A 55 kg baseball player slides into third base with an intial speed of 4.6 m/s.if the coefficient of kinetic friction between th e player and the ground is 0.46, what is the player's acceleration?
1 answer:
In the x-direction, the frictional force is the lone force pro
tem on the player
The formula is: Fnet = ma
f = ma
In the y-direction, N, the normal force turns up and mg, the force of gravity turns
down
N = mg (since there is no speeding up in the vertical direction)
f = -uN
=> f = -umg
-umg = ma
=> a = -ug
Using the kinematics equation:
d = (vf^2 - vi^2) / 2a
d = vi^2 / 2ug d = 4.6^2 / 2(4.508)
d = 2.35 m/s
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