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Finding energy via integration

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clm222
#1
Feb17-13, 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?
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bossman27
#2
Feb18-13, 01:34 AM
P: 204
Quote Quote by clm222 View Post
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
It's a little deceiving to write [itex] E= Fd [/itex] instead of [itex] W=Fd [/itex], since you're describing the work done on the object. Work is equal to the change in kinetic energy, and since an object could have KE before work is done, it's not quite accurate to say [itex] E= Fd [/itex].

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?
No, what that equation would be saying is that a force of 20N was applied for a distance of 4m. 80J is the additional kinetic energy the object now possesses. Assuming a generic case, it would be possible to apply only 10N for 4m, or 40N for 4m. The difference is that after those 4 meters, the object would have less KE in the first case, and more KE in the latter case. There isn't any particular "force required to move an object a distance d," except perhaps a minimum force needed if we need to overcome friction or something.

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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