A block of mass 2kg 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 0.7. The frictional force on the block is

Options

A. `9.8 N`
B. `0.7xx9.8xx sqrt(3)N`
C. `9.8xxsqrt(3)N`
D. `0.7xx9.8N`

Answer

A

Explanation

Angle of repose, `theta = tan^(-1)(mu_s)=tan^(-1)(0.7)`
or `tan theta_r=0.7`
Angle of plane is `theta = 30^@, tan theta= tan30^@ = 0.577`
Since `tan theta lt tan theta, theta lt theta`
Block will not slide or `f=mgsintheta ne mu mgcostheta`
or `f=(2)(9.8)sin 30^@ = 9.8N`