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Block $A$ of mass $2 \mathrm{~kg}$ is placed over block $B$ of mass $8 \mathrm{~kg}$. The combination is placed over a rough horizontal surface. Coefficient of friction between $B$ and the floor is 0.5 . Coefficient of friction between $A$ and $B$ is 0.4 . A horizontal force of $10 \mathrm{~N}$ is applied on block $B$. The force of friction between $A$ and $B$ is $\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$
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The correct answer is:
zero
Here, $m_A=2 \mathrm{~kg}, m_B=8 \mathrm{~kg}, \mu_1=0.4, \mu_2=0.5$, $F=10 \mathrm{~N}$
The frictional force between block $B$ and surface is
$f=\mu_2 N=\mu_2\left(m_A+m_B\right) g=0.5 \times(2+8) \times 10=50 \mathrm{~N}$
As applied force $F(=10 \mathrm{~N}) < f(=50 \mathrm{~N})$, the system will not move.
Hence, the force of friction between $A$ and $B$ is zero.
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