In the arrangement shown in the figure, the coefficient of friction between two blocks is 0.5. The force of friction between the two blocks is (Assume that the \(4 \mathrm{~kg}\) block is placed on a smooth horizontal surface.) (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\).)

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Aicharya · Accepted Answer

According to the question, the given condition is shown in the figure,

If \(f\) be the friction force between \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) block then the static friction force on \(2 \mathrm{~kg}\) block.

\(\begin{aligned} & \left.f^{\prime}=\mu R_1=0.5 \times 2 \mathrm{~g}=0.5 \times 2 \times 10 \quad \text { [Given, } \mu=0.5\right] \\ & f^{\prime}=10 \mathrm{~N} \end{aligned}\) Since, the static friction on \(2 \mathrm{~kg}\) block is more than force applied on it. \(\text {i.e, } \quad f^{\prime} > 2 \mathrm{~N}\) Hence, \(2 \mathrm{~kg}\) body will move along the direction of \(4 \mathrm{~kg}\) body. Hence, net friction force on the block of \(2 \mathrm{~kg}\), \(\begin{aligned} f & =f^{\prime}-2 \mathrm{~N} \\ & =10-2=8 \mathrm{~N} \end{aligned}\)

In the arrangement shown in the figure, the coefficient of friction between two blocks is 0.5. The force of…

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