Question

A small block of mass $m$ is projected on a larger block of mass $10m$ and length $l$ with a velocity $v$ as shown in the figure. The coefficient of friction between the two blocks is ${\mu }_2$ while that between the lower block and the ground is ${\mu }_1.$ Given that ${\mu }_2>11{\mu }_1.$ Find the minimum value of $v$, such that the mass m falls off the block of mass $10m$.

• A. $V_{min}=\sqrt{\frac{22({\mu }_2-{\mu }_1)gl}{10}}$
• B. $V_{min}=\sqrt{\frac{11({\mu }_2-{\mu }_1)gl}{10}}$
• C. $V_{min}=\sqrt{\frac{2({\mu }_2-{\mu }_1)gl}{10}}$
• D. $V_{min}=\sqrt{22({\mu }_2-{\mu }_1)gl}$

Answer 1

The correct option is D $V_{min}=\sqrt{22({\mu }_2-{\mu }_1)gl}$

Here, the K.E of the block of mass 'm' is equal to P.E due to gravitational force of the large block.

$\therefore \frac{1}{2}m{v}_{min}^2=\mu mgl$

Where,

$\mu =$Co-efficient of friction (Resultant)

$m=$mass of block

$g=$ accleration due to gravity

$l=$length of larger block

$\therefore \frac{1}{2}m{v}^2=11\left({\mu }_2-{\mu }_1\right)\times m\times gl$

$\therefore v^2=22\left({{\mu }_2-\mu }_1\right)gl$

$v_{min}=\sqrt{22\left(\right({\mu }_2-{\mu }_1\left)gl\right)}$