finding energy via integrationby clm222 Tags: calcuations, energy, theoretical, work, work & energy 

#1
Feb1713, 10:09 PM

P: 35

Hello, I havnt done physics in quite a while and I just want to ask a question about basic energy that i know how to deal with in algebraic terms, but not through means of calculus. I also don't really get the theory of the equation [itex]E=FD[/itex], where E=energy, F=force, and D=distance
is the F the force it took to move the body distance D? So, for example, applying a force of 20N on a body moves it 4m, [itex]E=FD=(20N)(4m)=80J[/itex]? if thats the case then how can we express this with derivatives or integrals? 



#2
Feb1813, 01:34 AM

P: 204

To give you a more general definition: In mechanics, the work done by a force F on an object that travels along a curve C is given by the line integral: [itex]W_{C} = \int_{C} F dx = \int_{C} F\cdot v dt [/itex] Where [itex] F\cdot v [/itex] (the dot product) is the instantaneous power [itex] P(t) [/itex], which is essentially the rate at which energy is being transferred to the object, in this case. And actually, it's usually defined the other way around: power is defined as the time derivative of work. In the case where the force is directed along the path of motion, this simplifies to [itex]W=Fd [/itex] Look up the definition of the dot product, if you're not already familiar, and note that what we've shown here is that work is the time integral of the component of the force in the direction of motion times the magnitude of the velocity. Hence if an object is moving straight along the positive x axis, and we're pushing it towards the positive y direction (without changing it's path), no work is being done. It's the force in the direction of motion that counts. So in the case where we have a constant force not directed along the path of motion, we can write: [itex] W=Fdcos \theta [/itex] 


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