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Near to the point where I am standing on the surface of Planet \(X\), the gravitational force on a mass \(m\) is vertically down but has magnitude \(m \gamma y^{2}\) where \(\gamma\) is a constant and \(y\) is the mass's height above the horizontal ground. (a) Find the work done by gravity on a mass \(m\) moving from \(\mathbf{r}_{1}\) to \(\mathbf{r}_{2}\), and use your answer to show that gravity on Planet \(X,\) although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height \(h\) above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it's clear what they are; don't worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height \(h\), how fast will it be going when it reaches the ground?

Short Answer

Expert verified
(a) Work is \( W = m \gamma (y_2^3 - y_1^3)/3 \) and energy is conservative; \( U(y) = -m \gamma y^3/3 + C \). (b) Forces: gravity (conservative), normal (non-conservative). (c) Speed at ground is \( \sqrt{2 \gamma h^3/3} \).

Step by step solution

01

Identifying Work Done by Gravity

The work done by gravity when a body moves from point \( \mathbf{r}_1 \) to \( \mathbf{r}_2 \) is given by the integral: \[ W = \int_{y_1}^{y_2} m \gamma y^2 \ dy \] where \(y_1\) and \(y_2\) are the initial and final heights, respectively. We need to evaluate this integral to find \( W \).
02

Calculating the Integral

To solve the integral for work, we compute: \[ W = m \gamma \int_{y_1}^{y_2} y^2 \ dy = m \gamma \left[ \frac{y^3}{3} \right]_{y_1}^{y_2} \] This simplifies to: \[ W = m \gamma \left( \frac{y_2^3}{3} - \frac{y_1^3}{3} \right) \].
03

Confirming Conservativeness of the Force

A force is conservative if the work done only depends on the initial and final positions. In this case, \( W = m \gamma \left( \frac{y_2^3}{3} - \frac{y_1^3}{3} \right) \) depends only on \( y_1 \) and \( y_2 \), confirming that the gravitational force on Planet \( X \) is conservative.
04

Finding the Potential Energy

For a conservative force, the potential energy, \( U \), associated with the gravitational force can be found such that \( U(y) = -\int m \gamma y^2 \ dy \). Evaluating this gives: \[ U(y) = -m \gamma \frac{y^3}{3} + C \] where \( C \) is a constant of integration.
05

Forces on a Bead on a Wire

When the bead is on the wire, it experiences a normal force perpendicular to the wire, and the gravitational force, which we've seen acts downward with magnitude \( m \gamma y^2 \). The gravitational force is conservative, and the normal force is non-conservative because it does no work.
06

Finding Speed at Ground Level

Using the conservation of energy, the total mechanical energy at the top \( E_{ ext{top}} = U(h) \), and at the bottom, \( E_{ ext{bottom}} = \frac{1}{2} m v^2 \). From conservation of energy principle: \[ U(h) = \frac{1}{2} m v^2 \] Since \( U(h) = -m \gamma \frac{h^3}{3} \), solving for \( v \) gives \[ v = \sqrt{\frac{2 \gamma h^3}{3}} \].

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Key Concepts

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

Gravitational Potential Energy
Imagine you're on the mysterious Planet X. Here, gravitational potential energy behaves in a unique way compared to Earth. When we discuss gravitational potential energy, we are referring to the energy an object possesses due to its position under the influence of gravity. On Planet X, the gravitational force acting on a mass is not constant like on Earth, but depends on the height above the ground, expressed as \( m \gamma y^2 \), where \( \gamma \) is a constant and \( y \) is the height.

This means that the gravitational potential energy, \( U(y) \), can be derived from the work done by gravity. For conservative forces, this work can be visualized by calculating the integral \(-\int m \gamma y^2 \ dy\). This results in a potential energy function \( U(y) = -m \gamma \frac{y^3}{3} + C \), where \( C \) is a constant determined by the context or boundary conditions of the problem.

Understanding this equation helps us know how energy changes with height in this peculiar environment.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that connects the work done by all forces acting on a body to its kinetic energy. On Planet X, we can see this principle in action when looking at the movement of an object under the influence of gravity. Here, the work done by gravitational force as an object moves from one point to another is calculated by integrating the force \( m \gamma y^2 \) over the distance traveled, which gives us the work \( W = m \gamma \left( \frac{y_2^3}{3} - \frac{y_1^3}{3} \right) \).

This result signifies that the work done is directly related to the change in the object's potential energy. When an object moves solely under the influence of this conservative force, the loss in potential energy equates to a gain in kinetic energy. This beautifully illustrates the work-energy principle, as the total energy in the system remains constant.
Conservation of Mechanical Energy
The conservation of mechanical energy involves understanding how energy transfers between different forms. In a system with only conservative forces, like gravitational force on Planet X, the total mechanical energy remains constant. This principle is observed when a bead is released from a height \( h \) and descends under the influence of gravity alone.

Initially, the bead has potential energy given by \( U(h) = -m \gamma \frac{h^3}{3} \). As it falls, this potential energy converts into kinetic energy until the bead reaches the ground level, where its kinetic energy is \( \frac{1}{2} m v^2 \). By setting the initial potential energy equal to the final kinetic energy, we can derive the bead's speed as it reaches the ground, calculated from the equation \( v = \sqrt{\frac{2 \gamma h^3}{3}} \).

By understanding the conservation of mechanical energy, students can grasp how energy is neither created nor destroyed, but simply transitions from one form to another within a closed system.

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Most popular questions from this chapter

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):(\mathbf{a}) f=\ln (r),\) (b) \(f=r^{n}\), (c) \(f=g(r),\) where \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) and \(g(r)\) is some unspecified function of \(r .\) [Hint: Use the chain rule.]

Verify that the gravitational force \(-G M m \hat{\mathbf{r}} / r^{2}\) on a point mass \(m\) at \(\mathbf{r},\) due to a fixed point mass \(M\) at the origin, is conservative and calculate the corresponding potential energy.

If a particle's potential energy is \(U(\mathbf{r})=k\left(x^{2}+y^{2}+z^{2}\right),\) where \(k\) is a constant, what is the force on the particle?

In one dimension, it is obvious that a force obeying Hooke's law is conservative (since \(F=-k x\) depends only on the position \(x,\) and this is sufficient to guarantee that \(F\) is conservative in one dimension). Consider instead a spring that obeys Hooke's law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass \(m\) at its other end, for instance.) Write down the force \(\mathbf{F}(\mathbf{r})\) exerted by the spring in terms of its length \(r\) and its equilibrium length \(r_{\mathrm{o}} .\) Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a x^{2}+b x y+c y^{2},(\mathbf{b}) g(x, y, z)=\sin \left(a x y z^{2}\right),(\mathbf{c}) h(x, y, z)=a e^{x y / z^{2}},\) where \(a, b,\) and \(c\) are constants. Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

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