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A 96.3-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.659. (a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.80 s, what is his initial speed?
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μ=coefficient of kinetic friction
N=normal force
F=frictional force
a, F=μxN=.659x96.3x9.8=622 newtons
b, the frictional force is the only force that acted on the player while he was sliding, so
F=ma
-622=96.3xa
a=-6.46m/s^2
Vi=?
Vf=0
t=1.8 sec
a=-6.46m/s^2
Vf=Vi+at
0=Vi+(-6.46)x1.8
Vi=11.6m/s