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Question

The two blocks $$m=10kg$$ are free to move as shown. The coefficient of static friction between the blocks is $$0.5$$ and there is no friction between $$M$$ and the ground. A minimum horizontal force $$F$$ is applied to hold in against $$M$$ that is equal to
1282354_a27c69e9e2064e008d76af56d113461c.png


A
100N
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B
50N
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C
240N
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D
180N
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Solution

The correct option is A $$100N$$
$$\begin{array}{l} F=\left( { M+m } \right) a \\ =20a \\ N=10a \\ N=\mu mg \\ =100\times \dfrac { 5 }{ { 10 } } =50 \\ \Rightarrow 50=10a \\ \Rightarrow a=5 \\ \therefore F=100N \\ Hence, \\ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$$

Physics

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