A block of mass M1 is placed on a slab of mass M2. The slab lies on a frictionless horizontal surface. The coefficient of static friction between the block and slab is μ1 and that of dynamic friction is μ2. A force F acts on the block M1. Take g = 10 ms⁻² . If M1 = 10 kg, M2 = 30 kg, 1 = 0.5, μ2 = 0.15 and F = 40 N, what will be the acceleration with which the slab will move

The correct answer is Zero.

Here is the step-by-step physical analysis and derivation:

Step 1: Identify the given parameters
Mass of the block, $M_1=10\text{kg}$
Mass of the slab, $M_2=30\text{kg}$
Coefficient of static friction between block and slab, ${\mu }_1=0.5$
Coefficient of dynamic/kinetic friction between block and slab, ${\mu }_2=0.15$
Applied force on block $M_1$, $F=40\text{N}$
Acceleration due to gravity, $g=10{\text{ms}}^{-2}$

Step 2: Calculate the limiting (maximum) static friction force between the block and the slab
The normal reaction force ($N_1$) acting on the block $M_1$ is:

$N_1=M_1\cdot g=10\text{kg}\times 10{\text{ms}}^{-2}=100\text{N}$

The maximum static friction force ($f_{s,\text{max}}$) that can act between the block and the slab is:

$f_{s,\text{max}}={\mu }_1\cdot N_1=0.5\times 100\text{N}=50\text{N}$

Step 3: Analyze the effect of the applied force
The external horizontal force applied to the block $M_1$ is $F=40\text{N}$.

Comparing the applied force with the limiting static friction:

$F=40\text{N}<f_{s,\text{max}}=50\text{N}$

Since the applied force $F$ is less than the limiting static friction force between the block and the slab, the applied force is insufficient to overcome the static friction barrier required to initiate relative motion or slip. Therefore, under this threshold condition, no motion is transferred to cause acceleration of the slab, and the acceleration of the slab remains Zero.