91Ó°ÊÓ

Kinetic friction is the resistive force that acts against the movement of an object sliding across a surface. It depends on two main factors:

• The coefficient of kinetic friction (\( \mu_k \))
• The normal force (\( N \))

In our exercise, the crate is sliding on the floor, so we must address the kinetic friction acting on it. This frictional force can be calculated using the formula:\[ f_k = \mu_k \times N \]Given that the coefficient of kinetic friction is 0.200, we need the normal force to ascertain the frictional force. Friction acts to oppose the pushing force (\( \overrightarrow{\mathbf{P}} \)) applied on the box. By balancing these forces, we achieve the condition where the work done by the applied force and the frictional force are equal but opposite, leading to zero net work.

The normal force is the support force exerted upon an object in contact with another stable object. In this case, it's the force that the floor exerts upward on the crate. It is crucial for calculating kinetic friction as it is one of the determinants of the frictional force.

In the scenario where a force (\( \overrightarrow{\mathbf{P}} \)) is applied at an angle to push the crate, the normal force isn't just the weight of the crate (mass times gravitational acceleration). Instead, it is adjusted by the vertical component of the pushing force. Mathematically, the normal force can be represented as:\[ N = mg + P \sin(\theta)\]Here:

• \( m \) is the mass of the crate (100 kg)
• \( g \) is the acceleration due to gravity (9.8 m/s²)
• \( P \) is the magnitude of the applied force
• \( \theta \) is the angle of the applied force with the horizontal (30°)

By accurately gauging all these elements, we can determine the normal force, critical for understanding the friction at play.

The work-energy principle states that the total work done on an object is equal to its change in kinetic energy. However, in this problem, the focus is on ensuring that the net work done by the force \( \overrightarrow{\mathbf{P}} \) and kinetic friction is zero.

For the net work done to be zero, the work done by the applied force \( \overrightarrow{\mathbf{P}} \) in moving the crate must exactly counterbalance the work done by the frictional force resisting the motion. Therefore, the condition to be achieved is:\[ P \cos(\theta) \times d - f_k \times d = 0\]Where:

• \( d \) is the distance over which the forces are applied. As long as the forces are constant over this distance, we can cancel it out from both terms.
• \( P \cos(\theta) \) represents the horizontal component of the applied force.
• \( f_k \) is the frictional force defined earlier.

This balancing act ensures no net gain or loss in the crate's kinetic energy, maintaining stable motion.