$A$ small bar starts sliding down an inclined plane forming an angle $\theta$ with the horizontal. The friction coefficient depends on the distance $x$ covered as $\mu = kx$,where $k$ is a constant. Find the distance covered by the bar until it stops.

$A$ body is placed on a rough inclined plane of inclination $\theta$. As the angle $\theta$ is increased from $0^o$ to $90^o$,the contact force between the block and the plane

The time taken by a block of mass $m$ to slide down from the highest point to the lowest point on a rough inclined plane is $50\%$ more compared to the time taken by the same block on an identical inclined smooth plane. Both inclined planes are at $45^\circ$ with the horizontal. The coefficient of kinetic friction between the rough inclined surface and the block is . . . . . . .

$A$ block is released from rest on a $45^\circ$ smooth incline and slides a distance '$d$'. The time taken to slide the same distance '$d$' on a rough incline is '$n$' times the time taken on the smooth incline. The coefficient of kinetic friction is

The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If the coefficient of friction is $0.25$,the angle of inclination of the plane is ...... $^o$.

$A$ block of base $10 \ cm \times 10 \ cm$ and height $15 \ cm$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$. Then