|May3-12, 08:30 PM||#1|
Finding work when displacement is a function of force
Hi, I've taken introductory calculus and am doing algebra based physics, but I'm trying to understand how to relate calculus to physics. This is more of a theoretical question than a practical one, so I might be rambling on about something that doesn't even work.
I know that W=∫F(d)Δd, but what would you do if you have displacement as a function of force instead? Could you say that W=∫d(F)ΔF ? Would this give you a correct answer?
Another question, back to the original equation W=∫F(d)Δd, if displacement is a function on time, will this make any difference when determining work? I think that it does not, but I am not sure. My reasoning is that if you had d(t) = 5t^2, force is still a function of displacement, so when integrating, the displacement should still increase at a constant rate regardless of time. Therefore it would not matter if displacement changes with respect to time since force is independent of time. Is this correct (if you can even understand what I'm asking )? I'm looking for a mathematical reasoning more than a "physics" reasoning. Anyways, thanks for your help!
|May4-12, 03:35 AM||#2|
Your second paragraph suggests that you may have a misconception. To calculate work you need to multiply the whole of the force (I'm uneasy about ΔF) by the distance it moves through. This assumes these are in the same direction; in general you take the dot product of the force and the displacement (though I know this is not the issue which worries you).
It's commonly the case that force and/or displacement is/are functions(s) of time. This doesn't affect the basic definition of work as dW = F.dr. Regard time, if you like, as a parameter in terms of which the force, F, on the body and/or, r, the body's displacement, can be expressed.
Simple example: suppose a body has velocity v, and is acted upon by a constant force F. In time dt the work done on it will be dtF.v.
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