A body of mass M starts sliding down on the inclined plane, where the critical angle is angle ACB =30° as shown in figure. The coefficient of kinetic friction will be

Question

A body of mass M starts sliding down on the inclined plane, where the critical angle is angle ACB =30° as shown in figure. The coefficient of kinetic friction will be (A) Mg / √3 (B) √3 Mg (C) √3 (D) None of these

Tardigrade Admin · Accepted Answer

We now that, θ= angle of repose μ= tan θ= tan 30°=(1/√3) Therefore, coefficient of kinetic friction = Mg × (1/√3)=( Mg /√3)

Q. A body of mass M starts sliding down on the inclined plane, where the critical angle is $∠ A CB = 3 0^{\circ}$ as shown in figure. The coefficient of kinetic friction will be

$M g / 3$

$3 M g$

$3$

None of these

We now that,
$θ =$ angle of repose
$μ = tan θ = tan 3 0^{\circ} = \frac{1}{3}$
Therefore, coefficient of kinetic friction
$= M g \times \frac{1}{3} = \frac{M g}{3}$

Q. A body of mass M starts sliding down on the inclined plane, where the critical angle is ∠ACB=30∘ as shown in figure. The coefficient of kinetic friction will be

Mg/3​

3​Mg

3​

None of these

We now that, θ= angle of repose μ=tanθ=tan30∘=3​1​ Therefore, coefficient of kinetic friction =Mg×3​1​=3​Mg​

Q. A body of mass M starts sliding down on the inclined plane, where the critical angle is $∠ A CB = 3 0^{\circ}$ as shown in figure. The coefficient of kinetic friction will be

$M g / 3$

$3 M g$

$3$

We now that,
$θ =$ angle of repose
$μ = tan θ = tan 3 0^{\circ} = \frac{1}{3}$
Therefore, coefficient of kinetic friction
$= M g \times \frac{1}{3} = \frac{M g}{3}$