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 w=FdrcosΘ

w=rcosΘF

w=rcosΘF2/2

w=rcosΘ(F2/2(tf)F2/2(to))

 

to is t initial and tf is t final

 

 

 

So is this right?

 Nov 29, 2016
 #1
avatar+33657 
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No.  The integral is with respect to r not F.

 

.

 Nov 29, 2016
 #2
avatar+245 
0

Could you help go through the problem?

JonathanB  Nov 29, 2016
 #3
avatar+33657 
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You need to know if F (and/or theta) is a function of r.  If it is constant (independent of r) then the result is just w = Frcos(theta).

 

I assume to and tf are times, but you need to supply more information about the problem before we can see where they fit in.

.

Alan  Nov 29, 2016

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