  Question

Two blocks of mass $$m_1$$ and $$m_2$$ are connected by a non deformed light spring on a horizontal table. The coefficient of friction between the blocks and table is $$\mu$$. The minimum force applied on block 1 which will move the block 2 also is given as $$\displaystyle \mu g \left ( m_1 + \frac{m_2}{x} \right )$$. Find $$x$$ Solution

Let the compression of the spring when the body of mass $${m}_{2}$$ starts moving=$$x$$The instant when the mass $${m}_{2}$$ starts moving spring force should be equal to frictional force on mass $${m}_{2}$$Therefore, $$k\times x=\mu \times { m }_{ 2 }\times g\quad \Rightarrow x=\dfrac { \mu { m }_{ 2 }g }{ k }\quad \longrightarrow \left( equation\quad 1 \right)$$Now, from $$principle\ of\ conservation\ of\ energy\ theorem,$$Work done by force F will be equal to energy dissipated by friction and the energy stored in springTherefore, $$F\times x=\mu \times { m }_{ 1 }\times g\times x+\frac { 1 }{ 2 } \times k\times { x }^{ 2 }\quad \Rightarrow F=\mu \times { m }_{ 1 }\times g+\frac { 1 }{ 2 } \times k\times x\\ (using\ equation\ 1)\\ F=\mu \times { m }_{ 1 }\times g+\frac { 1 }{ 2 } \times k\times \dfrac { \mu { m }_{ 2 }g }{ k } \quad \Rightarrow F=\mu g({ m }_{ 1 }+\dfrac { { m }_{ 2 } }{ 2 } )\\ So,\ x=2$$Physics

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