CATAGORY: Mechanics
QUESTION: A 10 Kg crate is pulled up a rough incline with an initial speed
of 1.5 m/s.  The pulling force is 100 N parallel to the incline, which
makes an angle of 20 degrees with the horizontal.  If the coefficient of
kinetic friction is .4 and the crate is pulled a distance of 5 meters. a)
what is the change in kinetic energy of the crate? b) what is the speed of
the crate after it is pulled 5 meters?

The text says the answer for a) is 150 Joules and for b)5.6 m/s.  The
wierd
thing is that when I calculated these questions excluding friction those
are the answers I got.  When I formulated an equation including friction I
got smaller numbers. Please help, I'm taking this independant study and
have to figure everything out alone.Thanks.

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The figure below shows a SIDEWAYS  view of the incline

            100N
              \
               \
           rope \ \         
                 \ \        
                  \ \       
blocks final       O \----> gravity force = mg
  position            \b      
                       \    
                        \   
                         \  
                 10kg     \a=20 degrees
                 block-->O \
                            |
                     initial position
                            |
                            |
                         
The angle b = 90 degrees - 20 degrees = 70 degrees.  The force of
gravity pulling the block down the incline is mgcos(b) = mgcos(70
degrees) = mg sin(20) and is directed down the incline.  The force of
friction is also directed down the incline and is equal to the normal
force which here is mg cos(20) times the coefficient of friction.  The
net force on the block in the direction of the incline is

 (1) F(net) = 100N - mg sin(20) - Uk mg cos(20)
 
            = 100N -10kg 9.8m/s^2 sin(20) - .4 10kg 9.8m/s^2 cos(20)
            
            = 100N - 33.52N - 36.84N 
            
            = 29.91N
             
The work done on moving the block 5 meters up the
incline is then

 (2) Work = F(net) x 5.00m = 149.55J                  
                                

thus we have

(3) KE(initial) + PE(initial) - Non conservative work 

                          = KE(final) + PE(final)

But 

(4)   PE(final) - PE(initial) + Non conservative work 

          = Work = 149.55J ~ 150 J
          
so          
     
(5)  KE(final)-KE(initial) =

     PE(final) - PE(initial) + Non conservative work = 150 J
     
 the final speed is given by solving (5) for KE(final) and then
 for v(final)
 
 (6) KE(final) = 150J -KE(initial)
 
               = 150J - 1/2 10kg(1.5m/s)^2
               
               = 138.75 J
               
thus since KE(final) = 1/2 m (v(final))^2 with m =10Kg 

(7)  1/2 10Kg (v(final))^2 = 138.75 J 

we have 

(8) v(final) = 5.26 m/s  

       ********************************
       *                              *
       * KE(final)-KE(initial) = 150J *
       *                              *
       *  v(final) = 5.26m/s          *
       *                              *
       ********************************