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Finding the coefficient of friction of baseball player?
A basebally player running with a speed of 9.00m/s slides into second base through a distance of 4.00m, find the coefficient of friction. His speed is zero when he touches the base.
I have no idea what to do here. The formula we use is Fk = (uk)(N), but I see no other equations that could help with this.
V^2 = Vo^2 + 2a (X - Xo)
0 = 9.00^2 + 2a (4.00 - 0)
-20.3 = 2a
a = -10.1 m/s^2
The acceleration is all I know so far. Help please.
2 Answers
- FiremanLv 71 decade agoFavorite Answer
By work energy theorem:-
=>W = ∆KE
=>Ff x s = 1/2mv^2
=>µ x mg x s = 1/2mv^2
=>µ = v^2/2gs
=>µ = (9)^2/[2 x 9.8 x 4]
=>µ = 1.03
- Johnny DLv 71 decade ago
Draw force diagram and you'll see that friction causes the acceleration since it's the only force in the x direction while he slides.
Fnet=ma=-ukN=-(uk)(mg)
a=-(uk)g
uk=a/-g=1.03
