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(a) Consider a mass \(m\) in a uniform gravitational field \(\mathbf{g},\) so that the force on \(m\) is \(m \mathbf{g},\) where \(\mathbf{g}\) is a constant vector pointing vertically down. If the mass moves by an arbitrary path from point 1 to point \(2,\) show that the work done by gravity is \(W_{\mathrm{grav}}(1 \rightarrow 2)=-m g h\) where \(h\) is the vertical height gained between points 1 and 2. Use this result to prove that the force of gravity is conservative (at least in a region small enough so that \(\mathrm{g}\) can be considered constant). (b) Show that, if we choose axes with \(y\) measured vertically up, the gravitational potential energy is \(U=m g y\) (if we choose \(U=0\) at the origin).

Short Answer

Expert verified
Work done by gravity: \( W_{\mathrm{grav}} = -mgh \). Gravity is conservative; potential energy \( U = mgy \).

Step by step solution

01

Understand Work Done by Gravity

The work done by a force is given by the integral of the force along the path of motion. For gravity, which acts downwards, the work done when moving from point 1 to point 2 is calculated using the formula: \[ W = \int_{1}^{2} \mathbf{F} \cdot d\mathbf{r} \]where \( \mathbf{F} = m \mathbf{g} \) is the force due to gravity and \( d\mathbf{r} \) is the differential path element.
02

Expressing the Force and Path Integral

Since \( \mathbf{g} \) is a constant vertical vector, we can express it and the differential path vector in component form. Let the vertical component of the path be \( y \) with changes \( dy \). Then the work integral becomes:\[ W_{\mathrm{grav}} = \int_{1}^{2} -m g\, dy \]taking into account that gravity does negative work when moving against its direction.
03

Evaluate the Integral

Integrate the expression:\[ W_{\mathrm{grav}} = -mg \int_{y_1}^{y_2} dy = -mg[y]_1^2 = -mg(y_2 - y_1) \]Here, \( h = y_2 - y_1 \) is the height change in the vertical direction. Therefore, the work done by gravity simplifies to:\[ W_{\mathrm{grav}} = -mgh \]
04

Prove Gravity is Conservative

A force is conservative if the work done in moving between two points is independent of the path taken. Since the work done \( W_{\mathrm{grav}} = -mgh \) depends only on the vertical displacement \( h \), and not on the specific path, gravity is a conservative force.
05

Define Gravitational Potential Energy

Define potential energy \( U \) such that the work done by gravity equals the negative change in potential energy. Therefore, for a small displacement:\[ dU = -dW_{\mathrm{grav}} = mg\, dy \]Integrating this for a change from point at height \( y_1 \) to height \( y_2 \):\[ U = mg(y_2 - y_1) \]Choosing \( U=0 \) at the origin \( y=0 \) leads to \( U = mgy \).
06

Conclusion

We have shown that for a mass \( m \) in a uniform gravitational field, the work done between two points depends solely on the vertical height difference, thus demonstrating it's a conservative force. The gravitational potential energy is defined as \( U = mgy \) when \( y \) is measured vertically upwards.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is a form of energy related to the position of a mass in a gravitational field. When a mass is located at a certain height above the ground, it possesses potential energy due to the gravitational force. This energy is given by the expression:\[ U = mgy \]where:
  • \( U \) is the gravitational potential energy,
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity, and
  • \( y \) is the height of the object above a reference point.
To calculate this energy, it's crucial to define a reference point where the potential energy is zero. Commonly, this reference point is chosen at ground level or the lowest point in a given problem. The significance of gravitational potential energy is that it represents the work that can be done by gravity if the object were allowed to fall freely back to the reference height.
Work Done by Gravity
The work done by gravity refers to the energy transferred by the gravitational force when an object moves within a gravitational field. When a mass travels from one point to another under the influence of gravity, the work done by gravity can be calculated using the formula:\[ W_{\text{grav}} = -mgh \]where:
  • \( W_{\text{grav}} \) is the work done by gravity,
  • \( m \) is the mass of the object,
  • \( g \) is the gravitational acceleration, and
  • \( h \) is the change in height between the two points.
In this expression, the negative sign indicates that gravity does negative work when an object moves upward, because it opposes the motion direction. Conversely, when an object falls downward, gravity does positive work. Importantly, the work done by gravity only depends on the initial and final heights— not on the path taken—signifying that gravity is a conservative force. Such forces allow potential energy to be easily converted into kinetic energy and vice versa.
Uniform Gravitational Field
A uniform gravitational field is a theoretical model where gravitational force is constant in magnitude and direction throughout the space in question. This model greatly simplifies calculations and is applicable in many practical scenarios, like near the surface of the Earth. Here, the gravitational field \( \mathbf{g} \) is a vector pointing orthogonally downward with constant magnitude:\[ \mathbf{g} = 9.81 \, \text{m/s}^2 \]Understanding this field provides insights into the behavior of objects within it. A few key points to consider:
  • The force due to gravity on a mass \( m \) is \( m\mathbf{g} \).
  • The potential energy depends only on vertical height \( y \), thus given by \( U = mgy \).
  • In a uniform field, all points at the same height have the same potential energy, regardless of the horizontal position.
This approximation allows for seamless integration into problems requiring analysis of motion and energy where small variations in gravity are negligible, demonstrating fundamental concepts like conservation of energy.

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Most popular questions from this chapter

A particle of mass \(m_{1}\) and speed \(v_{1}\) collides with a second particle of mass \(m_{2}\) at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that \(m_{1} \ll m_{2}\) and that \(m_{2} \ll m_{1}\).

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):\) (a) \(f=x^{2}+z^{3} .\) (b) \(f=k y\), where \(k\) is a constant. (c) \(f=r \equiv \sqrt{x^{2}+y^{2}+z^{2}} .\) [Hint: Use the chain rule.] (d) \(f=1 / r\).

(a) Consider an electron (charge \(-e\) and mass \(m\) ) in a circular orbit of radius \(r\) around a fixed proton (charge \(+e\) ). Remembering that the inward Coulomb force \(k e^{2} / r^{2}\) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\). (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius \(r\) around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy \(T_{2} .\) When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius \(r^{\prime} .\) (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy \(E\) long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables \(T_{2}, r,\) and \(r^{\prime}\).

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.

Consider the Atwood machine of Figure \(4.15,\) but suppose that the pulley has radius \(R\) and moment of inertia \(I\). (a) Write down the total energy of the two masses and the pulley in terms of the coordinate \(x\) and \(\dot{x}\). (Remember that the kinetic energy of a spinning wheel is \(\frac{1}{2} I \omega^{2}\).) (b) Show (what is true for any conservative one-dimensional system) that you can obtain the equation of motion for the coordinate \(x\) by differentiating the equation \(E=\) const. Check that the equation of motion is the same as you would obtain by applying Newton's second law separately to the two masses and the pulley, and then eliminating the two unknown tensions from the three resulting equations.

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