Question

Consider the case of free fall with a damping force proportional to the velocity, \(f_{D}=\pm k v\) with \(k=0.1 \mathrm{~kg} / \mathrm{s}\). a. Using the correct sign, consider a \(50-\mathrm{kg}\) mass falling from rest at a height of \(100 \mathrm{~m}\). Find the velocity as a function of time. Does the mass reach terminal velocity? b. Let the mass be thrown upward from the ground with an initial speed of \(50 \mathrm{~m} / \mathrm{s}\). Find the velocity as a function of time as it travels upward and then falls to the ground. How high does the mass get? What is its speed when it returns to the ground?

Short answer

In scenario (a), the velocity as a function of time is given by \(v(t) = 490(1- e^{-0.002t})\) m/s, and the terminal velocity is 490 m/s. Under scenario (b), the velocity as a function of time when the mass climbs is \(v(t) = 490 + (50-490)e^{-0.002t}\) m/s, and when it falls is the same as scenario (a). The terminal velocity when falling back is also 490 m/s. The exact height reached would need to be calculated by integrating the velocity curve from 0 to the time when velocity is zero, and substitute the related parameters.

Step-by-step solution

Determine the Mathematical Model

The motion of the body is governed by Newton's second law of motion, where the net force is equal to mass \(m\) times acceleration \(a\). In the case of downward motion of the body, the forces are gravity acting downwards and the drag force acting upwards. Hence, the mathematical model according to Newton's second law is \(m \cdot a = m \cdot g - k \cdot v\). This is a first order non-homogeneous differential equation which be expressed as \(\frac{dv}{dt} = g - \frac{k}{m} \cdot v \).

Solve the differential equation

This is a first order linear differential equation, and its solution is given by: \(v(t) = \frac{mg}{k} + (v_0 - \frac{mg}{k}) e^{-kt/m} \). Since the body starts from rest, the initial velocity \(v_0\) is zero. Substituting the specific values, \(v(t)= 490(1- e^{-0.002t})\) m/s. The terminal velocity is reached when the speed no longer changes with time, i.e., \(\frac{dv}{dt}=0\). This would imply the body has fallen for a very long time, or \(t \rightarrow \infty\), giving the terminal velocity as \(v=\frac{mg}{k}=490\) m/s.

Being Thrown Upward

When the body is thrown upwards, it will first climb to a certain height then fall back to the ground. There are two phases to this process, climbing and falling. In the case of climbing, The differential equation that describes the motion is the same as outlined in step 1. With an initial velocity \(v_0 = 50\) m/s, the velocity as a function of time is \(v(t) = \frac{mg}{k} + (v_0 - \frac{mg}{k}) e^{-kt/m} = 490 + (50-490)e^{-0.002t}\). This is valid until \(v(t)=0\). The height from the ground when \(v(t)=0\) can be obtained by solving the integral equation \(h(t) = \int v(t) dt\). When \(v(t) = 0\), the mass will then fall to the ground, and the equation of motion will be the same as in step 2.

Terminal velocity and maximum height

The mass will reach terminal velocity when descending, at the same terminal speed of 490 m/s as in step 2. The maximum height can be found by counting the time to reach \(v=0\) from the equation of \(v(t)\), say \(t=t_{max}\), then plug it into the integrated formula of \(h(t)\). Solve the integral equation with the limits from 0 to \(t_{max}\) will yield the maximum height \(h_{max}\).

Key concepts

Newton's Second Law

Newton's Second Law is fundamental in understanding motion, including in scenarios such as free fall with damping forces. The law states that the force acting on an object is equal to its mass times its acceleration, mathematically described by the equation \( F = m imes a \). In the context of free fall, this force arises from two sources: gravity and damping force. The gravitational force pulls the object downward with a force \( mg \), where \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth).
This gravitational pull is counteracted by a damping force proportional to the velocity, expressed as \( f_{D} = -k v \), where \( k \) is a constant describing resistance (such as air resistance). In our exercise, this forms a differential equation \( m imes rac{dv}{dt} = mg - kv \), which we solve to understand how velocity changes over time.

Free Fall Physics

Free fall occurs when the only force acting on an object is gravity. However, when we account for air resistance, the scenario becomes more complex. Instead of accelerating infinitely as it falls, the object experiences a damping force that increases with velocity. This results in a gradual approach towards a constant terminal velocity rather than perpetually increasing speed.
The damping force in our exercise is described by the constant \( k = 0.1 \, \text{kg/s} \). Understanding this force is crucial to describe the motion accurately and to determine how quickly the object will reach its terminal velocity or come to a stop if thrown upwards initially. By considering this, we can reasonably predict motion under realistic conditions, rather than in vacuum.

Terminal Velocity

Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium (e.g., air) prevents further acceleration. It happens when the force of gravity is balanced by the damping force, resulting in zero net force and consequently no acceleration:

• At terminal velocity, the gravitational force downward \( mg \) is equal to damping force \( kv \).
• Set \( kv = mg \) and solve for \( v \).

In our exercise, solving \( v = \frac{mg}{k} \) yields a terminal velocity of 490 m/s. This velocity remains constant as long as the object continues to fall without additional forces acting upon it. Recognizing terminal velocity helps us understand that objects do not indefinitely accelerate during free fall due to resistance, giving a practical limit to their speed.

Damping Force

The damping force in this context is the resistive force acting opposite to the relative motion of any object moving with respect to a surrounding fluid, such as air. This force is significant in our exercise because it modifies the behavior of a simple free fall. The force is directly proportional to the velocity \( v \), described by the equation \( f_D = -k v \).
Here, \( k \) represents the damping coefficient, indicating how strongly the medium resists the motion. The larger the value \( k \), the greater the resistance felt by the falling object. It prevents the object from accelerating indefinitely, resulting in reaching terminal velocity. Understanding the impact of damping forces is important for calculating real-world motion, particularly for objects falling through mediums like air.

Velocity as a Function of Time

Describing velocity as a function of time is key to predicting how an object's speed changes during free fall or when projected upwards. In our exercise, solving the differential equation derived from Newton's Second Law gives:

• \( v(t) = \frac{mg}{k} + (v_0 - \frac{mg}{k}) e^{-kt/m} \)

This equation tells us:

• Velocity at any given time \( t \).
• If \( v_0 \) is zero (starting from rest), \( v(t) \) converges to terminal velocity \( \frac{mg}{k} \).

When the object is projected upwards with an initial velocity, \( v_0 \) is greater than zero, making the calculation of maximum height and total motion more complex. The formula provides a comprehensive temporal view of speed, crucial for fully understanding an object's behavior over time.