(a) Apply work-energy theorem
Wnc = ∆ME
Wfk = ∆K (Kf – Ki) Kf = 0– fkd = – ½mvi2 – fk(0.12) = – ½ (0.030) (500)2 fk = 31250N
(b) Find find acceleration
– fk = ma – (31250) = (0.03)a
a = – 1.04 x 106m/s2
Then use kinematics vf = vi + at
0 = 500 + (– 1.04 x 106)t
t = 4.8 x 10– 4sec