(A) The initial momentum of the vehicle is $P = Mv$,so the initial velocity is $u = \frac{P}{M}$.The final velocity $v = 0$ as the vehicle comes to a

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(A) The initial momentum of the vehicle is $P = Mv$,so the initial velocity is $u = \frac{P}{M}$.The final velocity $v = 0$ as the vehicle comes to a stop.The frictional force acting on the vehicle is $f = \mu N = \mu Mg$,where $N = Mg$ is the normal reaction.Using Newton's second law,the retardation $a$ is given by $a = -\frac{f}{M} = -\frac{\mu Mg}{M} = -\mu g$.Using the kinematic equation $v^{2} - u^{2} = 2as$,where $s$ is the stopping distance:$0^{2} - (\frac{P}{M})^{2} = 2(-\mu g)s$$-\frac{P^{2}}{M^{2}} = -2\mu gs$$s = \frac{P^{2}}{2\mu g M^{2}}$.

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

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$A$ vehicle of mass $M$ is moving with momentum $P$ on a rough horizontal road. The coefficient of friction between the tyres and the horizontal road is $\mu$. The stopping distance is ($g$ $=$ acceleration due to gravity).

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

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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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$A$ vehicle of mass $M$ is moving with momentum $P$ on a rough horizontal road. The coefficient of friction between the tyres and the horizontal road is $\mu$. The stopping distance is ($g$ $=$ acceleration due to gravity).

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$A$ body of mass $2\, kg$ is being dragged with uniform velocity of $2\, m/s$ on a rough horizontal plane. The coefficient of friction between the body and the surface is $0.20$. The amount of heat generated in $5\, s$ is ........ $cal$ $(J = 4.2\, J/cal$ and $g = 9.8\, m/s^2)$

$A$ block of mass $2 \,kg$ is being pushed against a wall by a force $F=90 \,N$ as shown in the figure. If the coefficient of friction is $0.25$, then the magnitude of acceleration of the block is (Take $g=10 \,ms^{-2}$, $\sin 37^{\circ}=\frac{3}{5}$). (in $\,ms^{-2}$)

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