A large box of mass M is moving on a horizontal floor at speed v0. A small box of mass m is sitting on top of the large box. The coefficient of static friction between the two boxes is μs and coefficient of kinetic friction between the large box and floor is μk. Find an expression for the shortest distance dmin in which the large box can stop without the small box slipping.

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)}

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