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Two graphs of U in Joules on the vertical axis as a function of x in meters on the horizontal axis. In figure a, U of x is an upward opening parabola whose vertex is marked with a black dot and is at x=0, U=0. The region of the graph to the left of x=0 is labeled with a red arrow pointing to the right and the equation F equals minus the derivative of U with respect to x is greater than zero. The region of the graph to the right of x=0 is labeled with a red arrow pointing to the left and the equation F equals minus the derivative of U with respect to x is less than zero. Below the graph is a copy of the dot between copies of the red arrows and the force relations, F equals minus the derivative of U with respect to x is greater than zero on the left and F equals minus the derivative of U with respect to x is less than zero on the right. In figure b, U of x is an increasing function with an inflection point that is marked with a half filled circle at x=0, U=0. The region of the graph to the left of x=0 is labeled with a red arrow pointing to the left and the equation F equals minus the derivative of U with respect to x is less than zero. The region of the graph to the right of x=0 is also labeled with a red arrow pointing to the left and the equation F equals minus the derivative of U with respect to x is less than zero. Below the graph is a copy of the circle between copies of the red arrows, both of which point to the left, and the force relations, F equals minus the derivative of U with respect to x is less than zero on the left and F equals minus the derivative of U with respect to x is less than zero on the right.
Two examples of a potential energy function. The force at a position is equal to the negative of the slope of the graph at that position. (a) A potential energy function with a stable equilibrium point. (b) A potential energy function with an unstable equilibrium point. This point is sometimes called half-stable because the force on one side points toward the fixed point.

A practical application of the concept of stable equilibrium points is the force between two neutral atoms in a molecule. If two molecules are in close proximity, separated by a few atomic diameters, they can experience an attractive force. If the molecules move close enough so that the electron shells of the other electrons overlap, the force between the molecules becomes repulsive. The attractive force between the two atoms may cause the atoms to form a molecule. The force between the two molecules is not a linear force and cannot be modeled simply as two masses separated by a spring, but the atoms of the molecule can oscillate around an equilibrium point when displaced a small amount from the equilibrium position. The atoms oscillate due the attractive force and repulsive force between the two atoms.

Consider one example of the interaction between two atoms known as the van Der Waals interaction. It is beyond the scope of this chapter to discuss in depth the interactions of the two atoms, but the oscillations of the atoms can be examined by considering one example of a model of the potential energy of the system. One suggestion to model the potential energy of this molecule is with the Lennard-Jones 6-12 potential :

U ( x ) = 4 ε [ ( σ x ) 12 ( σ x ) 6 ] .

A graph of this function is shown in [link] . The two parameters ε and σ are found experimentally.

An annotated graph of E in Joules on the vertical axis as a function of x in meters on the horizontal axis. The Lennard-Jones potential, U, is shown as a blue curve that is large and positive at small x. It decreases rapidly, becomes negative, and continues to decrease until it reaches a minimum value at a position marked as the equilibrium position, F=0, then gradually increases and approaches E=0 asymptotically but remains negative. A horizontal green line of constant, negative value is labeled as E total. The green and blue E total and U curves cross at two places. The x value of the crossing to the left of the equilibrium position is labeled turning point, minus A, and the crossing to the right of the equilibrium position is labeled turning point, plus A. The region of the graph to the left of the equilibrium position is labeled with a red arrow pointing to the right and the equation F equals minus the derivative of U with respect to. The region of the graph to the right of the equilibrium position is labeled with a red arrow pointing to the left and the equation F equals minus the derivative of U with respect to x.
The Lennard-Jones potential energy function for a system of two neutral atoms. If the energy is below some maximum energy, the system oscillates near the equilibrium position between the two turning points.

From the graph, you can see that there is a potential energy well, which has some similarities to the potential energy well of the potential energy function of the simple harmonic oscillator discussed in [link] . The Lennard-Jones potential has a stable equilibrium point where the potential energy is minimum and the force on either side of the equilibrium point points toward equilibrium point. Note that unlike the simple harmonic oscillator, the potential well of the Lennard-Jones potential is not symmetric. This is due to the fact that the force between the atoms is not a Hooke’s law force and is not linear. The atoms can still oscillate around the equilibrium position x min because when x < x min , the force is positive; when x > x min , the force is negative. Notice that as x approaches zero, the slope is quite steep and negative, which means that the force is large and positive. This suggests that it takes a large force to try to push the atoms close together. As x becomes increasingly large, the slope becomes less steep and the force is smaller and negative. This suggests that if given a large enough energy, the atoms can be separated.

Questions & Answers

when I click on the links in the topics noting shows. what should I do.
Emmanuel Reply
can we regard torque as a force?
Emmanuel Reply
Torque is only referred a force to rotate objects.
SHREESH
thanks
Emmanuel
I need lessons on Simple harmonic motion
Emmanuel
what is the formulae for elastic modulus
Ark Reply
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
AFLAX Reply
At time to = 0 the current to the DC motor is reverse, resulting in angular displacement of the motor shafts given by angle = (198rad/s)t - (24rad/s^2)t^2 - (2rad/s^3)t^3 At what time is the angular velocity of the motor shaft zero
Princston Reply
3s
Basit
what is angular velocity
Sadiku
In three experiments, three different horizontal forces are ap- plied to the same block lying on the same countertop. The force magnitudes are F1 " 12 N, F2 " 8 N, and F3 " 4 N. In each experi- ment, the block remains stationary in spite of the applied force. Rank the forces according to (a) the
Sadiku
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
AFLAX
ty
Sharath
CD=5.83 n direction is NE
Ark
state Hooke's law of elasticity
Aarti Reply
Hooke's law states that the extension produced is directly proportional to the applied force provided that the elastic limit is not exceeded. F=ke;
Shaibu
thanks
Aarti
You are welcome
Shaibu
thnx
Junaid
what is drag force
Junaid
A backward acting force that tends to resist thrust
Ian
solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
Ian Reply
please help me solve this 👆👆👆
Ian
The weight inside the tank is lesser due to the buoyancy force by the water displaced. Weight of water displaced = His weight outside - his weight inside tank = 720 - 34.3 = 685.7N Now, the density of water = 997kg/m³ (this is a known value) Volume of water displaced = Mass/Density (next com)
Sharath
density or relative density
Shaibu
density
Ian
Upthrust =720-34.3=685.7N mass of water displayed = 685.7/g vol of water displayed = 685.7/g/997 hence, density of man = 720/g / (685.7/g/997) =1046.6 kg/m3
1046.8
R.d=weight in air/upthrust in water =720/34.3=20.99 R.d=density of substance/density of water 20.99=x/1 x=20.99g/cm^3
Shaibu
Kg /cubic meters
how please
Shaibu
Upthrust = 720-34.3=685.7N vol of water = 685.7/g/density of water = 685.7/g/997 so density of man = 720/g /(685.7/g/997) =1046.8 kg/m3
is there anyway i can see your calculations
Ian
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
okay i got it thanks
Ian
u welcome
Shaibu
HELLO kindly assist me on this...(MATHS) show that the function f(x)=[0 for xor=0]is continuous from the right of x->0 but not from the left of x->0
Duncan Reply
I do not get the question can you make it clearer
Ark
Same here, the function looks very ambiguous. please restate the question properly.
Sharath
please help me solve this problem.a hiker begins a trip by first walking 25kmSE from her car.she stops and sets her tent for the night . on the second day, she walks 40km in a direction 60°NorthofEast,at which she discovers a forest ranger's tower.find components of hiker's displacement for each day
Liteboho Reply
Take a paper. put a point (name is A), now draw a line in the South east direction from A. Assume the line is 25 km long. that is the first stop (name the second point B) From B, turn 60 degrees to the north of East and draw another line, name that C. that line is 40 km long. (contd.)
Sharath
Now, you know how to calculate displacements, I hope? the displacement between two points is the shortest distance between the two points. go ahead and do the calculations necessary. Good luck!
Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
Sharath
cheers!
Liteboho
cheers too
Duncan
hii
Lakshya
hii too
Siciid
haye
Siciid
yes
Siciid
yes
Lakshya
shggggg
Lakshya
you mean
Siciid
solution problems
Siciid
what is the definition of model
matthew Reply
please is there any way that i can understand physics very well i know am not support to ask this kind of question....
matthew
yes
Duncan
prove using vector algebra that the diagonals of a rhombus perpendicular to each other.
Baijnath Reply
A projectile is thrown with a speed of v at an angle of theta has a range of R on the surface of the earth. For same v and theta,it's range on the surface of moon will be
Roshani Reply
0
Keshav
what is soln..
Keshav
o
Duncan
Using some kinematics, time taken for the projectile to reach ground is (2*v*g*Sin (∆)) (here, g is gravity on Earth and ∆ is theta) therefore, on Earth, R = 2*v²*g*Sin(∆)*Cos(∆) on moon, the only difference is the gravity. Gravity on moon = 0.166*g substituting that value in R, we get the new R
Sharath
Some corrections to my old post. Time taken to reach ground = 2*v*Sin (∆)/g R = (2*v²*Sin(∆)*Cos(∆))/g I put the g in the numerator by mistake in my old post. apologies for that. R on moon = (R on Earth)/(0.166)
Sharath
state Newton's first law of motion
Awal Reply
Every body will continue in it's state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external force.
Kumaga
if you want this to become intuitive to you then you should state it
Shii
changing the state of rest or uniform motion of a body
koffi
if a body is in rest or motion it is always rest or motion, upto external force appied on it. it explains inertia
Omsai
what is a vector
smith
a ship move due north at 100kmhr----1 on a River flowing be due east on at 25kmperhr. cal the magnitude of the resultant velocity of the ship.
Emmanuel Reply
The result is a simple vector addition. The angle between the vectors is 90 degrees, so we can use Pythagoras theorem to get the result. V magnitude = sqrt(100*100 + 25*25) = 103.077 km/hr. the direction of the resultant vector can be found using trigonometry. Tan (theta) = 25/100.
Kumar
103.077640640442km/h
Peter
state Newton's first law of motion
Kansiime Reply
An object continues to be in its state of rest or motion unless compelled by some external force
Alem
First law (law of inertia)- If a body is at rest, it would remain at rest and if the body is in the motion, it would be moving with the same velocity until or unless no external force is applied on it. If force F^=0 acceleration a^=0 or v^=0 or constant.
Govindsingh
Practice Key Terms 3

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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