Question

A $40\text{kg}$ slab rests on a frictionless floor. A $10\text{kg}$ block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient of friction is 0.40. The $10\text{kg}$ block is acted upon by a horizontal force of $100\text{N}$ . The resulting acceleration of the slab will be

• A. $1.5{\text{ms}}^{-2}$
• B. $2.0{\text{ms}}^{-2}$
• C. $10{\text{ms}}^{-2}$
• D. $1.0{\text{ms}}^{-2}$

Answer 1

The correct option is D $1.0{\text{ms}}^{-2}$

Considering block of mass $10\text{kg}$
$N=mg=10\times 10=100\text{N}$
So to get static friction
$f_s={\mu }_{s}N=0.6\times 100=60\text{N}$
and friction on this block will be in opposite direction of the force applied.
So net force on $10\text{kg}$ block apart from friction is
$F_{net}=100\text{N}$
So if net force is greater than static friction, then it is a case of kinetic friction.
So,
$f_k={\mu }_{k}N\Rightarrow f_k=0.4\times 100=40\text{N}$
Now taking $40\text{kg}$ block into consideration
As $f_k$ was acting in opposite direction of force for $10\text{kg}$ block, it will act in reverse direction for $40\text{kg}$ block.
So only force acting on $40\text{kg}$ block will be $f_k$
$\Rightarrow ma=f_k\Rightarrow 40a=40\Rightarrow a=1{\text{ms}}^{-2}$