The mysterious sliding stones.Along the remote Racetrack Playa in Death Valley, California, stones sometimes gouge out prominent trails in the desert floor, as if the stones had been migrating (Fig. 6-18). For years, curiosity mounted about why the stones moved. One explanation was that strong winds during occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playground is about.What horizontal force must act on astone (a typical mass) to maintain the stone’s motion once a gust has started it moving? (Story continues with Problem 37)

Mass of the stone,$m=20kg$

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

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:

$F_{net}=\sum ma$

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,

$$\begin{gathered} \sum _{}^{}{F}_y=0 \\ N-mg=0 \\ N=mg \end{gathered}$$

Kinetic frictional force is defined as:

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

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.