$A$ block of mass $m$ is at rest with respect to a lift when placed on an inclined plane of inclination $\theta$ inside the lift. If the lift moves upward at a constant velocity $v$,then the work done by the friction force on the block in time $t$ is:

$A$ body is sliding down an inclined plane having a coefficient of friction $0.5$. If the normal reaction is twice that of the resultant downward force along the incline,the angle between the inclined plane and the horizontal is ....... $^o$

$A$ 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: $\left[g=10 \ m/s^2, \mu_s = 0.1\right]$

When the angle of inclination of an inclined plane is $\theta$,an object slides down with uniform velocity. If the same object is pushed up with an initial velocity $u$ on the same inclined plane,it goes up the plane and stops at a certain distance on the plane. Thereafter,the body:

$A$ block slides down an inclined plane with an acceleration $g/2$ as shown in the figure. Then the coefficient of kinetic friction is

$A$ box is placed on an inclined plane and has to be pushed down. The angle of inclination is