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For a rock thrown downward with speed \(v_{0}\), use conservation of energy to find the form of the dimensionless function \(f\) in \(v / \sqrt{g h}=f\left(v_{0} / \sqrt{g h}\right)\)

Short Answer

Expert verified
The dimensionless function \(f\) in the relation \( \frac{v}{\sqrt{gh}} = f \left(\frac{v_0}{\sqrt{gh}}\right) \) using conservation of energy is given by: \( f(x) = \sqrt{x^2 + 2} \) where \(x = \frac{v_0}{\sqrt{gh}}\).

Step by step solution

01

Write down the initial and final energies

To use the conservation of energy, we first need to write down the initial and final energies of the rock. The initial energy includes the kinetic energy (due to its initial speed, v_0) and the potential energy (due to its height, h). The final energy includes only the kinetic energy (since it reaches the ground, its potential energy is considered zero). Initial energy (E_i) = Kinetic energy (K_i) + Potential energy (U_i) Final energy (E_f) = Kinetic energy (K_f)
02

Define energy terms with their respective formulas

Next, we write the formulas for kinetic and potential energy. Kinetic energy: \( K = \frac{1}{2} mv^2 \) Potential energy: \( U = mgh \) Using these formulas, we will rewrite the initial and final energy expressions. E_i = \( \frac{1}{2} mv_0^2 + mgh \) E_f = \( \frac{1}{2} mv^2 \)
03

Apply conservation of energy

According to the conservation of energy, the initial and final energies must be equal. So, \( \frac{1}{2} mv_0^2 + mgh = \frac{1}{2} mv^2 \) Now, we will solve the equation for the final velocity v.
04

Solve for final velocity v

First, divide both sides of the equation by mass m. \( \frac{1}{2} v_0^2 + gh = \frac{1}{2} v^2 \) Multiply both sides by 2. \( v_0^2 + 2gh = v^2 \) Now, take the square root of both sides to isolate v. \( v = \sqrt{v_0^2 + 2gh} \)
05

Find the form of the dimensionless function f

Using the relation provided in the problem statement \( \frac{v}{\sqrt{gh}} = f \left(\frac{v_0}{\sqrt{gh}}\right) \), we will find the function f. Substitute the expression of v which we got in Step 4 into the relation. \( \frac{\sqrt{v_0^2 + 2gh}}{\sqrt{gh}} = f \left(\frac{v_0}{\sqrt{gh}}\right) \) Now, let's make a substitution to simplify the equation. Let \( x = \frac{v_0}{\sqrt{gh}} \) . So, \( v_0 = x\sqrt{gh} \) Substitute this into the equation. \( \frac{\sqrt{(x\sqrt{gh})^2 + 2gh}}{\sqrt{gh}} = f(x) \) Simplify the expression inside the square root. \( \frac{\sqrt{x^2 gh + 2gh}}{\sqrt{gh}} = f(x) \) Factor out gh from the expression inside. \( \frac{\sqrt{gh(x^2 + 2)}}{\sqrt{gh}} = f(x) \) Cancel out the common factor \(\sqrt{gh}\) . \( \sqrt{x^2 + 2} = f(x) \) Thus, the form of the dimensionless function f is: \( f(x) = \sqrt{x^2 + 2} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is a form of energy that an object possesses due to its motion. It's a way of quantifying how much work the object can do as a result of its motion. The formula for kinetic energy (KE) is
\[\begin{equation}KE = \frac{1}{2}mv^2\end{equation}\] where m represents the mass of the object and v its velocity.
In the context of our exercise, we start with a rock that's thrown downward with a certain initial speed, v_0. This speed contributes to the rock's initial kinetic energy. As the rock falls, gravity does work on it, which changes the kinetic energy.
In the exercise, applying conservation of energy involves equating the sum of the rock's initial kinetic and potential energy to its final kinetic energy just before it hits the ground. The result of this calculation gives us insight into how the rock's velocity changes as it falls, an elegant way to observe energy transformation in action.
Potential Energy
Potential energy, on the other hand, is the stored energy of an object because of its position or state. For an object near Earth's surface, this is often gravitational potential energy which can be described by the formula:
\[\begin{equation}PE = mgh\end{equation}\] where m is the mass of the object, g is the acceleration due to gravity, and h is the height above a reference level.
When the rock from our problem is held at height h before its descent, it has potential energy — a result of gravity acting on it. As the rock falls and h decreases, this potential energy is converted into kinetic energy. The sum of the kinetic and potential energy at any point during the fall is a constant if we disregard air resistance and other non-conservative forces—in essence, illustrating the conservation of energy principle.
Dimensionless Function
A dimensionless function is a mathematical expression that provides a relationship between variables but does not involve any units of measurement. These functions are unitless, making it easier to compare different scenarios without worrying about the scale of measurement.

The function f in our exercise gives us a way to nondimensionally relate the velocity of the rock at any point during its fall to its initial velocity. By expressing both velocities in terms of \[\begin{equation}\sqrt{gh}\end{equation}\], we essentially normalize the velocities by gravitational acceleration and the height, which are common to all free-falling objects near Earth's surface.
The form of the dimensionless function f derived in the solution, \[\begin{equation}f(x) = \sqrt{x^2 + 2}\end{equation}\], encapsulates the transformed energies and shows the relationship in a simplified, general form. It's these dimensionless comparisons that often reveal the underlying physical principles, free from the particulars of a given situation.

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