Question

Two bodies m₁and m₂ are kept on a table with coefficient of friction '$\mu$' and are joined by a spring initially, the spring is in its relaxed state. Find out the minimum force F which will make the other block $m_2$ move.(K is the spring constant).

• A. $2\mu g.m_1+\frac{m_2}{2}\mu g$
• B. $\mu g.m_1+\frac{m_2}{2}\mu g$
• C. $\mu g.m_1+\frac{m_2}{3}\mu g$
• D. $\mu g.m_1+\frac{3m_2}{2}\mu g$

Answer 1

The correct option is B $\mu g.m_1+\frac{m_2}{2}\mu g$

Motion of $m_2$ starts when,
$kx=\mu .m_2.g$
Where x = elongation in the spring.
$x=\mu .m_2.\frac{g}{k}$
The minimum force will be such that $m_1$ has no kinetic energy. Applying work energy principle for $m_1$.
${\int }_0^x(F-\mu m_{1}g-kx)dx=0\Rightarrow F=[\mu m_{1}g+\frac{1}{2}kx]=[\mu m_{1}g+\frac{\mu m_{2}g}{2}]\Rightarrow F_{min}=\mu g.m_1+\frac{m_2}{2}\mu g$