An object falling under the pull of gravity is acted upon by...

An object falling under the pull of gravity is acted upon by a frictional force of air resistance. The magnitude of this force is approximately proportional to the speed of the object, which can be written as $f=b v .$ Assume $b=15 \mathrm{kg} / \mathrm{s}$ and $m=$ 50 $\mathrm{kg}$ . (a) What is the terminal speed the object reaches while falling? (b) Does your answer to part (a) depend on the initial speed of the object? Explain.

Key Concepts

Newton's Second Law in Equilibrium

Newton's second law, which states that the net force on an object equals its mass times its acceleration (F = ma), underpins the analysis of motion. At terminal velocity, the acceleration decreases to zero because the upward drag force equals the downward gravitational force, allowing for the derivation of the terminal speed without the influence of initial conditions.

Independence from Initial Conditions

The terminal speed is determined solely by the balance of forces acting on the object (gravitational force and drag force) and is independent of the object's initial speed. Regardless of the starting velocity, the object eventually reaches the same constant terminal speed when these forces balance.

Terminal Velocity

Terminal velocity is the constant speed attained by an object when the force of gravity is exactly balanced by the resistive drag force, resulting in zero net acceleration. This concept is critical in understanding motion under gravitational and resistive forces since it represents the steady-state (equilibrium) condition for a falling object.

Drag Force Proportional to Velocity

A drag force proportional to velocity implies that the resistive force opposing the motion increases linearly with the object's speed, often expressed as f = b*v, where b is a constant that characterizes the medium's resistance. This proportionality is essential in modeling how air resistance affects falling objects.

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