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Question

A ball is rolling on ice with velocity of 4.9m/s, stop after travelling 4m.What is the coefficient of friction

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Solution


Ice surface provides retardation
u=4.9m/s. s=4m. v=0.
then by third equation of motion
v2=u2+2as
after solving we get
a=-3.00125m/s2......(Negative sign only means it is retardation)
then u*=f/R=ma/mg=0.30625m/s (g=9.8m/s2)

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