$A$ 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 $= 10 \text{ ms}^{-2}$)

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 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

If $250 \,J$ of work is done in sliding a $5 \,kg$ block up an inclined plane of height $4 \,m$,the value of work done against friction is ............. $J$ $\left(g=10 \,ms^{-2}\right)$.

$A$ body is moving up an inclined plane of angle $\theta$ with an initial kinetic energy $E$. The coefficient of friction between the plane and the body is $\mu$. The work done against friction before the body comes to rest is

As shown in the figure,a body of mass $m$ moving vertically downward with a speed of $3\, m/s$ hits a smooth fixed inclined plane and rebounds with a velocity $v_f$ in the horizontal direction. If the angle of the inclined plane is $30^{\circ}$,the velocity $v_f$ will be: