A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is: [Take g = 10 m/s2]

Question

A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is: [Take g = 10 m/s2] (A) (2/3) (B) (√3/2) (C) (√3/4) (D) (1/2)

Tardigrade Admin · Accepted Answer

2+m g sin 30=μ m g cos 30° 10=m g sin 30+μ m g cos 30° =2 μ m g cos 30-2 6=μ m g cos 30 4=m g sin 30 (3/2)=μ × √3 μ=(√3/2)

$\frac{2}{3}$

$\frac{3}{2}$

$\frac{3}{4}$

$\frac{1}{2}$

$2 + m g sin 30 = μ m g cos 3 0^{\circ}$
$10 = m g sin 30 + μ m g cos 3 0^{\circ}$
$= 2 μ m g cos 30 - 2$
$6 = μ m g cos 30$
$4 = m g sin 30$
$\frac{3}{2} = μ \times 3$
$μ = \frac{3}{2}$

32​

23​​

43​​

21​

2+mgsin30=μmgcos30∘ 10=mgsin30+μmgcos30∘ =2μmgcos30−2 6=μmgcos30 4=mgsin30 23​=μ×3​ μ=23​​

$\frac{2}{3}$

$\frac{3}{2}$

$\frac{3}{4}$

$\frac{1}{2}$

$2 + m g sin 30 = μ m g cos 3 0^{\circ}$
$10 = m g sin 30 + μ m g cos 3 0^{\circ}$
$= 2 μ m g cos 30 - 2$
$6 = μ m g cos 30$
$4 = m g sin 30$
$\frac{3}{2} = μ \times 3$
$μ = \frac{3}{2}$