Asked by anonymous
                A .45 kg shuffleboard puck is given an initial velocity down the playing surface of 4.5 m/2. If the coefficient of friction between the puck and the surface is .2, how far will the puck slide before coming to rest?
            
            
        Answers
                    Answered by
            bobpursley
            
    vf^2=vi^2+2ad  where a= f/m= -.2mg/m=-.2g
d= vi^2/2a= 4.5^2/.4g
    
d= vi^2/2a= 4.5^2/.4g
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