Potential Energy Graphs and Motion

Where are you the most stable? At the top of a building that is a thousand meters tall, or just above the surface on the ground floor? The answer seems logical and obvious. When you are on the ground floor, you have low potential energy. Low potential energy systems are stable systems. Most chemical reactions occur because they are thermodynamically favorable. This means that the process goes from a state of high energy to low energy, from being less stable to more stable. Here, we will discuss the relationship between potential energy and stability, as well as all the information that can be revealed from a system just by analyzing potential energy and graphs.

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In a Potential Energy vs Position graph, the total mechanical energy of the system would be represented by a ___ line in the graph.

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The potential energy will be always ___ the total energy.

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The correct expression for velocity in terms of kinetic energy and potential energy is ___.

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Points where the slope is ___ are considered equilibrium points.

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The locations with local maximums are locations of ___ equilibrium, while local minimums indicate locations of ___ equilibrium.

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The ___ is the negative of the slope when we visualize the potential energy as a function of the object's position in a graph.

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By definition, if the potential energy is increasing then ___

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If \(\frac{\operatorname dU}{\operatorname dx}\) is positive, ___.

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If we have a stable equilibrium point, ___.

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If two atoms are very close there is a ___ force, but at a distance of an atomic diameter there are ___ forces that bond them. This happens because at a distance of an atomic diameter the ___ is overcome by ___.

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The transition state is represented as a ___ in the potential energy as a function of the reaction coordinate graph. 

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In a Potential Energy vs Position graph, the total mechanical energy of the system would be represented by a ___ line in the graph.

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The potential energy will be always ___ the total energy.

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The correct expression for velocity in terms of kinetic energy and potential energy is ___.

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Points where the slope is ___ are considered equilibrium points.

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The locations with local maximums are locations of ___ equilibrium, while local minimums indicate locations of ___ equilibrium.

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The ___ is the negative of the slope when we visualize the potential energy as a function of the object's position in a graph.

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By definition, if the potential energy is increasing then ___

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If \(\frac{\operatorname dU}{\operatorname dx}\) is positive, ___.

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If we have a stable equilibrium point, ___.

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    Potential and Kinetic Energy Graphs

    By plotting the potential energy as a function of position, we can learn various physical properties of a system. First, we take a look at the most simple case, a free-falling object. We know that the change of potential energy \(\Delta{U}\) of this system will be given by the expression below,

    $$\Delta{U}=mg\Delta{y},$$

    where \(m\) is the object's mass in kilograms, \(\mathrm{kg}\), \(g\) is the acceleration due to gravity in meters per second squared, \(\frac{\mathrm m}{\mathrm s^2}\), and \(\Delta{y}\) is the object's position or altitude in meters, \(\mathrm{m}\).

    Potential Energy and Graphs Potential Energy vs Position Free-Fall StudySmarterFig. 1 - Potential energy as a function of position for an object that is free-falling. The force mg is the slope, above the slope, we have kinetic energy, and below we have potential energy. The total energy of the system is a constant horizontal line.

    We know that the total mechanical energy of an isolated system is conserved and is constant. It is represented by a horizontal line on the graph, as know that the potential energy \(U\) and the kinetic energy \(K\) are interchanging values such that the total mechanical energy \(E\) remains constant. For example, take a look at the point \(y_A\). The energy below the line corresponds to potential energy, while the energy above the line is kinetic energy.

    When you throw an object and it reaches its highest position, we know that its velocity will be zero as its motion changes direction and it begins to fall. In the graph, we see that when the object reaches \(y_max\), the potential energy equals the total energy of the system, meaning that the kinetic energy at this moment will be zero. The kinetic energy will always be zero or positive, such that the potential energy will be always equal to or less than the total energy,

    $$\begin{align*}K&=E-U,\\K&\geq 0,\\U&\leq E.\end{align*}$$

    The velocity of the object can also be determined by knowing its potential energy and the total energy of the system:

    $$\begin{align*}E&=K+U,\\E&=\frac12mv^2+U,\\v&=\pm\sqrt{\frac2m(E-U)}.\end{align*}$$

    Now we will consider the case of a spring-mass system. If we examine the energy of the system, we see that the potential energy looks like a parabola, as it depends on the square of the position,

    $$U=\frac12kx^2,$$

    where \(k\) is the spring constant that determines the stiffness of the spring in Newtons per meter, \(\frac{\mathrm N}{\mathrm m}\), and \(x\) is the object's displacement from the equilibrium position in meters \(\mathrm m\).

    Potential Energy and Graphs Potential Energy vs Position Spring-Mass system StudySmarterFig. 2 - Potential energy as a function of position for a spring-mass system. The force \(kx\) is the slope, above the slope, we have kinetic energy, and below we have potential energy. The turning points indicate points where the potential energy is maximum. This happens when the spring is fully compressed or stretched. The total energy of the system is a constant horizontal line.

    Stable and Unstable Equilibrium

    The points on a potential energy against position graph where the slope is zero are considered equilibrium points. The locations with local maximums are locations of unstable equilibrium, while local minimums indicate locations of stable equilibrium.

    Potential Energy and Graphs Stable and Unstable Equilibrium StudySmarter OriginalsFig. 3 - The graph of potential energy against position indicates the different types of stability.

    An equilibrium position for any object is one in which the object would be at rest naturally when there are no net forces on it. For example, an equilibrium position for a marble allowed to roll around the sides of a glass bowl would be at the bottom of the bowl. It would naturally come to rest at the bottom if no other external forces move the marble or bowl around. This position is known as a stable equilibrium.

    An object is in stable equilibrium if it is given a displacement from the equilibrium position and a force acts on it, in the opposite direction, to return it to that equilibrium position. The potential energy of the object increases momentarily, before returning to its value at equilibrium. Lift the marble slightly and release it, and it will return to the equilibrium position at the bottom of the bowl.

    An object is in unstable equilibrium if it is given a slight displacement from the equilibrium position and a force acts on it, in the same direction, pushing it further away from that equilibrium position. The potential energy of the object changes rapidly once displaced. Imagine the marble is made to rest on the lip of the bowl in a position of unstable equilibrium. The marble is then given a slight nudge, which will cause it to roll down the side of the bowl into the center, or oppositely, it's forced out of the bowl entirely if pushed in the other direction.

    An object is in neutral equilibrium if it is given a slight displacement from the equilibrium position and this does not affect its equilibrium. The potential energy of the object is unchanged after it is displaced. Imagine the marble has been displaced by a few centimeters on a flat, horizontal surface for an example of this.

    Potential Energy and Force in Graphs

    When we visualize the potential energy as a function of the object's position in a graph, we find that the force is the negative of the slope. This is due to the relationship between potential energy and work (recall that work is equal to the product of force and displacement):

    $$\begin{align*}\Delta U&=-W,\\\Delta U&=-F\Delta x,\\F&=-\frac{\Delta U}{\Delta x},\\F&=\lim\limits_{\Delta x\to 0}-\frac{\Delta U}{\Delta x},\\F&=-\frac{\operatorname dU}{\operatorname dx}.\end{align*}$$

    Alternatively, we can use calculus and integrals to find the expression for the potential energy. The following equation applies to all conservative forces, forces that only depend on the initial and final position of the object. In other words, conservative forces are independent of the path taken by the object,

    $$\Delta U=-\int_{x_i}^{x_f\;}\vec{F}_{cons}\cdot\operatorname d\vec{x}.$$

    If we know the expression for the potential energy we can determine the force applied. We know that the potential energy stored in a spring is \(U=\frac12kx^2\), so we can determine the force that causes the system to oscillate by taking the derivative of the potential energy with respect to the position, or in other words the rate of change of the potential energy with distance:

    $$\begin{align*}F&=-\frac{\operatorname dU}{\operatorname dx},\\F&=-\frac{\operatorname d({\displaystyle\frac12}kx^2)}{\operatorname dx},\\F&=-\frac12(2kx^{2-1}),\\F&=-kx.\end{align*}$$

    This corresponds to Hooke's Law, which experimentally proves the description of the motion for a spring-mass system.

    The relationship between the potential energy and force tells us a lot about the stability of the system. By definition, if the potential energy is increasing then \(\frac{\operatorname dU}{\operatorname dx}\) is positive, which means that the force would be negative. Similarly, if the potential energy is decreasing, then the force is positive. When \(x=0\), we see that the slope, the force, and the acceleration are zero. This is what an equilibrium point looks like. On either side of stable equilibrium points, there is a force that points back to equilibrium. We see that local minimums indicate locations of stable equilibrium.

    Potential Energy and Graphs Potential Energy vs Position to visualize equilibrium points StudySmarterFig. 4 - Visual representation of how forces point back to equilibrium around a point of stable equilibrium.

    On the other hand, if the force points away from the equilibrium point, there is an unstable equilibrium. Points of unstable equilibrium are located in a potential energy graph as local maximums. In the image below, we see the potential energy graph for a system that has stable and unstable equilibrium points.

    Potential Energy and Graphs Visualization the relation between the force and the potential energy StudySmarterFig. 5 - If the potential energy is increasing, then the force must be negative. If it is decreasing, the force must be positive. A point of unstable equilibrium is a maximum. We see that around unstable equilibrium points, the forces point away from the equilibrium point. In stable equilibrium points, forced point back to the equilibrium point.

    An object of mass \(m=4\;\mathrm{kg}\) has a potential energy function,

    $$U(x)=(x-2)-{(2x-3)}^3,$$

    where \(x\) is the displacement measured in meters and \(U\) is the potential energy measured in joules. The following graph is a sketch of the potential energy function.

    Potential Energy and Graphs Potential Energy vs Position Example StudySmarterFig. 6 - Potential energy as a function of the position to find equilibrium points.

    Questions

    (a) Determine the positions of points \(\text{A}\) and \(\text{B}\), the equilibrium points.

    (b) If the object is released from rest at point \(\text{B}\) with a small force applied, can it reach point \(\text{A}\) or \(\text{C}\)? Explain.

    (c) The particle is released from rest at point \(\text{C}\). Determine its speed as it passes point \(\text{A}\).

    Solutions

    (a) Points \(\text{A}\) and \(\text{B}\) are points where the slope/force is zero, so they are equilibrium points. At these points, the rate of change of the potential energy with distance will also be zero. First, we take the derivative of the potential energy with respect to the position,

    $$\begin{align*}\frac{\operatorname dU}{\operatorname dx}&=1-3{(2x-3)}^2(2),\\\frac{\operatorname dU}{\operatorname dx}&=-24x^2+72x-53.\end{align*}$$

    Now we look for the points where the rate of change of the potential energy with distance is zero:

    $$\begin{align*}\frac{\operatorname dU}{\operatorname dx}&=0,\\0&=-24x^2+72x-53,\\x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a},\\x&=\frac{-72\pm\sqrt{72^2-4(-24)(-53)}}{2(-24)},\\x&=\frac{-72\pm\sqrt{5,184-5,088}}{-48},\\x&=\frac{-72\pm\sqrt{96}}{-48},\\x&=\frac{-72\pm9.80}{-48},\\\mathrm x&=1.30\;\mathrm m\;\mathrm{and}\;1.70\;\mathrm m.\end{align*}$$

    (b) It is possible that if the object is released from rest at point \(\text{B}\) it can reach point \(\text{A}\). Point \(\text{B}\) is a point of unstable equilibrium, so the force applied in the correct direction could move the object away such that it gets to point \(\text{A}\).

    It's impossible for the object to go to point \(\text{C}\), as it would need to pass through point \(\text{A}\) before going to \(\text{C}\). Point \(\text{A}\) is a point of stable equilibrium and the forces at either side make the object return to equilibrium position \(\text{A}\), so it will never reach to \(\text{C}\).

    Additionally, at point \(\text{B}\) the system has total energy that is negative. Its velocity and therefore kinetic energy is zero at that point, which means that the total energy is equal to the potential energy. Point \(\text{C}\) has a higher and positive total energy, so this is another reason why it's impossible to go from point \(\text{B}\) to point \(\text{C}\) without work being done on the object.

    (c) We first find the potential energies at both points and use conservation of energy to find the particle's speed at point \(\text{A}\):

    $$\begin{align*}U_C\left(0.5\;\mathrm m\right)&=\left(\left(0.5\;\mathrm m\right)-2\right)-{\left(2\left(0.5\;\mathrm m\right)-3\right)}^3,\\U_C(0.5\;\mathrm m)&=6.5\;\mathrm J,\\U_A\left(1.3\;\mathrm m\right)&=\left(\left(1.3\;\mathrm m\right)-2\right)-{\left(2\left(1.3\;\mathrm m\right)-3\right)}^3,\\U_A\left(1.3\;\mathrm m\right)&=-0.636\;\mathrm J.\end{align*}$$

    We know that the particle is at rest, \(K_C=0\). By conservation of energy,

    $$\begin{align*}\cancelto{0}{K_C}+U_C&=K_A+U_A,\\6.5\mathrm J&=K_A-0.636\mathrm J,\\6.5\;\mathrm J&=-0.636\;\mathrm J\;+\frac12(4\;\mathrm{kg}){\mathrm v}_{\mathrm A}^2,\\v_A&=1.89\;\frac{\mathrm m}{\mathrm s}.\end{align*}$$

    Graph Between Potential Energy and Internuclear Distance

    Graphs of potential energy as a function of position are useful in understanding the properties of a chemical bond between two atoms. The energy of a system made up of two atoms depends on the distance between their nuclei. At large distances, the energy is zero, meaning that the two atoms are not bonded and are separate from each other. If the two atoms are very close, there is a repulsive force, but at a distance of an atomic diameter, there are attractive forces that bond them. This happens because, at a distance of an atomic diameter, the electromagnetic force is overcome by the strong nuclear force. The local minimum in the curve represents the distance where attractive and repulsive forces are balanced. This distance between atoms is called the bond length. The energy at this distance is called the bond energy.

    Potential Energy and Graphs Potential Energy as a function of internuclear distance between two atoms StudySmarterFig. 7 - Potential Energy as a function of the internuclear distance (picometers or \(10^{-12}\,\mathrm{m}\)) between two atoms. The minimum indicates the bond energy and the distance between atoms at the point where repulsive and attractive forces balance each other.

    Potential Energy and Reaction Coordinate Graph

    A reaction coordinate graph shows how the energy of a system changes during a chemical reaction. During a reaction, reactants transform into products. Energy must be added to the system in order to reach the transition state.

    Potential Energy and Graphs Potential Energy as a function of reaction coordinates StudySmarterFig. 8 - Potential Energy as a function of reaction coordinates. The difference between the energy of the reactant and the maximum is the activation energy. The difference between the reactant’s energy and the product’s energy is what indicates if a reaction is exothermic or endothermic.

    This transition state is represented as a maximum in the potential energy as a function of the reaction coordinate graph. The difference between the maximum and the energy of the reactant at the beginning of the reaction is called the activation energy \(E_act\). For a reaction to reach the transition state, bonds in the reactants must be stretched or broken. The energy required to provoke these changes is the activation energy. The difference between the reactant's energy and the product's energy is \(\triangle E\). This energy difference lets us know if the reaction is exothermic (releases heat) or endothermic (absorbs heat). In the above case, we see that the product's energy is lower than the reactant's energy, so the excess energy is released as heat and the reaction is exothermic.

    Potential Energy Graphs and Motion - Key takeaways

    • When we visualize the potential energy as a function of the object's position in a graph, we find that the force is the negative of the slope, \(\Delta U=-F\Delta x\).
    • The energy below the line corresponds to potential energy, while the energy above the line is kinetic energy.
    • The total energy is represented by a horizontal line in the graph, meaning that it is constant and conserved.
    • The relationship between the potential energy and force, \(F=-\frac{\operatorname dU}{\operatorname dx}\), tells us a lot about the stability of the system. By definition, if the potential energy is increasing then \(\frac{\operatorname dU}{\operatorname dx}\) is positive, which means that the force would be negative. Similarly, if the potential energy is decreasing, then the force is positive.
    • At a stable equilibrium point, on either side of the equilibrium point, there is a force that points back to equilibrium.
    • At an unstable equilibrium point, the force points away from the equilibrium point.
    Frequently Asked Questions about Potential Energy Graphs and Motion

    What is the potential energy in a graph?

    The potential energy is the energy related to the position of an object. A graph of Potential Energy vs Position will show how much potential energy an object has at different positions.

    How is potential energy related to motion?

    The potential energy is related to an object's position, while the kinetic energy is related to an object's motion. An object will be in motion and still have potential energy. At maximum potential energy the object is not in motion and the kinetic energy is zero.

    How do you read a potential energy graph in physics?

    The potential energy curve will depend on the expression for the position. For example, for a free-falling object the graph will be a line as it depends linearly on the position. For a spring-mass system, the graph will be a parabola as it depends on the square of the position. Points below the curve refer to potential energy, while points above the curve refer to kinetic energy. The horizontal line is the total energy of the system. Maximums in this graph will be points of unstable equilibrium, while minimums represent points of stable equilibrium.

    What is the slope of a potential energy graph?

    The negative of slope of a potential energy graph is the force that causes the object's motion.

    What is the main difference between the kinetic energy and potential energy graphs?

    The main difference is that the kinetic energy and potential energy are always interchanging such that the total energy of the system is constant. When a graph shows that potential energy is maximum, the same graph for kinetic energy will show a minimum, and vice versa.

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