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
    $20$
  • B
    $25$
  • C
    $16$
  • D
    $30$

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Similar Questions

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 of mass $m$ slides down an inclined plane at an angle of $30^{\circ}$ with an acceleration of $\frac{g}{4}$. The value of the coefficient of kinetic friction will be:

$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \text{ N}$, the mass of the block is $\left(g=10 \text{ ms}^{-2}\right)$ (in $\text{ kg}$)

$A$ block is lying on an inclined plane which makes $60^\circ$ with the horizontal. If the coefficient of friction between the block and the plane is $0.25$ and $g = 10\,m/s^2$,then the acceleration of the block when it moves along the plane will be ........ $m/s^2$.

$A$ block of mass $2 \, kg$ rests on a plane inclined at an angle of $30^o$ with the horizontal. The coefficient of friction between the block and the surface is $0.7$. The frictional force acting on the block is . . . . . . $N$.

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