(D) When a block is on the verge of sliding down an inclined plane,the angle of inclination $\theta$ is equal to the angle of repose.At this condition

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(D) When a block is on the verge of sliding down an inclined plane,the angle of inclination $\theta$ is equal to the angle of repose.At this condition,the coefficient of static friction $\mu$ is given by $\mu = \tan \theta$.Given $\sin \theta = \frac{3}{5}$.In a right-angled triangle,if the perpendicular is $3$ and the hypotenuse is $5$,then the base is $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.Therefore,$\tan \theta = \frac{\text{perpendicular}}{\text{base}} = \frac{3}{4}$.Thus,$\mu = 0.75$.

Class 11 Physics Newton's Laws of Motion and Friction Motion (or rest) on Rough Inclined Surface

Medium

$A$ block of mass $4\, kg$ rests on an inclined plane. The inclination of the plane is gradually increased. It is found that when the inclination is $3$ in $5$ $\left( \sin \theta = \frac{3}{5} \right)$,the block just begins to slide down the plane. The coefficient of friction between the block and the plane is

A
$0.4$
B
$0.6$
C
$0.8$
D
$0.75$

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Newton's Laws of Motion and Friction

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Motion (or rest) on Rough Inclined Surface

$A$ 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:

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$A$ block of wood resting on an inclined plane of angle $30^o$ just starts moving down. If the coefficient of friction is $0.2$,its velocity (in $ms^{-1}$) after $5\, s$ is: $(g = 10\, ms^{-2})$

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An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and the surface is $1/3$. If the line joining the centre of the hemispherical surface to the insect makes an angle $\alpha$ with the vertical,the maximum possible value of $\alpha$ so that the insect does not slip is given by

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$A$ body is sliding down an inclined plane (angle of inclination $45^{\circ}$). If the coefficient of friction is $0.5$ and $g = 9.8\, m/s^2$,then the acceleration of the body downwards in $m/s^2$ is

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When a body is placed on a rough plane inclined at an angle $\theta$ to the horizontal,its acceleration is

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$A$ 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:

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