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The coefficient of friction between $4kg$ and $5\, kg$ blocks is $0.2$ and between $5kg$ block and ground is $0.1$ respectively. Choose the correct statements
Minimum force needed to cause system to move is $17N$
When force is $4N$ static friction at all surfaces is $4N$ to keep system at rest
Maximum acceleration of $4kg$ block is $2m/s^2$
Slipping between $4kg$ and $5\, kg$ blocks start when $F$ is $17N$
Solution
So block $Q$ is moving due to force and block $P$ due to friction.
Friction direction on both blocks is shown in the figure.
The first block $"Q"$ will move and $"P"$ will move with $Q$ so by taking p and
$\mathrm{q}$ as a system.
$F-9=0 \Rightarrow F=9 N$
When $9 \mathrm{N}$ force is applied then acceleration $\mathrm{ap}=\mathrm{aq}=0$
$4Kg$ block is koving due to friction and maximanum friction force is $8 \mathrm{N}$.
So acceleration $=8 / 4=2 \mathrm{m} / \mathrm{s}^ 2$
Block Q has positive acceleeration equal to maximum acceleration of block P i.e. $2 \mathrm{m} / \mathrm{s}^ 2$
