91Ó°ÊÓ

Understanding Newton's second law of motion is fundamental to analyzing most physical systems, including the motion of an object against friction. According to this law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is commonly expressed as the equation:
\[ F = ma \]
where \(F\) is the net force applied to the object, \(m\) is the mass of the object, and \(a\) is the acceleration.

In our case of a stockroom worker pushing a box across a surface, the law becomes particularly applicable. As the problem states that the box is moving with a constant velocity, we infer that the acceleration (\(a\)) is zero because constant velocity implies no change in speed or direction. This leads us to conclude that the net force is also zero, which means the pushing force exerted by the worker is equal and opposite to the force of friction. This is a direct application of Newton's second law.

When any additional force is removed, and only friction acts on the box, the second law tells us that the box will decelerate (negative acceleration) and eventually come to a stop. The amount of force due to friction, which causes this deceleration, can be calculated through the law as well.

The coefficient of kinetic friction (\( \( \mu_k \) \)) is a dimensionless scalar value that describes the ratio of the force of friction between two bodies and the force pressing them together. It has no units because it is a ratio of two forces. The force of kinetic friction can be calculated using the equation:
\[ F_{friction} = \mu_k F_{normal} \]
where \(F_{friction}\) is the force of friction and \(F_{normal}\) is the normal force, which in this case equals to the gravitational force on the box (\(m*g\)).

In our textbook exercise, the coefficient of kinetic friction is a crucial factor in determining how much force the worker must exert and how far the box will slide after the force is removed. The smaller the coefficient, the less force is needed to maintain the box's motion, and the further it will slide when the force is removed. The constant value of 0.20 in this case shows us the resistance to motion the box faces on the surface.

The scenario described in the problem demonstrates constant velocity motion, which is a key concept in kinematics. Constant velocity means that the object in question is moving in a straight line at an unchanging speed. Moreover, if velocity is constant, there is no acceleration (\(a = 0\)) since acceleration is defined as the rate of change of velocity.

Understanding this allows us to solve for the forces at play intuitively. For an object to move at constant velocity, the net force acting on the object must be zero; hence the forces acting on the object are in balance. In this context, the constant velocity tells us that the applied force by the worker is exactly balancing the kinetic frictional force.

Additionally, the concept is used when the force is removed. Even though the box will eventually come to a stop due to kinetic friction, the motion between the removal of the force and stopping is also a constant velocity motion, for each infinitesimal moment until the box stops, as the velocity decreases uniformly due to the constant frictional force.