Question

What should be the maximum value of $M$ (in $kg$ ) so that the $4kg$ block does not slip over the $5kg$ block? (Take $g=10m/s^2$ )

Answer 1

Answer: 6

Common acceleration of system is
$a=\frac{Mg}{M+9}$
For $4kg$ block we use $f=4a=\frac{4Mg}{M+9}$
$=16N$
$\Rightarrow \frac{4Mg}{M+9}=16$
$\Rightarrow 40M=16M+144$
$\Rightarrow M=6kg$