A $2 \mathrm{~kg}$ block is placed over a $4 \mathrm{~kg}$ block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is $0.20$. Find the acceleration of the two blocks if a horizontal force of $12 \mathrm{~N}$ is applied to (a) the upper block, (b) the lower block. Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$.
(a)
Normal contact for $2 \mathrm{Kg}=2 \mathrm{~g}=20 \mathrm{~N}$.
Maximum friction force between $2 \mathrm{Kg}$ and $4 \mathrm{Kg}=\mu N$
$=02(20)$
$=4 \mathrm{~N}$
Since $F=12 N>$ ff bond will break
for $2 \mathrm{Kg}$ block,
12-4=2a
$8=2 a$
$a=4 \mathrm{~m} / \mathrm{s}^{2}$
for $4 \mathrm{Kg}$ block,
$4-0=4^{a}$
$a^{\prime}=1 \mathrm{~m} / \mathrm{s}^{2}$
(D)
Individual acc.
for $4 \mathrm{Kg}$ block,
$12-4=4 a$
$a=2 \mathrm{~m} / \mathrm{s}^{2}$
for $2 \mathrm{Kg}$ block,
$4-0=2^{a}$
$a^{\prime \prime}=2 m / s^{2}$
So, acceleration of both blocks are $2 \mathrm{~m} / \mathrm{s}^{2}$.