Re: work done by friction

[Robert Carlson <Raacc@AOL.COM>]

Leigh,

Neglect my last message.  I have thought of a way where rotation need not be
a factor.  But, I am wondering why you are limiting the work done by the
frictional force to just the change in kinetic energy.  You write:


                   1           2      2
              WA = - MA * ( VAf  - VAi  )
                   2

and:


                   1           2      2
              WB = - MB * ( VBf  - VBi  )
                   2

Where I would have written:


                   1           2      2
              WA = - MA * ( VAf  - VAi  ) + CA*(TA - Ti)
                   2
and:


                   1           2      2
              WB = - MB * ( VBf  - VBi  ) + CB*(TB - Ti)
                   2

Bob Carlson