Question 174 of 210

Two blocks $A$ and $B$ of masses ${m}_{A}=1kg$ and ${m}_{B}=3kg$ are kept on the table as shown in figure. The coefficients of friction between $A$ and $B$ is $0.2$ and between $B$ and the surface of the table is also $0.2.$ The maximum force $F$ that can be applied on $B$ horizontally, so that the block $A$ does not slide over the block $B$ is : [Take $g=10m/{s}^{2}$ ]

A$16N$B$12N$C$40N$D$8N$

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Answer: (A) $16N$

Maximum possible acceleration of $1kg$ block $a=\dfrac{µmg}{m}=µg=2m/{s}^{2}$

For $1g$ block not to slide the maximum acceleration of both blocks should be

${F}_{max}-µmg=m{a}_{max}$
${F}_{max}-0.2\times 4\times 10=4\times 2$
${F}_{max}-8=8$
${F}_{max}=16N$