussion Status

The discussion includes confirmations of the original poster's calculation, with some participants providing additional context about the nature of work in physics. There appears to be an understanding of the method used, though the original poster expresses a desire for guidance rather than direct answers.

Contextual Notes

The original poster requests not to be told why their answer may be incorrect, indicating a preference for exploratory discussion rather than direct solutions. This reflects a learning-focused approach within the homework context.

Jbreezy
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The work done (in Joules) by a force in moving a body is calculated by the dot product of the free vector representing the force (in Newtons) by the free vector representing the displacement of the object (in metres).
a. If the magnitude of the force is 4i + 2j -3k Newtons and the displacement vector is <.25,.5,1> metres what is the total work done by the force?

Homework Statement



Solution

I just dotted them.

<4,2,-3> dot <.<.25,.5,1> = 1+ 1 - 3 = -1
Is this correct? The work is - 1?

If this is not correct do not tell me why it isn't rather propose a thought. I don't want the answer given to me. Thanks very much :)
 
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This is correct. It is really just a dot product.
 
Hi Jbreezy! :smile:
Jbreezy said:
Is this correct? The work is - 1?

Yes, that's correct, and perfectly normal …

for example, work done by friction is always negative. :wink:
 
Thanks very much. People:)
 

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