Question 151 of 210

A block of mass $40kg$ slides over a surface, when a mass of $4kg$ is suspended through an inextensible massless string passing over frictionless pulley as shown below.
The coefficient of kinetic friction between the surface and block is $0.02$. The acceleration of block is: (Given $g=10{ms}^{-2}$.)

A$\dfrac{8}{11}m{s}^{-2}$B$1m{s}^{-2}$C$\dfrac{1}{5}m{s}^{-2}$D$\dfrac{4}{5}m{s}^{-2}$

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Answer: (A) $\dfrac{8}{11}m{s}^{-2}$

For hanging block,
$40-T=4a\cdots \left(1\right)$
Now, maximum value of friction force will be,
${f}_{max}=\mu N=0.02\times 40\times 10=8N$
For block on table, $T-8=40a\cdots \left(2\right)$
From equation(1) and equation (2), we get$32=44a\Rightarrow a=\dfrac{32}{44}=\dfrac{8}{11}m{s}^{-2}$