$A$ block of mass $15\,kg$ is resting on a rough inclined plane as shown in the figure. The block is tied up by a horizontal string which has a tension of $50\,N$. The coefficient of friction between the surface of contact is: $(g = 10\,m/s^2)$

$A$ 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

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 $\text{ kg}$)

$A$ block of mass $5 \text{ kg}$ is placed on a rough inclined surface as shown in the figure. If $\vec{F}_1$ is the force required to just move the block up the inclined plane and $\vec{F}_2$ is the force required to just prevent the block from sliding down,then the value of $|\vec{F}_1|-|\vec{F}_2|$ is: [Use $g=10 \text{ m/s}^2$]

$A$ block of mass $5 \text{ kg}$ starts up a $45^{\circ}$ inclined plane with an initial kinetic energy of $100 \text{ J}$. If the coefficient of friction between the block and the plane is $0.5$,then the distance covered by the block before it stops is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

$A$ block slides down on an inclined surface of inclination $30^{\circ}$ with the horizontal. Starting from rest,it covers $8 \ m$ in the first two seconds. Find the coefficient of friction. (Take $g = 10 \ m/s^2$)