Q126 / 160 · AP EAMCET 2019 (22 Apr Shift 1) · Physics
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 \(=10 \mathrm{~ms}^{-2}\).)
- Correct answerA\(8 \mathrm{~N}\)
- B\(10 \mathrm{~N}\)
- C\(6 \mathrm{~N}\)
- D\(4 \mathrm{~N}\)
Solution
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 \(2 \mathrm{~kg}\), \(\begin{aligned} f & =f^{\prime}-2 \mathrm{~N} \\ & =10-2=8 \mathrm{~N} \end{aligned}\)
Hence, the correct answer is option: 'A'.