$A$ block of mass $10\, kg$ is placed on an inclined plane. When the angle of inclination is $30^\circ$,the block just begins to slide down the plane. The force of static friction is ....... $kg\, wt$.

The blocks are in equilibrium. The friction force acting on the $10 \, kg$ block is:

An insect is crawling in a hemispherical bowl of radius $R$. If the coefficient of friction between the insect and the bowl is $\mu$,then the maximum height to which the insect can crawl in the bowl is

$A$ block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is $3$ times the force required to just prevent it from sliding down. If we define $N=10 \mu$,then $N$ is

$A$ block of mass $1 \ kg$ is pushed up a surface inclined to the horizontal at an angle of $60^{\circ}$ by a force of $10 \ N$ parallel to the inclined surface as shown in the figure. When the block is pushed up by $10 \ m$ along the inclined surface,the work done against the frictional force is: $\left[g=10 \ m/s^2, \mu_s = 0.1\right]$

$A$ particle of mass $m$ slides from rest down a plane inclined at $30^{\circ}$ to the horizontal. The force of resistance acting on the particle during motion is $ms^2$,where $s$ is the displacement of the particle from its initial position. The velocity (in $m/s$) of the particle when $s = 1\,m$ is $v$. The value of $\frac{3v^2}{14}$ is: