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 weighing $20 \ kg$ just slides down a rough inclined plane that rises $5$ in $12$. The coefficient of friction 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 ............ $N$ $(g=10 \,m/s^2)$.

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F_1)$ required to move the block up the inclined plane will be

$A$ block of mass $2 \, kg$ slides down an inclined plane of inclination $30^{\circ}$. The coefficient of friction between the block and the plane is $0.5$. The contact force between the block and the plane is:

$A$ block of mass $5 \ kg$ is kept against an accelerating wedge with a wedge angle of $45^{\circ}$ to the horizontal. The coefficient of friction between the block and the wedge is $\mu = 0.4$. What is the minimum absolute value of the acceleration of the wedge to keep the block steady? Assume $g = 10 \ m \ s^{-2}$.