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Byju's Answer

Standard XII

Physics

1st Equation of Motion

A block of ma...

Question

A block of mass 2 kg rests on a rough inclned plane making am angle of $30^{o}$ with the horizontal. The coefficient of static friction between the block and the plans is 0.7. The frictional force on the block is $(g = 9.8 m / s^{2})$ :

9.8 N

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0.7 \times 9.8 N

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9.8 \times 7 N

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0.8 \times 9.8 N

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Solution

The correct option is A $9.8 N$
From above image
Since the static friction is greater than the force
Show only the required friction force $= 2 g s i n 30^{\circ}$
$= g$
$f = 9.8 N$

Hence (A) option is the correct answer

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Similar questions

A block of mass 4 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: $(g = 9.8 m / s^{2})$

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A block of mass 2 kg rests on a rough inclned plane making am angle of 30o with the horizontal. The coefficient of static friction between the block and the plans is 0.7. The frictional force on the block is (g=9.8m/s2) :

The correct option is A 9.8NFrom above imageSince the static friction is greater than the forceShow only the required friction force =2g sin30∘ =g f=9.8NHence (A) option is the correct answer

A block of mass 2 kg rests on a rough inclned plane making am angle of $30^{o}$ with the horizontal. The coefficient of static friction between the block and the plans is 0.7. The frictional force on the block is $(g = 9.8 m / s^{2})$ :

The correct option is A $9.8 N$
From above image
Since the static friction is greater than the force
Show only the required friction force $= 2 g s i n 30^{\circ}$
$= g$
$f = 9.8 N$

Hence (A) option is the correct answer