A 40kg slab rests on a frictionless floor as shown in the figure. A 10kg block rests on the top of the slab. The static coefficient of friction between the block and slab is 0.60 , while the coefficient of kinetic friction is 0.40. The 10kg block is acted upon by a horizontal force 100N. If $g = 9.8 m s^{- 2}$ , the resulting acceleration of the slab will be

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$1.47 m / s^{2}$

$0.98 m / s^{2}$

$1.52 m / s^{2}$

$6.1 m / s^{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} Limiting friction between block and slab \\ = μ_{s} m_{A} g = 0.6 \times 10 \times 9.8 = 58.8 N \\ But applied force on block A is 100 N. \\ So the block will slip over the slab . \\ Now kinetic friction works between block and slab \\ F_{k} = μ_{k} m_{A} g = 0.4 \times 10 \times 9.8 = 39.2 N \\ This kinetic friction helps to move the slab \\ ∴ A c c l e r a t i o n o f s l a b = \frac{39.2}{m_{B}} = \frac{39.2}{40} = 0.98 m / s^{2} \end{array}$