$A$ block of mass $10 \,kg$ is held at rest against a rough vertical wall $[\mu=0.5]$ under the action of a force $F$ as shown in the figure. The minimum value of $F$ required for it is ............ $N$ $(g=10 \,m/s^2)$.

$A$ block of mass $m$ is placed on a wedge of angle $\theta$. The wedge is moving with a constant velocity $v$ on a horizontal surface. If the block does not slide on the wedge,find the work done by the friction force on the block in time $t$.

If an inclined plane is made slowly horizontal by reducing its inclination with the horizontal,the component of weight parallel to the plane of a block resting on the inclined plane:

In the given arrangement of a doubly inclined plane,two blocks of masses $M$ and $m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is $0.25$. The value of $m$,for which $M=10 \text{ kg}$ will move down with an acceleration of $2 \text{ m/s}^2$,is: (take $g=10 \text{ m/s}^2$ and $\tan 37^{\circ}=3/4$) (in $\text{ kg}$)

$A$ body of mass $2 \ kg$ is at rest at the bottom of an inclined plane of length $8 \ m$ and height $1 \ m$ as shown in the figure. If the coefficient of friction is $0.2$,what is the work done in moving the body from the bottom to the top of the incline (in $J$)? (Take $g = 10 \ m/s^2$)

The acceleration of a body of mass $m$ sliding down an inclined plane with an angle of inclination $\theta$ and a coefficient of kinetic friction $\mu$ is: