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27. The coefficient of friction of a box on a level table is 0.054.  If a 10.0 kg box is moving to the right at 
30.0 m/s, how far does the box slide before it stops?
        
A. 45.9 m C. 456 m
B. 91.8 m D. 851 m
     
Ffriction = m a = coefficient friction x Fnormal  = coefficient friction x m g
    

          cf m g
a  =  ----------- =  coefficient friction x g = 0.054 x 9.80 m/s2 = 0.529 m/s2
            m

   
Forward is positive thus acceleration is negative

vf = 0   (box stops)
   

          vf 2  -  vi 2        02  -  (30.0 m/s)2
X =  ---------------- =  -------------------------- = 851 meters
               2a              2  x ( - 0.529 m/s2)