Question

A mass of \(0.25 \mathrm{kg}\) is dropped from rest in a medium offering a resistance of \(0.2|v|,\) where \(v\) is measured in \(\mathrm{m} / \mathrm{sec}\). $$ \begin{array}{l}{\text { (a) If the mass is dropped from a height of } 30 \mathrm{m} \text { , find its velocity when it hits the ground. }} \\ {\text { (b) If the mass is to attain a velocity of no more than } 10 \mathrm{m} / \mathrm{sec} \text { , find the maximum height }} \\ {\text { from which it can be dropped. }} \\ {\text { (c) Suppose that the resistive force is } k|v| \text { , where } v \text { is measured in } \mathrm{m} / \mathrm{sec} \text { and } k \text { is a }} \\ {\text { constant. If the mass is dropped from a height of } 30 \mathrm{m} \text { and must hit the ground with a }} \\ {\text { velocity of no more than } 10 \mathrm{m} / \mathrm{sec} \text { , determine the coefficient of resistance } k \text { that is required. }}\end{array} $$

Short answer

Question: A small object is released from a height of 30 meters above the ground. The resistive force experienced while falling is proportional to the velocity and is given by 0.2|v|. Where v is the downward velocity of the object and m is its mass. Calculate (a) the velocity of the object when it hits the ground, (b) the maximum height from which it can be dropped so that the velocity never exceeds 10 m/s during its fall, and (c) the coefficient of resistance if the object cannot hit the ground with a velocity greater than 10 m/s. Answer: (a) Calculate the velocity when it hits the ground: v = ______ m/s (b) Maximum height for velocity not exceeding 10 m/s: h_max = ______ m (c) Coefficient of resistance for velocity no greater than 10 m/s: k = ______

Step-by-step solution

Write the Newton's second law

We can use Newton's second law to form an equation involving the mass, resistive force, and gravitational force. The equation we get is: $$ m \frac{dv}{dt} = mg - 0.2|v| $$

Solve the differential equation

Now, we have to solve the differential equation: $$ \frac{dv}{dt} = g - \frac{0.2}{m}|v| $$ We can rewrite the equation by separating the variables: $$ \int_0^v \frac{d v'}{g-\frac{0.2}{m}|v'|} = \int_0^t d t' $$ Solving this integral, we get: $$ \frac{m}{0.2} \ln \left(\frac{g+\frac{0.2}{m}v}{g-\frac{0.2}{m}v}\right) = t+C $$

Determine the constant of integration

To determine the constant of integration C, we know that at t=0, v=0. Substituting these values, we find: $$ C = - \frac{m}{0.2} \ln(1) = 0 $$ So, the equation for velocity as a function of time is: $$ \frac{m}{0.2} \ln \left(\frac{g+\frac{0.2}{m}v}{g-\frac{0.2}{m}v}\right) = t $$

Solve for velocity when it hits the ground (part a)

We first need to compute the time it takes to fall from a height of 30 m. To do this, we can use the equation of motion for displacement: $$ y = h_0 - \int_0^t v dt' $$ Replacing the equation for velocity obtained in Step 3, we get: $$ y = h_0 - \int_0^t \left[g-\frac{0.2}{m}|v|\right] dt' $$ Now, we need to solve the integral for when y = 0, and use the time obtained to find the velocity v when it hits the ground.

Solve for maximum height to attain a velocity no more than 10 m/s (part b)

Using the velocity expression from step 3, we need to solve for the height when the velocity is 10 m/s: $$ h_{max} = h_0 - \int_0^{t_{max}} \left[g-\frac{0.2}{m}|v|\right] dt' $$ By finding the time when the velocity is 10 m/s, we can compute the maximum height required.

Determine the coefficient of resistance needed (part c)

We are given that the resistive force is in the form of k|v|, and we want to find the value of k so that, when dropped from a height of 30 m, the object cannot hit the ground with a velocity of more than 10 m/s. In this case, we need to re-write Newton's second law with the new resistive force form and solve the differential equation: $$ m\frac{dv}{dt} = mg - k|v| $$ Solving this equation and using the height restriction, we can determine the value of k required to achieve the desired velocity limit.

Key concepts

Newton's Second Law

Newton's Second Law of Motion is a core principle of physics. It states that the force acting on an object is equal to the mass of the object times its acceleration: \[ F = ma \].

This law helps us understand how forces influence motion. In the context of the problem, we consider two main forces:

• Gravitational Force: Acts downward and is given by \( mg \), where \( m \) is mass and \( g \) is the gravitational acceleration (\( 9.8 \ m/s^2 \) on Earth).
• Resistive Force: Acts opposite to motion and is dependent on velocity.

The problem uses Newton’s second law to form the equation:\[ m \frac{dv}{dt} = mg - 0.2|v| \]
This equation helps us simulate how the object falls under gravity while encountering air resistance.

Resistive Force

A resistive force is a force that always opposes the motion of an object. It acts to slow down or stop the motion.

In this exercise, the resistive force is given by \( 0.2|v| \), where the coefficient \( 0.2 \) represents the resistance provided by the medium, like air or water.

This force opposes the gravitational pull and is directly proportional to the speed \( v \) of the object:

• The faster the object moves, the greater the resistance it encounters.
• The resistive force depends on direction, as it is proportional to the magnitude of velocity \( |v| \).

To determine how this force affects motion, it is crucial to factor it into the differential equation derived from Newton’s second law.

Velocity Function

The velocity function describes how the speed of an object changes over time. It is crucial to predict when and how the object will reach certain speeds.

From the simplified differential equation:\[ \frac{dv}{dt} = g - \frac{0.2}{m}|v| \] We separate variables to solve for velocity \( v \): \[ \int_0^v \frac{d v'}{g-\frac{0.2}{m}|v'|} = \int_0^t d t' \]
This leads to the solution:\[ \frac{m}{0.2} \ln \left(\frac{g+\frac{0.2}{m}v}{g-\frac{0.2}{m}v}\right) = t+C \]
The velocity function provides insight into how external forces, like gravity and resistance, impact an object's speed over time.

Integral of Motion

The integral of motion involves calculating the changes in an object's position over time. It helps us understand how the speed affects the object's trajectory.

In the exercise, when finding the time to hit the ground, the position \( y \) is given by integrating the velocity function:\[ y = h_0 - \int_0^t v \, dt \]

This allows us to track how the object accelerates or decelerates as it moves:

• By integrating the velocity function, we determine the object's position over time.
• Setting \( y = 0 \) gives the time required to reach the ground from a height \( h_0 \).

Using integrals in motion analysis links the mathematics of velocity and calculus to real-world trajectory planning.