A block is lying on an inclined plane which makes an angle of 60° with the horizontal. If coefficient of friction between block and plane is 0.25 and g=10 ms -2, then the acceleration of block when it moves along the plane will be [Take √3=1.7]

Question

A block is lying on an inclined plane which makes an angle of 60° with the horizontal. If coefficient of friction between block and plane is 0.25 and g=10 ms -2, then the acceleration of block when it moves along the plane will be [Take √3=1.7] (A) 2.50 ms -2 (B) 5.00 ms -2 (C) 7.25 ms -2 (D) 8.66 ms -2.

Tardigrade Admin · Accepted Answer

Along the inclined plane, m g sin θ-μ m g cos θ=m a a=g( sin θ-μ cos θ) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0f234c323ca476631512bd0a1fc8bb9e-.png" /> ⇒ a=10 ( m / s 2)( sin 60°-0.25 cos 60°) =10[(√3/2)-(0.25) (1/2)]=(10(1.7-0.25)/2)=7.25 ( m / s 2)

Q. A block is lying on an inclined plane which makes an angle of $6 0^{\circ}$ with the horizontal. If coefficient of friction between block and plane is $0.25$ and $g = 10 m s^{- 2},$ then the acceleration of block when it moves along the plane will be [Take $3 = 1.7]$

$2.50 m s^{- 2}$

$5.00 m s^{- 2}$

$7.25 m s^{- 2}$

$8.66 m s^{- 2}$ .

Along the inclined plane,
$m g sin θ - μ m g cos θ = ma$
$a = g (sin θ - μ cos θ)$

$\Rightarrow a = 10 \frac{m}{s ^{2}} (sin 6 0^{\circ} - 0.25 cos 6 0^{\circ})$
$= 10 [\frac{3}{2} - (0.25) \frac{1}{2}] = \frac{10 ( 1.7 - 0.25 )}{2} = 7.25 \frac{m}{s ^{2}}$

Q. A block is lying on an inclined plane which makes an angle of 60∘ with the horizontal. If coefficient of friction between block and plane is 0.25 and g=10ms−2, then the acceleration of block when it moves along the plane will be [Take 3​=1.7]

2.50ms−2

5.00ms−2

7.25ms−2

8.66ms−2.