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 .......... N (take g = 10 m/s²).

Question

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 .......... N (take g = 10 $m / s^{2}$ ).

Smart Achievers · Accepted Answer

(10)

Since $μ m g \cos θ > m g \sin θ$

So block is in rest.

Force of friction is $f = m g \sin θ$

$= 2 \times 10 \times \sin 30 ° = 10 N$

Solution

Q.
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 .......... N (take g = 10 $m / s^{2}$ ).

Ans.
(10)

Since $μ m g \cos θ > m g \sin θ$

So block is in rest.

Force of friction is $f = m g \sin θ$

$= 2 \times 10 \times \sin 30 ° = 10 N$

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Solution

Q. 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 .......... N (take g = 10 m/s2).

Ans. (10) Since μmgcosθ>mgsinθ So block is in rest. Force of friction is f=mgsinθ =2×10×sin30°=10 N

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Q.
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 .......... N (take g = 10 $m / s^{2}$ ).

Ans.
(10)

Since $μ m g \cos θ > m g \sin θ$

So block is in rest.

Force of friction is $f = m g \sin θ$

$= 2 \times 10 \times \sin 30 ° = 10 N$