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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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Verified Answer
The correct answer is:
\(8 \mathrm{~N}\)
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 \(2 \mathrm{~kg}\),
\(\begin{aligned}
f & =f^{\prime}-2 \mathrm{~N} \\
& =10-2=8 \mathrm{~N}
\end{aligned}\)
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 \(2 \mathrm{~kg}\),
\(\begin{aligned}
f & =f^{\prime}-2 \mathrm{~N} \\
& =10-2=8 \mathrm{~N}
\end{aligned}\)
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