Question

Two bodies of masses $m_1$ and $m_2$ are placed on a horizontal table with coefficient of friction $\mu$ and are joined by a spring. Initially, the spring has its natural length. If $F$ is minimum force which, when continuously applied on $m_1$, will make the other block $m_2$ just move ($k$ is the spring constant) and $x$ is elongation in spring at that instant

• A. $x=\frac{\mu m_{2}g}{k}$
• B. $x=\frac{2\mu m_{2}g}{k}$
• C. $F=\mu m_{2}g+\frac{\mu m_{1}g}{2}$
• D. $F=\mu m_{1}g+\frac{\mu m_{2}g}{2}$

Answer 1

Answer: (A), (D)

$F=\mu m_{1}g+\frac{\mu m_{2}g}{2}$

Motion of $m_2$ starts when $kx=\mu .m_{2}g$
where $x=$ extension in the string
$x=\mu \frac{m_{2}g}{k}$
The minimum force will be such that $m_1$ has no kinetic energy. Applying work-energy principle on body of mass $m_1$
${\int }_0^x(F_{\text{min}}-\mu m_{1}g-kx)dx=0$
$\Rightarrow F_{\text{min}}x-\mu m_{1}gx-\frac{1}{2}k{x}^2=0$
$\Rightarrow F_{\text{min}}=[\mu m_{1}gh+\frac{1}{2}kx]=[\mu m_{1}g+\frac{\mu m_{2}g}{2}]$
$\Rightarrow F_{\text{min}}=\mu m_{1}g+\frac{\mu m_{2}g}{2}$