(C) The initial momentum of the block is $P = Mu$,where $u$ is the initial velocity. Thus,$u = \frac{P}{M}$.Since the surface is rough,the frictional

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(C) The initial momentum of the block is $P = Mu$,where $u$ is the initial velocity. Thus,$u = \frac{P}{M}$.Since the surface is rough,the frictional force $f = \mu N = \mu Mg$ acts on the block.The retardation (deceleration) $a$ is given by $a = \frac{f}{M} = \frac{\mu Mg}{M} = \mu g$.Using the kinematic equation $v^{2} = u^{2} - 2as$,where $v = 0$ (final velocity when it stops):$0 = u^{2} - 2as$$2as = u^{2}$$s = \frac{u^{2}}{2a}$Substituting $u = \frac{P}{M}$ and $a = \mu g$:$s = \frac{(\frac{P}{M})^{2}}{2 \mu g} = \frac{P^{2}}{M^{2} \cdot 2 \mu g} = \frac{P^{2}}{2 \mu M^{2} g}$.

Class 11 Physics Newton's Laws of Motion and Friction Kinetic Friction and Motion on Rough Horizontal Surface

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$A$ block of mass $M$ is moving on a rough horizontal surface with momentum $P$. The coefficient of friction between the block and the surface is $\mu$. The distance covered by the block before it stops is $[g = \text{acceleration due to gravity}]$.

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A
$\frac{2 \mu Mg}{P}$
B
$\frac{P}{2 \mu Mg}$
C
$\frac{P^{2}}{2 \mu M^{2} g}$
D
$\frac{2 \mu M^{2} g}{P^{2}}$

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Newton's Laws of Motion and Friction

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Kinetic Friction and Motion on Rough Horizontal Surface

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$A$ block of mass $M$ is moving on a rough horizontal surface with momentum $P$. The coefficient of friction between the block and the surface is $\mu$. The distance covered by the block before it stops is $[g = \text{acceleration due to gravity}]$.

Similar Questions

You are holding a vertical cylindrical rod of mass $1 \, kg$ by pressing it (with a force of $12 \, N$) with two fingers. The coefficient of friction between the rod and the fingers is $\mu = 0.5$. The force of friction on the rod will be ........ $N$.

$A$ block of mass $m$ slides along a floor while a force of magnitude $F$ is applied to it at an angle $\theta$ as shown in the figure. The coefficient of kinetic friction is $\mu_{K}$. Then,the block's acceleration $a$ is given by: ($g$ is acceleration due to gravity)

$A$ man of mass $60 \, kg$ slides down a pole by applying a force of $600 \, N$ on it. If the coefficient of friction between the hands and the pole is $0.5$,then the acceleration of the man is ............ $m/s^2$ $(g = 10 \, m/s^2)$.

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