91Ó°ÊÓ

Mass of the stone,$m=20kg$

Coefficient of kinetic friction force,${\mathrm{μ}}_k=0.80$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use Newton's 2nd law of motion along vertical and horizontal direction.

Consider the formula for the net force as:

$F_{net}=∑ma$

Here, F is the net force, mis mass and ais an acceleration

Free body diagram of the stone:

Kinetic frictional force must act on the stone during its motion,

To find its magnitude, use Newton’s 2nd law along y direction,

Stone is not moving along y.So, the acceleration is 0, it gives,

$$\begin{gathered} \underset{}{\overset{}{∑}}{F}_y=0 \\ N-mg=0 \\ N=mg \end{gathered}$$

Kinetic frictional force is defined as:

$\begin{array}{rcl}{f}_k& =& {\mathrm{μ}}_{k}N\\ & =& {\mathrm{μ}}_k\left(mg\right)\\ & & \end{array}$

Substitute the values and solve as:

$\begin{array}{rcl}{f}_k& =& \left(0.80\right)\left(\left(20\right)\left(9.81\right)\right)\\ & =& 156.96 \text{N}\\ & & \end{array}$

Therefore, the horizontal force must act on a $20kg$ stone (a typical mass) to maintain the stone’s motion once a gust has started it moving is 156.96 N.