Question

A block of mass 2 kg slides down an incline plane of inclination ${30}^0$. The coefficient of friction between block and plane is 0.5. The contact force between block and plank is:

• A. 20 N
• B. $10\sqrt{3}$ N
• C. $5\sqrt{7}$ N
• D. $5\sqrt{15}$ N

Answer 1

The correct option is A 20 N

Given,

$m=2kg$

$\theta ={30}^0$

$\mu =0.5$

$g=10m/s^2$

The contact force is $F_c$is the resultant of frictional force and normal force.

From the free body diagram,

Normal force, $N=mgcos\theta$

$N=2\times 10\times cos{30}^0$

$N=17.320N$

The friction force, $F_r=\mu mgcos\theta$

$F_r=0.5\times 2\times \times 10\times cos{30}^0$

$F_r=8.660N$

$N$ and $F_r$ are perpendicular to each other.

The contact force is, $F_c=\sqrt{N^2+F_r^2}$

$F_c=\sqrt{(17.320{)}^2+(8.699{)}^2}$

$F_c=19.381N$

We can say approximately 20 N