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. $F=\mu (2m_1+\frac{m_2}{2})g$
• B. $F=\mu (2m_1+m_2)g$
• C. $F=\mu (m_2+\frac{m_1}{2})g$
• D. $F=\mu (m_1+\frac{m_2}{2})g$

Answer 1

Answer: (D)

$F=\mu (m_1+\frac{m_2}{2})g$

$f_1=\mu m_{1}g$

$f_2=\mu m_{2}g$

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 ;

$f_2=k\left(2x\right)=\mu m_{2}g$

$kx=\frac{\mu m_2}{2}g$

Hence from eq. (1);

$F-f_1=kx=\mu m_{1}g+\mu \frac{m_2}{2}g=\mu (m_1+\frac{m_2}{2})g$