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PE s = 1 2 kx 2 , size 12{"PE" rSub { size 8{s} } = { {1} over {2} } ital "kx" rSup { size 8{2} } } {}

where k size 12{k} {} is the spring’s force constant and x size 12{x} {} is the displacement from its undeformed position. The potential energy represents the work done on the spring and the energy stored in it as a result of stretching or compressing it a distance x size 12{x} {} . The potential energy of the spring PE s size 12{"PE" rSub { size 8{s} } } {} does not depend on the path taken; it depends only on the stretch or squeeze x size 12{x} {} in the final configuration.

An undeformed spring fixed at one end with no potential energy. (b) A spring fixed at one end and stretched by a distance x by a force F equal to k x. Work done W is equal to one half k x squared. P E s is equal to one half k x squared. (c) A graph of force F versus elongation x in the spring. A straight line inclined to x axis starts from origin. The area under this line forms a right triangle with base of x and height of k x. Area of this triangle is equal to one half k x squared.
(a) An undeformed spring has no PE s size 12{"PE" rSub { size 8{s} } } {} stored in it. (b) The force needed to stretch (or compress) the spring a distance x size 12{x} {} has a magnitude F = kx size 12{F= ital "kx"} {} , and the work done to stretch (or compress) it is 1 2 kx 2 size 12{ { {1} over {2} } ital "kx" rSup { size 8{2} } } {} . Because the force is conservative, this work is stored as potential energy ( PE s ) size 12{ \( "PE" rSub { size 8{s} } \) } {} in the spring, and it can be fully recovered. (c) A graph of F size 12{F} {} vs. x size 12{x} {} has a slope of k size 12{k} {} , and the area under the graph is 1 2 kx 2 size 12{ { {1} over {2} } ital "kx" rSup { size 8{2} } } {} . Thus the work done or potential energy stored is 1 2 kx 2 .

The equation PE s = 1 2 kx 2 size 12{"PE" rSub { size 8{s} } = { {1} over {2} } ital "kx" rSup { size 8{2} } } {} has general validity beyond the special case for which it was derived. Potential energy can be stored in any elastic medium by deforming it. Indeed, the general definition of potential energy    is energy due to position, shape, or configuration. For shape or position deformations, stored energy is PE s = 1 2 kx 2 size 12{"PE" rSub { size 8{s} } = { {1} over {2} } ital "kx" rSup { size 8{2} } } {} , where k size 12{k} {} is the force constant of the particular system and x size 12{x} {} is its deformation. Another example is seen in [link] for a guitar string.

A six-string guitar is placed vertically. The left-most string is plucked in the left direction with a force F shown by an arrow pointing left. The displacement of the string from the mean position is d. The plucked string is labeled P E sub string, to represent the potential energy of the string.
Work is done to deform the guitar string, giving it potential energy. When released, the potential energy is converted to kinetic energy and back to potential as the string oscillates back and forth. A very small fraction is dissipated as sound energy, slowly removing energy from the string.

Conservation of mechanical energy

Let us now consider what form the work-energy theorem takes when only conservative forces are involved. This will lead us to the conservation of energy principle. The work-energy theorem states that the net work done by all forces acting on a system equals its change in kinetic energy. In equation form, this is

W net = 1 2 mv 2 1 2 mv 0 2 = Δ KE. size 12{W rSub { size 8{"net"} } = { {1} over {2} } ital "mv" rSup { size 8{2} } - { {1} over {2} } ital "mv" rSub { size 8{0} rSup { size 8{2} } } =Δ"KE" "." } {}

If only conservative forces act, then

W net = W c , size 12{W rSub { size 8{"net"} } =W rSub { size 8{c} } } {}

where W c is the total work done by all conservative forces. Thus,

W c = Δ KE. size 12{W rSub { size 8{c} } =Δ"KE"} {}

Now, if the conservative force, such as the gravitational force or a spring force, does work, the system loses potential energy. That is, W c = Δ PE size 12{W rSub { size 8{c} } = +- D"PE"} {} . Therefore,

Δ PE = Δ KE size 12{ - Δ"PE"=Δ"KE"} {}

or

Δ KE + Δ PE = 0 . size 12{Δ"KE"+Δ"PE"=0} {}

This equation means that the total kinetic and potential energy is constant for any process involving only conservative forces. That is,

KE + PE = constant     or KE i + PE i = KE f + PE f } (conservative forces only),

where i and f denote initial and final values. This equation is a form of the work-energy theorem for conservative forces; it is known as the conservation of mechanical energy    principle. Remember that this applies to the extent that all the forces are conservative, so that friction is negligible. The total kinetic plus potential energy of a system is defined to be its mechanical energy    , ( KE + PE ) size 12{ \( "KE"+"PE" \) } {} . In a system that experiences only conservative forces, there is a potential energy associated with each force, and the energy only changes form between KE size 12{"KE"} {} and the various types of PE size 12{"PE"} {} , with the total energy remaining constant.

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
how far
Abubakar
cool u
Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
Abegail Reply
hello
BenJay
hi
Method
I am eliacin, I need your help in maths
Rood
how can I help
Sir
hmm can we speak here?
Amoon
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
Arundhati Reply
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
Jakoiya Reply
how to reduced echelon form
Solomon Reply
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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