A 55.2 kg softball player moving 3.11 m/s slides across dirt with Hk =0.310. How far does she slide before coming to a stop?

Answer 1

Find the length of the slide (d)

m = 55.2 kg

I assume Hk is supposed to be coefficient of kinetic friction = µk = 0.31

vi = 3.11 m/s vf = 0

Weight (Fg) = mg = (55.2)(9.8) = 540.96 N

Because we assume this is on level ground the normal force (FN) is equal to the weight = 540.96 N

We calculate the friction force (Ff) = µkFN = (0.31)(540.96) = -167.7 N (it is negative because it is the opposite direction of the motion)

a = Ff/m = -167.7/55.2 = -3.04 m/s²

v²f = v²i + 2ad

0 = (3.11)² + 2(-3.04)d

d = 1.59 m