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

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 (A) 9.8 N (B) 0.7× 9.8× √3N (C) 9.8× √3N (D) 0.7× 9.8N

Tardigrade Admin · Accepted Answer

Since, m g cos θ > mg sin θ ∴ Force of friction is = mg sin θ

Q. A block of mass $2 k g$ rests on a rough inclined plane making an angle of $3 0^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$ . The frictional force on the block is

9.8 N

$0.7 \times 9.8 \times 3 N$

$9.8 \times 3 N$

$0.7 \times 9.8 N$

Since, $m g cos θ > m g sin θ$
$∴$ Force of friction is = $m g sin θ$

Q. A block of mass 2kg 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

9.8 N

0.7×9.8×3​N

9.8×3​N

0.7×9.8N

Since, mgcosθ>mgsinθ ∴ Force of friction is = mgsinθ

Q. A block of mass $2 k g$ rests on a rough inclined plane making an angle of $3 0^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$ . The frictional force on the block is

$0.7 \times 9.8 \times 3 N$

$9.8 \times 3 N$

$0.7 \times 9.8 N$

Since, $m g cos θ > m g sin θ$
$∴$ Force of friction is = $m g sin θ$