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Question

Two blocks of masses $${m}_{1}$$ and $${m}_{2}$$ are connected by spring constant $$K$$. Initially the spring is at its natural length. The coefficient of friction between the bars and the surface is $$\mu$$. What minimum constant force has to be applied in the horizontal direction on the block of mass $${m}_{1}$$, in order to shift the other block?
764610_13859b9d0e324d4ead964dfa468a8458.png


A
F=μ(2m1+m22)g
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B
F=μ(2m1+m2)g
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C
F=μ(m2+m12)g
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D
F=μ(m1+m22)g
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Solution

The correct option is C $$F=\mu \left( { m }_{ 1 }+\cfrac { { m }_{ 2 } }{ 2 } \right) g$$
$$f_1 = \mu m_1 g$$

$$f_2 = \mu m_2 g$$

where, $$m_2$$ is fixed and $$F - f_1 = F - \mu m_1 g >0$$

Under the influence of the spring $$m_1$$ moves like $$a$$ damped oscillation.

Let at point A,

$$ F - f_1 = kx$$         ..............(1)

Net force is zero then $$m_1$$ have $$E$$. So that $$m_1$$ move upto length $$2x$$ from their normal position where A point is mean position.

Then when m_1 at point B. End of first half ;

$$f_2 = k(2x) = \mu m_2 g$$

$$ kx = \cfrac{\mu m_2}{2} g$$

Hence from eq. (1);

$$ F - f_1 = kx = \mu m_1 g + \mu \cfrac{m_2}{2} g = \mu (m_1 + \cfrac{m_2}{2})g$$   

981605_764610_ans_3ebb0dd89d854c0fa4193e111781f6dc.png

Physics

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