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 $\mu g(m_1+\frac{m_2}{x})$. Find $x$

Answer 1

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, $k\times x=\mu \times m_2\times g\Rightarrow x=\frac{\mu m_{2}g}{k}⟶\left(equation1\right)$

Now, from $principleofconservationofenergytheorem,$

Work done by force F will be equal to energy dissipated by friction and the energy stored in spring

Therefore, $F\times x=\mu \times m_1\times g\times x+\frac{1}{2}\times k\times x^2\Rightarrow F=\mu \times m_1\times g+\frac{1}{2}\times k\times x\left(usingequation1\right)F=\mu \times m_1\times g+\frac{1}{2}\times k\times \frac{\mu m_{2}g}{k}\Rightarrow F=\mu g(m_1+\frac{m_2}{2})So,x=2$