$A$ 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 $10 \,kg$ 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 ............ $N$ $(g=10 \,m/s^2)$.

$A$ child weighing $25$ kg slides down a rope hanging from the branch of a tall tree. If the force of friction acting against him is $2$ $N$,the acceleration of the child is ........ $m/s^2$. (Take $g = 9.8$ $m/s^2$)

$A$ block slides down an inclined plane of slope angle $\theta$ with a constant velocity. It is then projected up the plane with an initial velocity $u$. The distance up to which it will rise before coming to rest is

When an object is shot from the bottom of a long smooth inclined plane kept at an angle $60^{\circ}$ with the horizontal,it can travel a distance $x_{1}$ along the plane. But when the inclination is decreased to $30^{\circ}$ and the same object is shot with the same velocity,it can travel $x_{2}$ distance. Then $x_{1}: x_{2}$ will be

The minimum force required to start pushing a body up a rough (frictional coefficient $\mu$) inclined plane is $F_{1}$,while the minimum force needed to prevent it from sliding down is $F_{2}$. If the inclined plane makes an angle $\theta$ with the horizontal such that $\tan \theta = 2\mu$,then the ratio $\frac{F_{1}}{F_{2}}$ is: