Question

Two bodies $m_1$ and $m_2$ 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. g ( + )
• B.
• C.
• D.

Answer 1

The correct option is B

For minimum force

$m_1$ will firse accelearate then decelerate and reach a distance $x_0$ from intial point.

Kx on $m_1$ will keep increasing with distance hence net force will be positive firse , the 0,

then negative.

This $x_0$ will be such that spring force on $m_2$ i.e ., K$x_0$ = $\mu$$m_2$g ...(1)

So $m_2$ will just start its motion.

Applying work energy theorem on $m_1$ initial and final velocity of $m_1$ = 0

$\Delta$ K.E = 0

All forces acting on $m_1$ while it moves a distance $x_0$ is spring force, friction, gravity, normal, force F min

$W_g$ = $W_N$ = 0 (As $\theta$ = ${90}^{\circ }$)

$W_{sp}$ + $W_{fr}$ + $W_f$ = $\Delta$ K.E = 0

$\Delta$ $U_{sp}$ - $\mu$$m_1$g $x_0$ = 0 ($W_{sp}$ = $\Delta$$U_{sp}$)

- $\frac{1}{2}$ K ${x_0}^2$ - $\mu$$m_1$g$x_0$ + $F_{min}$$x_0$ = 0

$F_{min}$ = $\mu$$m_1$g + $\frac{1}{2}$$x_0$

From (i) $x_0$ = $\frac{\mu m_{2}g}{k}$

$F_{min}$ = $\mu$$m_1$g + $\frac{\mu m_{2}g}{k}$