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A block of mass $$m=5$$ kg is attached with a mass-less spring of force constant $$k$$. The block is placed over a fixed rough inclined surface for which the coefficient of friction is $$0.75$$. The block of mass $$m$$ is initially at rest. The block of mass $$M$$ is released from rest with spring in un-deformed state. The minimum value of $$M$$ required to move the block up the plane is (neglect mass of string and pulley and friction in pulley.)

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A
3 kg
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B
4 kg
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C
10 kg
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D
5 kg
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Solution

The correct option is A 3 kg
As long as the block of mass m remains stationary, the block of mass M released from rest comes down by $$\dfrac{2Mg}{K}$$ (before coming it rest momentarily again).

Thus the maximum extension in spring is $$x=\dfrac{2Mg}{K}$$---------(1) for block of mass m to just move up the incline $$kx=mg sin \theta + \mu mg cos \theta$$-----------(2)

$$2 Mg = mg \times \dfrac{3}{5}+\dfrac{3}{4}mg \times \dfrac{4}{5} $$ or $$M=\dfrac{3}{5}m =\dfrac{6}{5}mg= 3 Kg$$

Physics

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