Determine the maximum acceleration in $m/s^2$ of the train in which a box lying on its floor will remain stationary, given that the co-efficient of static friction between the box and the train’s floor is $0.15.$

To avoid slipping while walking on ice, one should take smaller steps because of the

As shown in the figure, a block of mass $\sqrt{3}\, kg$ is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3 \sqrt{3}}$. The critical force to be applied on the vertical surface as shown at an angle $60^{\circ}$ with horizontal such that it does not move, will be $3 x$. The value of $3x$ will be

$\left[ g =10 m / s ^{2} ; \sin 60^{\circ}=\frac{\sqrt{3}}{2} ; \cos 60^{\circ}=\frac{1}{2}\right]$

A block of mass $m$ (initially at rest) is sliding up (in vertical direction) against a rough vertical wall with the help of a force $F$ whose magnitude is constant but direction is changing. $\theta  = {\theta _0}t$  where $t$ is time in sec. At $t$ = $0$ , the force is in vertical upward direction and then as time passes its direction is getting along normal, i.e., $\theta  = \frac{\pi }{2}$ .The value of $F$ so that the block comes to rest when $\theta  = \frac{\pi }{2}$ , is 

An army vehicle of mass $1000\, kg$ is moving with a velocity of $10 \,m/s$ and is acted upon by a forward force of $1000\, N$ due to the engine and a retarding force of $500 \,N$ due to friction. ........... $m/s$ will be its velocity after $10\, s$

$Assertion$ : Angle of repose is equal to the angle of limiting friction.
$Reason$ : When the body is just at the point of motion, the force of friction in this stage is called limiting friction.