A block slides down an inclined plane with an acceleration g/2 as shown in fig. Then coefficient of kinetic friction is :-

Question

A block slides down an inclined plane with an acceleration g/2 as shown in fig. Then coefficient of kinetic friction is :- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-alecpnmdp788fayt.jpg" /> (A) √3 (B) √3-1 (C) (√3/2) (D) (1/√3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-alecpnmdp788wuqx.jpg" /> mgsinθ -f=ma mgsinθ -μ mgcosθ =ma q=60° a=g/2 g(sin 60 ° - m cos 60 ° )=g/2 (√3/2)-μ ((1/2))=(1/2) √3-1=μ

Q. A block slides down an inclined plane with an acceleration $g /2$ as shown in fig. Then coefficient of kinetic friction is :-

$3$

$3 - 1$

$\frac{3}{2}$

$\frac{1}{3}$

$m g s in θ - f = ma$
$m g s in θ - μ m g cos θ = ma$
$q = 6 0^{\circ} a = g /2$
$g (s in 6 0^{\circ} - m cos 6 0^{\circ}) = g /2$
$\frac{3}{2} - μ (\frac{1}{2}) = \frac{1}{2}$
$3 - 1 = μ$

Q. A block slides down an inclined plane with an acceleration g/2 as shown in fig. Then coefficient of kinetic friction is :-

3​

3​−1

23​​

3​1​

mgsinθ−f=ma mgsinθ−μmgcosθ=ma q=60∘a=g/2 g(sin60∘−mcos60∘)=g/2 23​​−μ(21​)=21​ 3​−1=μ

Q. A block slides down an inclined plane with an acceleration $g /2$ as shown in fig. Then coefficient of kinetic friction is :-

$3$

$3 - 1$

$\frac{3}{2}$

$\frac{1}{3}$

$m g s in θ - f = ma$
$m g s in θ - μ m g cos θ = ma$
$q = 6 0^{\circ} a = g /2$
$g (s in 6 0^{\circ} - m cos 6 0^{\circ}) = g /2$
$\frac{3}{2} - μ (\frac{1}{2}) = \frac{1}{2}$
$3 - 1 = μ$