Question

A block of mass 1 kg is kept on a rough inclined plane making an angle of ${30}^0$ with the horizontal. If ${\mu }_s=0.5$ and ${\mu }_k=0.4$, the frictional force on the block is

• A. $1.96\sqrt{3}N$
• B. $0.4\times 9.8\sqrt{3}N$
• C. $9.8\times \sqrt{3}N$
• D. $0.4\times 9.8N$

Answer 1

Answer: (A)

$1.96\sqrt{3}N$

$\theta ={30}^o;m=1kg;g=9.8m/s^2$

Along plane of inclination force is

$mgsin\theta =0.5mg=4.9N$

Since downward force is greater than the force of static friction, so the kinetic friction will act.

Thus, maximum Friction force is ${\mu }_{k}mgcos\theta =1.96\sqrt{3}N$