The upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

When a body slides down from rest along a smooth inclined plane making an angle of $30^{\circ}$ with the horizontal,it takes time $T$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance,it takes time $\alpha T$,where $\alpha$ is a constant greater than $1$. The coefficient of friction between the body and the rough plane is $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$ where $x = .....$.

$A$ block is placed on a parabolic shape ramp given by the equation $y = \frac{x^2}{20}$. If the coefficient of static friction $\mu_s$ is $0.5$, then what is the maximum height above the ground at which the block can be placed without slipping (in $m$)?

$A$ body is moving down a long inclined plane of slope $37^{\circ}$. The coefficient of friction between the body and the plane varies as $\mu = 0.3x$,where $x$ is the distance travelled down the plane. The body will have maximum speed at:

The time taken by an object to slide down a $45^{\circ}$ rough inclined plane is $n$ times the time it takes to slide down a perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the inclined plane is:

If $300\,J$ of work is done to move a $2\,kg$ block up a rough inclined plane to a height of $10\,m$,find the work done against friction in $J$. (Take $g = 10\,m/s^2$)