A 96.3-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.619. (a) What is the magnitude of the frictional force? (b) If the player comes to rest after 1.00 s, what is his initial speed?

Answer 1

M*g = 96.3 * 9.8=943.7 N.=Wt. of player.
= Normal force(Fn).

a. Fk = u*Fn = 0.619 * 943.7 = 584.2 N.

b. a = u*g = 0.619 * (-9.8)=-6.07 m/s^2.

V = Vo + a*t = 0.
Vo = -a*t = -(-6.07)*1.0 = 6.07 m/s.