Calculating Work Done by Force: Vectors & Joules

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The work done by a force in moving an object is calculated using the dot product of the force vector and the displacement vector. For a force of 4i + 2j - 3k Newtons and a displacement of <0.25, 0.5, 1> metres, the calculation yields a total work of -1 Joule. This negative value indicates that the work done is against the direction of the force, which is typical in cases like friction. The discussion confirms that the calculation is correct and emphasizes the nature of negative work in physics. Understanding these concepts is essential for solving similar problems in mechanics.
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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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