Laws of Motion 3 Question 13
14. A block of mass $2 \mathrm{~kg}$ 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
$(1980,2 \mathrm{M})$
(a) $9.8 \mathrm{~N}$
(b) $0.7 \times 9.8 \times \sqrt{3} \mathrm{~N}$
(c) $9.8 \times \sqrt{3} \mathrm{~N}$
(d) $0.7 \times 9.8 \mathrm{~N}$
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Answer:
Correct Answer: 14. (a)
Solution:
- Since, $\mu m g \cos \theta>m g \sin \theta$
$\therefore$ Force of friction is $f=m g \sin \theta$
$ 2 \times 9.8 \times \frac{1}{2}=9.8 \mathrm{~N} $
