SATHEE: Laws of Motion 3 Question 13

Laws of Motion 3 Question 13

14. 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

$(1980, 2 M)$

(a) $9.8 N$

(b) $0.7 \times 9.8 \times \sqrt{3} N$

(d) $0.7 \times 9.8 N$

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Answer:

Correct Answer: 14. (a)

Solution:

Since, $μ m g \cos θ > m g \sin θ$

$∴$ Force of friction is $f = m g \sin θ$

$2 \times 9.8 \times \frac{1}{2} = 9.8 N$

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Laws of Motion 3 Question 13

14. A block of mass 2 kg rests on a rough inclined plane making an angle of 30∘ with the horizontal. The coefficient of static friction between the block and the plane is 0.7 . The frictional force on the block is

Answer:

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14. 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