A block of mass m is placed in equilibrium on a moving plank. The maximum horizontal acceleration of the plank for μ = 0.2 is:

If block does not slide w.r.t platformF = ma--------- (i)But f ≤ μN------- (ii)From (i) and (ii), ma ≤ μ mga ≤ μgHence maximum acceleration of the plank is0.2 x10 = 2 m/s2.******* a max=μsg=0.2×10=2 ms-2