Asked by Emma
A baseball player stealing second base runs at 7.2 m/s . If he slides the last 3.8 m , slowing to a stop at the base, what's the coefficient of kinetic friction between player and ground?
Answers
Answered by
soma sekharam
initial speed u=7.2m/s
final speed v=0
distance traveled s=3.8m
coefficient of kinetic friction muK =?
We know that v^2-U^2=2as
0^2-U^2=2*(-muk^g)s
muk=(U^2)/(2*g*s)
= (7.2*7.2)/(2*9.8*3.8)
=0.696(no units)
= 0.70 (n0 units and two significant figures)
final speed v=0
distance traveled s=3.8m
coefficient of kinetic friction muK =?
We know that v^2-U^2=2as
0^2-U^2=2*(-muk^g)s
muk=(U^2)/(2*g*s)
= (7.2*7.2)/(2*9.8*3.8)
=0.696(no units)
= 0.70 (n0 units and two significant figures)
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