Question

Two blocks of masses $m_1$ and $m_2$ are connected by spring of 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 to 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. $F=\mu (m_1+\frac{m_1}{3})g$
• D. $F=\mu (m_1+\frac{m_2}{2})g$

Answer 1

The correct option is D $F=\mu (m_1+\frac{m_2}{2})g$

Given: two blocks of masses $m_1$ and $m_2$ are connected by spring of spring constant K. Initially the spring is at its natural length. The coefficient of friction between the bars and the surface is $mu$.

To find the minimum constant force has to be applied in the horizontal direction to the block of mass $m_1$, in order to shift the other block

Solution:

When $m_2$is fixed and $F-K{m}_{1}g>0$

$F_1-K{m}_{1}g=bA$

Here $b$ is spring constant

$⟹f=2bA$

and thus we have

$2bA\le K{m}_{2}g$

So, the minimum value of F is

$2(F-K{m}_{1}g)=K{m}_{2}g⟹F=K(m_1+\frac{m_2}{2})g$

is the minimum constant force has to be applied in the horizontal direction to the block of mass $m_1$, in order to shift the other block