91Ó°ÊÓ

Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. It is different from static friction, which prevents surfaces from starting to slide. To calculate kinetic friction, we use the formula:

• Kinetic friction force \(F_k\) = coefficient of kinetic friction \( \mu_k \) \times\ normal force \(F_N\)

Here, \( \mu_k \) is the coefficient of kinetic friction, a dimensionless value that represents how 'slippery' two surfaces are with each other. In our exercise, \( \mu_k = 0.80 \). The coefficient depends on the materials in contact; rougher surfaces have higher coefficients. Kinetic friction always acts in the direction opposite to the displacement of the object, making it a resistive force. This is crucial because, to keep an object moving at a constant velocity, the applied force must overcome this kinetic friction.

Newton's Second Law describes how the velocity of an object changes when it is subjected to an external force. The law is often stated as \[ F = ma \], where \( F \) is the net force applied to an object, \( m \) is the mass of the object, and \( a \) is its acceleration. When an object moves at a constant velocity, like in our exercise, its acceleration \( a = 0 \), meaning the net force acting on it is zero. This implies that any horizontal force applied to maintain this constant velocity must exactly balance the kinetic friction force. Therefore, in calculating the necessary force to maintain motion, we use \( F = F_k \). In this problem, the stone's motion is analyzed under the condition of constant velocity, simplifying our focus to balancing forces without consideration of acceleration.

The normal force \( F_N \) is the perpendicular force exerted by a surface to support the weight of an object resting on it. It prevents objects from 'falling through' the surface and is always directed away from the surface. To calculate the normal force for an object resting on a horizontal surface, we use:

• \( F_N = mg \)

Here, \( m \) is the mass of the object (20 kg) and \( g \) is the acceleration due to gravity (9.8 m/s²). For the stone in our exercise, \[ F_N = 20 \text{ kg} \times 9.8 \text{ m/s}^2 = 196 \text{ N} \] The normal force directly affects the kinetic friction force, as the frictional force depends on \( F_N \). A larger normal force implies greater friction, making it harder to move the object. On uneven or inclined surfaces, the normal force calculation would involve angle considerations, but on a horizontal plane, it simplifies to \( mg \).

Calculating the force necessary to keep the stone moving involves understanding the balance of forces. Since the stone moves at a constant velocity, the applied force must counteract the kinetic friction force. Utilizing the given data:

• Mass of stone: 20 kg
• Coefficient of kinetic friction: 0.80
• Normal force: 196 N (calculated from \( F_N = mg \))

We use the relationship:

• Frictional force \( F_k = \mu_k \times F_N \)

Substituting the known values: \[ F_k = 0.80 \times 196 \text{ N} = 156.8 \text{ N} \] Thus, the horizontal force needed to maintain the stone's motion on the wet playa ground is 156.8 N. This calculation shows that to keep the stone moving steadily, we must apply a force specifically equal to the resistive frictional force. Changing conditions, such as a different coefficient of friction or changes in mass, would alter this needed force.