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?

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

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

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