Answer:
dmin = V0^2 / {2(9.8)(μs)}
Explanation:
According to free body diagram, the only forces acting on the small box are force of gravity and force of friction.
Therefore, the vector sum of the forces in the horizontal direction will be,
ma = F of friction which would be ma = mgμs
a = gμs ……. (1)
Kinematics equation states that, Vf^2 = Vi^2+ 2a (delta d)
Since the large box stops moving, which means Vf = 0 and initial velocity Vi=V0, then kinematics equation becomes
0 = V0^2 + 2a (delta d)
(delta d or dmin) = V0^2/2a …… (2)
Substitute the value of a from eqn.1 in eqn. 2, we should get
dmin = V0^2 / {2(9.8)(μs)}