posted by anonymous on .
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?
vf^2=vi^2+2ad where a= f/m= -.2mg/m=-.2g
d= vi^2/2a= 4.5^2/.4g