Two blocks $A$ and $B$ are released from the top of a rough inclined plane so that $A$ slides along the plane and $B$ falls down freely. Which will have higher velocity on reaching the ground?

The upper half of an inclined plane with inclination $\phi$ is perfectly smooth,while the lower half is rough. $A$ body starting from rest at the top will again come to rest at the bottom,if the coefficient of friction for the lower half is given by-

$A$ force of $750 \, N$ is applied to a block of mass $102 \, kg$ to prevent it from sliding on a plane with an inclination angle $30^{\circ}$ with the horizontal. If the coefficients of static friction and kinetic friction between the block and the plane are $0.4$ and $0.3$ respectively,then the frictional force acting on the block is...... $N$.

$A$ 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 $g = 10 \ m/s^2$)

An inclined plane is bent in such a way that the vertical cross-section is given by $y = \frac{x^2}{4}$,where $y$ is in the vertical direction and $x$ is in the horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction $\mu = 0.5$,the maximum height in $cm$ at which a stationary block will not slip downward is............$cm$.

$A$ block of mass $10 \ kg$, initially at rest, makes a downward motion on a $45^{\circ}$ inclined plane. The distance travelled by the block after $2 \ s$ is (Assume the coefficient of kinetic friction to be $0.3$ and $g=10 \ m/s^2$)