$A$ block of mass $5\,kg$ and surface area $2\,m^2$ just begins to slide down an inclined plane when the angle of inclination is $30^{\circ}$. Keeping the mass the same,the surface area of the block is doubled. The angle at which this starts sliding down is:
- A$30^{\circ}$
- B$60^{\circ}$
- C$15^{\circ}$
- DNone of these
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$A$ body of mass $2 \ kg$ slides down with an acceleration of $4 \ m/s^2$ on an inclined plane having a slope of $30^{\circ}$. The external force required to take the same body up the plane with the same acceleration will be (Acceleration due to gravity $= 10 \ m/s^2$) (in $N$)
$A$ cubic block of mass $m$ is sliding down on an inclined plane at $60^{\circ}$ with an acceleration of $\frac{g}{2}$. The value of the coefficient of kinetic friction is:
$A$ block of mass $m$ is at rest on an inclined plane with an angle of inclination $\theta$ and coefficient of friction $\mu$,as shown in the figure. The frictional force acting on the block is:
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View SolutionThe minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$,then the angle of the inclined plane is (in $^{\circ}$)
DifficultTS EAMCET 2005
View SolutionThe time taken by a block to slide down a rough inclined plane of angle $30^{\circ}$ is $n=2$ times the time taken to slide down a frictionless inclined plane of the same angle $30^{\circ}$. The coefficient of kinetic friction between the block and the plane is:
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