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Question: A block rests on a rough inclined plane making an angle of \(30 ^ { \circ }\) with the horizontal....

A block rests on a rough inclined plane making an angle of 30∘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 N, the mass of the block (in kg) is (take g=10 m/s2)\left. g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)

A

2.0

B

4.0

C

1.6

D

2.5

Answer

2.0

Explanation

Solution

Angle of repose

α=tan⁡−1(μ)=tan⁡−1(0.8)=38.6∘\alpha = \tan ^ { - 1 } ( \mu ) = \tan ^ { - 1 } ( 0.8 ) = 38.6 ^ { \circ }

Angle of inclined plane is given θ=30∘\theta = 30 ^ { \circ } . It means block is at rest therefore,

Static friction = component of weight in downward direction

=mgsin⁡θ=10N= m g \sin \theta = 10 N ∴\therefore m=109×sin⁡30∘=2 kgm = \frac { 10 } { 9 \times \sin 30 ^ { \circ } } = 2 \mathrm {~kg}