(A) The component of the weight acting down the inclined plane is $F_{applied} = mg \sin \theta = 2 \times 9.8 \times \sin 30^\circ = 2 \times 9.8 \ti

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(A) The component of the weight acting down the inclined plane is $F_{applied} = mg \sin \theta = 2 \times 9.8 \times \sin 30^\circ = 2 \times 9.8 \times 0.5 = 9.8 \, N$. The limiting friction is given by $F_l = \mu_s N = \mu_s mg \cos \theta$. $F_l = 0.7 \times 2 \times 9.8 \times \cos 30^\circ = 0.7 \times 19.6 \times \frac{\sqrt{3}}{2} \approx 11.88 \, N$. Since the applied force $(9.8 \, N)$ is less than the limiting friction $(11.88 \, N)$,the block remains at rest. Therefore,the static frictional force is equal to the component of the weight acting down the plane,which is $9.8 \, N$.

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

Medium

$A$ block of mass $2 \, kg$ 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 $0.7$. The frictional force on the block is ....... $N$.

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A
$9.8$
B
$0.7 \times 9.8 \times \sqrt{3}$
C
$9.8 \times \sqrt{3}$
D
$0.8 \times 9.8$

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Newton's Laws of Motion and Friction

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Motion (or rest) on Rough Inclined Surface

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$A$ block of mass $2 \, kg$ 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 $0.7$. The frictional force on the block is ....... $N$.

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$A$ small bar starts sliding down an inclined plane forming an angle $\theta$ with the horizontal. The friction coefficient depends on the distance $x$ covered as $\mu = kx$,where $k$ is a constant. Find the distance covered by the bar until it stops.

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