Two blocks m1 and m2 are resting on a rough inclined plane of inclination 37- as shown in figure- The contact force between the blocks is m1 - 4 kg- m2 - 2 kg- -1- 0-8- -2 - 0-5- g - 10 m-s2- sin 37- - 35

For a mass sliding down on a inclined plane , we know that ⇒a = g sinθ μg cosθ. Hence acceleration depends on the coefficient of friction , As μ_1 gt; μ_2 ...

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Standard XIII

Physics

Newton's Second Law

Two blocks m1...

Question

Two blocks $m_{1}$ and $m_{2}$ are resting on a rough inclined plane of inclination $37^{\circ}$ as shown in figure. The contact force between the blocks is $(m_{1} = 4 k g, m_{2} = 2 k g, μ_{1} = 0.8, μ_{2} = 0.5, g = 10 {m/s}^{2}, sin 37^{\circ} = \frac{3}{5})$

3.2 N

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3.6 N

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7.2 N

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Zero

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Solution

The correct option is D Zero
For a mass sliding down on a inclined plane , we know that
$\Rightarrow a = g sin θ - μ g cos θ$ .
Hence acceleration depends on the coefficient of friction ,
As $μ_{1} > μ_{2}$ $\Rightarrow a_{1} < a_{2}$ . (directly from the equation we can say)
Hence the acceleration of $m_{2}$ is greater than that of $m_{1}$ . Hence, there will not be any contact.
So, $N = 0$

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Two blocks m1 and m2 are resting on a rough inclined plane of inclination 37∘ as shown in figure. The contact force between the blocks is (m1=4 kg,m2=2 kg,μ1=0.8,μ2=0.5,g=10m/s2,sin37∘=35)

Two blocks $m_{1}$ and $m_{2}$ are resting on a rough inclined plane of inclination $37^{\circ}$ as shown in figure. The contact force between the blocks is $(m_{1} = 4 k g, m_{2} = 2 k g, μ_{1} = 0.8, μ_{2} = 0.5, g = 10 {m/s}^{2}, sin 37^{\circ} = \frac{3}{5})$