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 \(w = \int FdrcosΘ \)

\(w=rcosΘ \int F \)

\(w= rcosΘ\int F^2/2 \)

\(w= rcosΘ( F^2/2 ( tf) - F^2/2 (to))\)

 

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 
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Could you help go through the problem?

JonathanB  Nov 29, 2016
 #3
avatar+33657 
0

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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