A block 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 0.8 . If the frictional force on the block is 10 N , the mass of the block is (T a k e g = 10 m s- 2)

Question

A block 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 0.8 . If the frictional force on the block is 10 N , the mass of the block is (T a k e g = 10 m s- 2) (A) 1 kg (B) 2 kg (C) 3 kg (D) 4 kg

Tardigrade Admin · Accepted Answer

As the block is at the state of rest, ∴ R=mgcos θ , which is balanced. and f1= force of friction =10 N (given) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-flihumjrwqzwqcau.jpg" /> Hence, f1=mgsin θ 10=m× 10× sin 30o m=(10/10 × sin 30o)=2 kg

Q. A block 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 $0.8$ . If the frictional force on the block is $10 N$ , the mass of the block is $(T ak e g = 10 m s^{- 2})$

$1 k g$

$2 k g$

$3 k g$

$4 k g$

As the block is at the state of rest,
$∴ R = m g cos θ$ , which is balanced.
and $f_{1} =$ force of friction $= 10 N$ (given)

Hence, $f_{1} = m g s in θ$
$10 = m \times 10 \times s in 3 0^{o}$
$m = \frac{10}{10 \times s in 3 0 ^{o}} = 2 k g$

Q. A block 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 0.8 . If the frictional force on the block is 10N , the mass of the block is (Takeg=10ms−2)

1kg

2kg

3kg

4kg