Question

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Answer 1

$\bigstar$ Answer :

Option => 4.

$\sf \implies \dfrac{\mu}{2}\;(2M_1 + M_2)g$

$\bigstar$ Explanation :

Given that :

Two blocks are connected by a spring.

The blocks are placed on a horizontal rough surface with 'μ' as the coefficient of friction between the blocks and the surface.

To Find :

Minimum horizontal force on M₁ so as to move M₂.

Solution :

We have, to move M₂ , kx = μM₂g.

We know that :

Work Energy Theorem :

$\bigstar\;\boxed{\sf F_{min}\;x-\mu\;M_1\;gx- \dfrac{1}{2}\;kx^2 = 0-0}}$

Values in Equation :

$\sf \\ \\ \implies F_{min} = \mu\;M_1g +\dfrac{\mu\;M_2g}{2}\\ \\ \implies \dfrac{\mu}{2}\;(2M_1 +M_2)g\\ \\ \\ \\\bigstar\;\textbf{\underline{\underline{Hence, Option = 4 is the correct answer.}}}$

$\rule{300}{1.5}$

$\bigstar\;\textbf{\underline{\underline{Extra Information :}}}$

$\bigstar$ Work Energy Theorem :

The work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

$\bigstar$ Horizontal Surface :

The horizontal surface is the flat surface at right angles to a plumb line.

Answer 2

Answer :-

$F = \mug ( M_1 + M_2)$

Given :-

Two blocks of $M_1$ and $M_2$ are connected by a spring.

Where k is a spring constant.

$\mu$ is friction coefficient.

To find :-

The minimum horizontal force on $M_1$.

Solution:-

Let us consider block $M_1$ as system.

The force acting on block is :-

• Mg force downward.
• Normal force upward.
• Spring force in right side .

As spring force is given by :-

→$F = Kx$

where,

X is eleongation and compression.

→$\mathsf{ Kx = f_k}$

→$\mathsf{Kx = \mu M_2g----1) }$

• For block $M_1$

→$\mathsf{F = Kx + f_k}$

→$\mathsf{F = Kx + \mu M_1g}$

• Put the value of Kx.

→$\mathsf{F = \mu M_2g + \mu M_1g}$

→$\mathsf{ F = \mug ( M_1 + M_2)}$

hence,

The minimum force required to move $\mug ( M_1 + M_2)$.

Note :- From question it's not clear that acceleration is included or not but in ans there is no clue of acceleration.