Chapter 4: Problem 26
A mass \(m\) is in a uniform gravitational field, which exerts the usual force \(F=m g\) vertically down, but with \(g\) varying with time, \(g=g(t) .\) Choosing axes with \(y\) measured vertically up and defining \(U=m g y\) as usual, show that \(\mathbf{F}=-\nabla U\) as usual, but, by differentiating \(E=\frac{1}{2} m v^{2}+U\) with respect to \(t,\) show that \(E\) is not conserved.
Short Answer
Step by step solution
Define Potential Energy
Calculate the Gradient of Potential Energy
Set Up Force Equation
Differentiate the Energy Expression
Differentiate the Kinetic Energy
Differentiate the Potential Energy
Combine Energy Derivatives
Simplify and Conclude Non-conservation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Potential Energy
This formula assumes that the gravitational force acts vertically downward. Thus, we define the positive direction for potential energy as vertically upward.
An interesting aspect of this problem is that \( g \), the gravitational field strength, varies with time. This means the potential energy changes as time progresses, even if the position, \( y \), remains constant. Since potential energy depends not only on position but also on the strength of the gravitational field, the time-varying \( g \) directly impacts the potential energy value.
Gravitational Field
Usually, we consider gravitational field strength, denoted by \( g \), to be a constant value. However, in this exercise, the gravitational field strength changes over time, denoted as \( g(t) \).
The force exerted by a gravitational field on a mass is calculated using the formula \( F = mg \), meaning the force depends on both the mass and the field strength. In our situation, this force remains directed downward, parallel to the traditional setup where the gravitational field is uniform but constant.
Energy Conservation
In our problem, the total energy \( E \) is given by \( E = \frac{1}{2} mv^2 + U \), which includes both the kinetic energy and the potential energy.
Differentiating this expression with respect to time reveals that \( \frac{dE}{dt} \) is not zero unless \( \frac{dg}{dt} = 0 \). This insight tells us that energy is not conserved if \( g \) varies with time.
- When \( g \) is constant over time, potential and kinetic energies change in a way that keeps total energy the same.
- If \( g \)'s value changes, our system gains or loses energy in response to how \( g(t) \) changes, thus leading to energy non-conservation.