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| {{Redirect|Mechanical work|other uses of "Work" in physics|Work (electrical)|and|Work (thermodynamics)}}
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| {{Use British English|date=April 2013}}
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| {{Infobox physical quantity
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| | bgcolour={default}
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| | name = Work
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| | image = [[File:Baseball pitching motion 2004.jpg|250px]]
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| | caption = A [[baseball]] [[pitcher]] does positive work on the ball by applying a force to it over the distance it moves while in his grip.
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| | basequantities = 1 [[kilogram|kg]]·[[metre|m]]<sup>2</sup>/[[second|s]]<sup>2</sup>
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| | unit = [[joule]] (J)
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| | symbols = ''W''
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| | derivations = ''W'' = '''[[force (physics)|F]]''' · '''[[distance|d]]''' <br />
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| ''W'' = ''[[torque|τ]]'' ''[[angle|θ]]''
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| }}
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| {{Classical mechanics|cTopic=Fundamental concepts}}
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| In [[physics]], a [[force]] is said to do '''work''' when it acts on a body, and there is a displacement of the point of application in the direction of the force. The force does not need to cause the displacement. For example, when you lift a suitcase from the floor, there are two forces that do work: the [[normal force]] by your hand and the gravitational force.
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| The term ''work'' was introduced in 1826 by the French mathematician [[Gaspard-Gustave Coriolis]]<ref>{{cite book | last = Jammer | first = Max | title = Concepts of Force | publisher = Dover Publications, Inc. | year = 1957 | isbn = 0-486-40689-X |page=167; footnote 14 |url=http://books.google.com/?id=CZtEBcmOe6gC&printsec=frontcover#PPA167,M1 }}</ref><ref>Coriolis, Gustave. (1829). ''Calculation of the Effect of Machines, or Considerations on the Use of Engines and their Evaluation'' (''Du Calcul de l'effet des Machines, ou Considérations sur l'emploi des Moteurs et sur Leur Evaluation''). Paris: Carilian-Goeury, Libraire.</ref> as "weight ''lifted'' through a height", which is based on the use of early steam engines to lift buckets of water out of flooded ore mines. The [[SI unit]] of work is the newton-metre or [[joule]] (J).
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| The work done by a constant force of magnitude ''F'' on a point that moves a displacement (not distance) ''d'' in the direction of the force is the product,
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| :<math>W = Fd.</math>
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| For example, if a force of 10 newtons (''F'' = 10 N) acts along a point that travels 2 metres (''d'' = 2 m), then it does the work ''W'' = (10 N)(2 m) = 20 N m = 20 J. This is approximately the work done lifting a 1 kg weight from ground to over a person's head against the force of gravity. Notice that the work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance.
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| ==Units==
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| The SI unit of work is the [[joule]] (J), which is defined as the work expended by a force of one [[newton (unit)|newton]] through a distance of one [[metre]].
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| The dimensionally equivalent [[newton-metre]] (N·m) is sometimes used as the measuring unit for work, but this can be confused with the unit newton-metre, which is the measurement unit of [[torque]]. Usage of N·m is discouraged by the [[General Conference on Weights and Measures|SI authority]], since it can lead to confusion as to whether the quantity expressed in newton metres is a torque measurement, or a measurement of energy.<ref>http://www.bipm.org/en/si/si_brochure/chapter2/2-2/2-2-2.html</ref> | |
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| Non-SI units of work include the [[erg]], the [[foot-pound (energy)|foot-pound]], the [[foot-poundal]], the [[kilowatt hour]], the [[litre-atmosphere]], and the [[horsepower|horsepower-hour]]. Due to work having the same [[Dimensional analysis|physical dimension]] as [[heat]], occasionally measurement units typically reserved for heat or energy content, such as [[therm]], [[BTU]] and [[Calorie]], are utilized as a measuring unit.
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| ==Work and energy==
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| Work is closely related to [[energy]]. The [[law of conservation of energy]] states that the change in total internal energy of a system equals the added heat, minus the work performed by the system (see the [[first law of thermodynamics]]),
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| :<math>dE = \delta Q - \delta W,</math>
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| where the symbol <math>\delta</math> indicates that heat (''Q'') and work (''W'') are [[inexact differentials]].
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| From [[Newton's laws of motion|Newton's second law]], it can be shown that work on a free (no fields), rigid (no internal degrees of freedom) body, is equal to the change in kinetic energy of the velocity and rotation of that body,
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| :<math>W = \Delta KE.</math>
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| The work of forces generated by a potential function is known as [[potential energy]] and the forces are said to be conservative. Therefore work on an object that is merely displaced in a conservative force field, without change in velocity or rotation, is equal to ''minus'' the change of potential energy of the object,
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| :<math>W = -\Delta PE.</math>
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| These formulas demonstrate that work is the energy associated with the action of a force, so work subsequently possesses the [[Dimensional analysis|physical dimensions]], and units, of energy.
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| <!-- There is nothing obvious about this derivation. It is misleading. --- Let us work backwards to see why this is obvious. The change in total (internal)energy is the change of internal potential energy plus the change of internal kinetic energy, <math>\scriptstyle \Delta E = \Delta KE + \Delta PE</math> . Internal kinetic energy is [[heat]] (which is the random, disorganised jigglings of particles inside,=movement,=Kinetic energy), and work is ''minus'' the change of potential energy.So <math>\scriptstyle \Delta H = \Delta KE</math>, and <math>\scriptstyle W = -\Delta PE</math>. Substituting into the original equation,
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| :<math>\Delta E = \Delta PE + \Delta KE </math>
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| Which is the definition of energy. --->
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| The work/energy principles discussed here are identical to Electric work/energy principles.
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| ==Constraint forces==
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| Constraint forces determine the movement of components in a system, constraining the object within a boundary (in the case of a slope plus gravity, the object is ''stuck to'' the slope, when attached to a taut string it cannot move in an outwards direction to make the string any 'tauter'). They eliminate all movement in the direction of the constraint, thus constraint forces do not perform work on the system, as the velocity of that object is constrained to be 0 parallel to this force, due to this force.
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| For example, the [[centripetal force]] exerted ''inwards'' by a string on a ball in uniform [[circular motion]] ''sideways'' constrains the ball to circular motion restricting its movement away from the center of the circle. This force does zero work because it is perpendicular to the velocity of the ball.
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| Another example is a book on a table. If external forces are applied to the book so that it slides on the table, then the force exerted by the table constrains the book from moving downwards. The force exerted by the table supports the book and is perpendicular to its movement which means that this constraint force does not perform work.
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| The [[magnetic force]] on a charged particle is F = ''q'''''v''' × '''B''', where ''q'' is the charge, '''v''' is the velocity of the particle, and '''B''' is the [[magnetic field]]. The result of a [[cross product]] is always perpendicular to both of the original vectors, so '''F''' ⊥ '''v'''. The [[dot product]] of two perpendicular vectors is always zero, so the work ''W'' = '''F''' · '''v''' = 0, and the magnetic force does not do work. It can change the direction of motion but never change the speed.
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| ==Mathematical calculation==
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| For moving objects, the quantity of work/time is calculated as distance/time, or velocity. Thus, at any instant, the rate of the work done by a force (measured in joules/second, or '''watts''') is the [[scalar product]] of the force (a vector), and the velocity vector of the point of application. This scalar product of force and velocity is classified as instantaneous [[power (physics)|power]]. Just as velocities may be integrated over time to obtain a total distance, by the [[fundamental theorem of calculus]], the total work along a path is similarly the time-integral of instantaneous power applied along the trajectory of the point of application.<ref name="Resnick">Resnick, Robert and Halliday, David (1966), ''Physics'', Section 1-3 (Vol I and II, Combined edition), Wiley International Edition, Library of Congress Catalog Card No. 66-11527</ref>
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| Work is the result of a force on a point that moves through a distance. As the point moves, it follows a curve '''X''', with a velocity '''v''', at each instant. The small amount of work ''δW'' that occurs over an instant of time ''dt'' is calculated as
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| :<math> \delta W = \mathbf{F}\cdot d\mathbf{x} = \mathbf{F}\cdot\mathbf{v}dt </math>
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| where the '''F.v''' is the power over the instant ''dt''. The sum of these small amounts of work over the trajectory of the point yields the work,
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| :<math>W = \int_{t_1}^{t_2}\mathbf{F} \cdot \mathbf{v}dt = \int_{t_1}^{t_2}\mathbf{F} \cdot {\tfrac{d\mathbf{x}}{dt}}dt =\int_C \mathbf{F} \cdot d\mathbf{x},</math>
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| where ''C'' is the trajectory from '''x'''(t<sub>1</sub>) to '''x'''(t<sub>2</sub>). This integral is computed along the trajectory of the particle, and is therefore said to be ''path dependent''.
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| If the force is always directed along this line, and the magnitude of the force is ''F'', then this integral simplifies to
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| :<math>W = \int_C Fds</math>
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| where ''s'' is distance along the line. If '''F''' is constant, in addition to being directed along the line, then the integral simplifies further to
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| :<math>W = \int_C Fds = F\int_C ds = Fd</math>
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| where ''d'' is the distance travelled by the point along the line.
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| This calculation can be generalized for a constant force that is not directed along the line, followed by the particle. In this case the [[dot product]] '''F'''·d'''x''' = ''F''cos''θ''dx, where ''θ'' is the angle between the force vector and the direction of movement,<ref name="Resnick"/> that is
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| :<math>W = \int_C \mathbf{F} \cdot d\mathbf{x} = Fd\cos\theta.</math> | |
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| In the notable case of a force applied to a body always at an angle of 90 degrees from the velocity vector (as when a body moves in a circle under a central force), no work is done at all, since the cosine of 90 degrees is zero. Thus, no work can be performed by gravity on a planet with a circular orbit (this is ideal, as all orbits are slightly elliptical). Also, no work is done on a body moving circularly at a constant speed while constrained by mechanical force, such as moving at constant speed in a friction-less ideal centrifuge.
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| Calculating the work as "force times straight path segment" would only apply in the most simple of circumstances, as noted above. If force is changing, or if the body is moving along a curved path, possibly rotating and not necessarily rigid, then only the path of the application point of the force is relevant for the work done, and only the component of the force parallel to the application point [[velocity]] is doing work (positive work when in the same direction, and negative when in the opposite direction of the velocity). This component of force can be described by the scalar quantity called ''scalar tangential component'' (<math>\scriptstyle F\cos\theta</math>, where <math>\scriptstyle \theta</math> is the angle between the force and the velocity). And then the most general definition of work can be formulated as follows:
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| :''Work of a force is the line integral of its scalar tangential component along the path of its application point.''
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| <!-- commented out this section because it is redundant
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| Simpler (intermediate) formulas for work and the transition to the general definition are described in the text below.
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| ===Force and displacement===
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| If a force '''F''' that is constant with respect to time, acts on an object while the object is translationally displaced for a [[displacement vector]] '''d''', the work done by the force on the object is the [[dot product]] of the [[Vector (mathematics and physics)|vectors]] '''F''' and '''d''':<ref name="Resnick"/>
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| : <math>W = \bold{F} \cdot \bold{d} = F d \cos\theta </math> (1)
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| where <math>\textstyle\theta</math> is the angle between the force and the displacement vector.
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| Whereas the magnitude and direction of the force must remain constant, the object's path may have any shape: the work done is independent of the path and is determined only by the total displacement vector <math>\scriptstyle\bold{d}</math>. A most common example is the work done by gravity – see diagram. The object descends along a curved path, but the work is calculated from <math>\scriptstyle d \cos\theta = h </math>, which gives the familiar result <math>\scriptstyle mgh </math>.
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| More generally, if the force causes (or affects) rotation of the body, or if the body is not rigid, displacement of the point to which the force is applied (the application point) must be used to calculate the work. This is also true for the case of variable force (below) where, however, magnitude of <math>\scriptstyle \mathrm{d}\bold{x}</math> can equally be interpreted as differential displacement magnitude or differential length of the path of the application point. (Although use of displacement vector most frequently can simplify calculation of work, in some cases simplification is achieved by use of the path length, as in the work of torque calculation below.)
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| In situations where the force changes over [[time]], equation (1) is not generally applicable. But it is possible to divide the motion into small steps, such that the force is well approximated as being constant for each step, and then to express the overall work as the sum over these steps. This will give an approximate result, which can be improved by further subdivisions into smaller steps (numerical integration). The exact result is obtained as the mathematical limit of this process, leading to the general definition below.
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| The general definition of mechanical work is given by the following [[line integral]]:
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| : <math>W_C = \int_{C} \bold{F} \cdot \mathrm{d}\bold{x} = \int_{C}\bold{F}\cdot \bold{v}dt,</math> (2)
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| where:
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| :<math>\textstyle _C</math> is the path or [[curve]] traversed by the application point of the force;
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| :<math>\bold F</math> is the [[force]] vector;
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| :<math>\bold x</math> is the [[position vector]]; and
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| :<math>\bold{v} = d\bold{x}/dt</math> is its velocity.
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| The expression <math>\delta W = \bold{F} \cdot \mathrm{d}\bold{x}</math> is an [[inexact differential]] which means that the calculation of <math>\textstyle{ W_C}</math> is path-dependent and cannot be differentiated to give <math>\bold{F} \cdot \mathrm{d}\bold{x}</math>.
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| Equation (2) explains how a non-zero force can do zero work. The simplest case is where the force is always perpendicular to the direction of motion, making the [[integrand]] always zero. This is what happens during circular motion. However, even if the integrand sometimes takes nonzero values, it can still integrate to zero if it is sometimes negative and sometimes positive.
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| The possibility of a nonzero force doing zero work illustrates the difference between work and a related quantity, [[Impulse (physics)|impulse]], which is the integral of force over time. Impulse measures change in a body's [[momentum]], a vector quantity sensitive to direction, whereas work considers only the magnitude of the velocity. For instance, as an object in uniform circular motion traverses half of a revolution, its centripetal force does no work, but it transfers a nonzero impulse.
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| -->
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| ===Torque and rotation===
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| A torque results from equal, and opposite forces, acting on two different points of a rigid body. The sum of these forces cancel, but their effect on the body is the torque '''T'''. The work of the torque is calculated as
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| :<math> \delta W = \mathbf{T}\cdot\vec{\omega}\delta t,</math>
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| where the '''T.ω''' is the power over the instant δ''t''. The sum of these small amounts of work over the trajectory of the rigid body yields the work,
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| :<math>W = \int_{t_1}^{t_2}\mathbf{T}\cdot\vec{\omega}dt.</math>
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| This integral is computed along the trajectory of the rigid body with an angular velocity ω that varies with time, and is therefore said to be ''path dependent''.
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| If the angular velocity vector maintains a constant direction, then it takes the form,
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| :<math> \vec{\omega}= \dot{\phi}\mathbf{S},</math>
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| where φ is the angle of rotation about the constant unit vector '''S'''. In this case, the work of the torque becomes,
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| :<math>W = \int_{t_1}^{t_2}\mathbf{T}\cdot\vec{\omega}dt = \int_{t_1}^{t_2}\mathbf{T}\cdot \mathbf{S}\frac{d\phi}{dt}dt = \int_C\mathbf{T}\cdot \mathbf{S} d\phi,</math>
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| where ''C'' is the trajectory from ''φ(t<sub>1</sub>)'' to ''φ(t<sub>2</sub>)''. This integral depends on the rotational trajectory ''φ(t)'', and is therefore path-dependent.
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| If the torque '''T''' is aligned with the angular velocity vector so that,
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| :<math> \mathbf{T}=\tau\mathbf{S},</math>
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| and both the torque and angular velocity are constant, then the work takes the form,<ref name="Young">{{cite book
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| | author = Hugh D. Young and Roger A. Freedman
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| | title = University Physics
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| | edition = 12th
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| | publisher = Addison-Wesley
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| | year = 2008
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| | isbn = 978-0-321-50130-1
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| | page = 329}}</ref>
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| :<math>W = \int_{t_1}^{t_2}\tau \dot{\phi}dt = \tau(\phi_2-\phi_1).</math>
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| [[File:Work on the lever arm.jpg|right|thumb|A force of constant magnitude and perpendicular to the lever arm]]
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| This result can be understood more simply by considering the torque as arising from a force of constant magnitude ''F'', being applied perpendicularly to a lever arm at a distance ''r'', as shown in the figure. This force will act through the distance along the circular arc ''s=rφ'', so the work done is
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| :<math>W=Fs = Fr\phi .</math>
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| Introduce the torque ''τ=Fr'', to obtain
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| :<math> W=Fr\phi=\tau\phi,</math>
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| as presented above.
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| <!--This is redundant and does not make it clear that it is path dependent--
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| In the more general case of a variable torque but still in the direction of rotation, the work can be calculated with the integral
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| :<math>W=\int_{\theta_1}^{\theta_2}\tau d\theta</math>,
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| where the rotation is from angle <math>\theta_1</math> to <math>\theta_2</math>.
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| -->
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| Notice that only the component of torque in the direction of the angular velocity vector contributes to the work.
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| ==Work and potential energy==
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| The scalar product of a force '''F''' and the velocity '''v''' of its point of application defines the [[power (physics)|power]] input to a system at an instant of time. Integration of this power over the trajectory of the point of application, ''C''='''x'''(t), defines the work input to the system by the force.
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| ===Path dependence===
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| Therefore, the [[mechanical work|work]] done by a force '''F''' on an object that travels along a curve ''C'' is given by the [[line integral]]:
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| :<math>W = \int_C \mathbf{F} \cdot \mathrm{d}\mathbf{x} = \int_{t_1}^{t_2}\mathbf{F}\cdot \mathbf{v}dt,</math>
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| where '''x'''(''t'') defines the trajectory ''C'' and '''v''' is the velocity along this trajectory. In general this integral requires the path along which the velocity is defined, so the evaluation of work is said to be path dependent.
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| The time derivative of the integral for work yields the instantaneous power, | |
| :<math>\frac{dW}{dt} = P(t) = \mathbf{F}\cdot \mathbf{v} .</math>
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| ===Path independence===
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| If the work for an applied force is independent of the path, then the work done by the force is evaluated at the start and end of the trajectory of the point of application. This means that there is a function ''U'' ('''x'''), called a "potential," that can be evaluated at the two points '''x'''(t<sub>1</sub>) and '''x'''(t<sub>2</sub>) to obtain the work over any trajectory between these two points. It is tradition to define this function with a negative sign so that positive work is a reduction in the potential, that is
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| :<math>W = \int_C \mathbf{F} \cdot \mathrm{d}\mathbf{x} = \int_{\mathbf{x}(t_1)}^{\mathbf{x}(t_2)} \mathbf{F} \cdot \mathrm{d}\mathbf{x} = U(\mathbf{x}(t_1))-U(\mathbf{x}(t_2)).
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| </math>
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| The function ''U''('''x''') is called the [[potential energy]] associated with the applied force. Examples of forces that have potential energies are gravity and spring forces.
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| In this case, the partial derivative of work yields
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| :<math> \frac{\partial W}{\partial \mathbf{x}} = -\frac{\partial U}{\partial \mathbf{x}} = -\big(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z}\big) = \mathbf{F},</math>
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| and the force '''F''' is said to be "derivable from a potential."<ref>J. R. Taylor, [http://books.google.com/books?id=P1kCtNr-pJsC&pg=PA117&lpg=PA117#v=onepage&q&f=false ''Classical Mechanics,''] University Science Books, 2005.</ref>
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| Because the potential ''U'' defines a force '''F''' at every point '''x''' in space, the set of forces is called a [[force field (physics)|force field]]. The power applied to a body by a force field is obtained from the gradient of the work, or potential, in the direction of the velocity '''V''' of the body, that is
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| :<math>P(t) = -\frac{\partial U}{\partial \mathbf{x}} \cdot \mathbf{v} = \mathbf{F}\cdot\mathbf{v}.</math>
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| Examples of work that can be computed from potential functions are gravity and spring forces.
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| ===Work by gravity===
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| [[File:Work of gravity F dot d equals mgh.JPG|right|thumb|Gravity F=mg does work W=mgh along any descending path]]
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| Gravity exerts a constant downward force '''F<sub>g</sub>'''=(0, ''mg'', 0) on the center of mass of every body near the surface of the earth.
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| :<math>W = F_g\Delta y = mg\Delta y</math>
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| where the integral of the vertical component of velocity is the vertical distance. Notice that the work of gravity depends only on the vertical movement of the body.
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| Where ''F<sub>g</sub>'' is weight (pounds in imperial units, and newtons in SI units), and Δ''y'' is the change in height.
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| [[File:Analogie ressorts contrainte.svg|upright|left|thumb|Forces in springs assembled in parallel]]
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| ===Work by a spring===
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| A horizontal spring exerts a force '''F'''=(kx, 0, 0) that is proportional to its deflection in the ''x'' direction. The work of this spring on a body moving along the space curve '''X'''(t) = (''x''(t), ''y''(t), ''z''(t)), is calculated using its velocity, '''v'''=(''v''<sub>x</sub>, ''v''<sub>y</sub>, ''v''<sub>z</sub>), to obtain
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| :<math>W=\int_0^t\mathbf{F}\cdot\mathbf{v}dt =\int_0^tkx v_x dt = \frac{1}{2}kx^2. </math>
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| For convenience, consider contact with the spring occurs at t=0, then the integral of the product of the distance x and the x-velocity, xv<sub>x</sub>, is (1/2)x<sup>2</sup>.
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| ===Work by a gas===
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| :<math>W=\int_a^b{P}dV </math>
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| Where ''P'' is pressure, ''V'' is volume, and ''a'' and ''b'' are initial and final volumes.
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| ==Work-energy principle==
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| The principle of work and [[kinetic energy]] (also known as the '''work-energy principle''') states that ''the work done by all forces acting on a particle (the work of the resultant force) equals the change in the kinetic energy of the particle.''<ref>{{cite book
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| | author = Andrew Pytel, Jaan Kiusalaas
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| | title = Engineering Mechanics: Dynamics – SI Version, Volume 2
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| | edition = 3rd
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| | publisher = Cengage Learning,
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| | year = 2010
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| | isbn = 9780495295631
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| | page = 654}}</ref>
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| The work ''W'' done by the [[resultant force]] on a [[particle]] equals the change in the particle's kinetic energy <math>E_k</math>,<ref name="Young"/>
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| :<math>W=\Delta E_k=\tfrac12mv_2^2-\tfrac12mv_1^2</math>,
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| where <math>v_1</math> and <math>v_2</math> are the [[speed]]s of the particle before and after the change and ''m'' is its [[mass]].
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| If a resultant torque acts on a rigid body, the definition can be extended to equate the work done by the resultant torque and the change in rotational kinetic energy of the rigid body.
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| ===Overview===
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| The derivation of the ''work-energy principle'' begins with Newton's second law and the resultant force on a particle which includes forces applied to the particle and constraint forces imposed on its movement. Computation of the scalar product of the forces with the velocity of the particle evaluates the instantaneous power added to the system.<ref>B. Paul, [http://books.google.com/books?ei=JGMfULH-KK2r2AWIzICYAw&id=3UdSAAAAMAAJ&dq=kinematics+and+dynamics+of+planar+machinery&q=instantaneous+power#search_anchor ''Kinematics and Dynamics of Planar Machinery,''] Prentice-Hall, 1979.</ref>
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| Constraints define the direction of movement of the particle by ensuring there is no component of velocity in the direction of the constraint force. This also means the constraint forces do not add to the instantaneous power. The time integral of this scalar equation yields work from the instantaneous power, and kinetic energy from the scalar product of velocity and acceleration. The fact the work-energy principle eliminates the constraint forces underlies [[Lagrangian mechanics]].<ref>E. T. Whittaker, [http://books.google.com/books?id=OSckAAAAMAAJ&printsec=frontcover&dq=analytical+dynamics&source=bl&ots=AGqvlTnrw7&sig=x7DfbqV97-G73bWcuEFC-rU3h2s&hl=en&sa=X&ei=9lobUPz8KofU2AW9sIGIDw&ved=0CGcQ6AEwCA#v=onepage&q&f=false ''A treatise on the analytical dynamics of particles and rigid bodies,''] Cambridge University Press, 1904</ref>
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| <!-- This paragraph seems redundant and the reference is incomplete --If an object is pushed and [[friction]] also acts on the object, the change of kinetic energy is equal to the work done by the net force, which is friction subtracted from your push. In such a situation, only part of the total work that you do changes the object's kinetic energy. The rest is soaked up by friction, which goes into heat. If the force of friction is equal and opposite to your push, the net force on the object is zero and no net work is done. Thus, there is zero change in the object's kinetic energy.<ref name="Hewitt 2006, page 115">Hewitt (2006), page 115</ref>-->
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| <!--- This section discusses in general words of the formulas presented above ---The work-energy theorem applies to decreasing speed. When the brakes of a car are slammed and the car skids, the road does work on the car. This work is the friction force multiplied by the distance over which the friction force acts. The maximum friction that the brakes can supply is nearly the same whether the car moves slowly or quickly. In a short stop with antilock brakes, the only way for the brakes to do more work is to act over a longer distance. A car moving at twice the speed of another takes four times (2<sup>2</sup> = 4) as much work to stop. As the frictional force is nearly the same for both cars, the faster one takes four times as much distance to stop. For brakes that can lock the wheels, the force of friction on a skidding tire is also nearly independent of speed. An automobile going 100 kilometers per hour, with four times the kinetic energy that it would have at 50 kilometers per hour, skids four times as far with its wheels locked as it would from a speed of 50 kilometers per hour.-->
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| This section focusses on the work-energy principle as it applies to particle dynamics. In more general systems work can change the [[potential energy]] of a mechanical device, the heat energy in a thermal system, or the [[electrical energy]] in an electrical device. Work transfers energy from one place to another or one form to another.
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| ===Derivation for a particle moving along a straight line===
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| In the case the [[resultant force]] '''F''' is constant in both magnitude and direction, and parallel to the velocity of the particle, the particle is moving with constant acceleration ''a'' along a straight line.<ref>[http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html Work-energy principle]</ref> The relation between the net force and the acceleration is given by the equation ''F'' = ''ma'' ([[Newton's laws of motion#Newton's second law|Newton's second law]]), and the particle [[displacement (vector)|displacement]] ''d'' can be expressed by the equation
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| :<math>d = \frac{v_2^2 - v_1^2}{2a}</math>
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| which follows from <math>v_2^2 = v_1^2 + 2ad</math> (see [[Equations of motion#Uniform acceleration|Equations of motion]]).
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| The work of the net force is calculated as the product of its magnitude and the particle displacement. Substituting the above equations, one obtains:
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| :<math>W = Fd = mad = ma \left(\frac{v_2^2 - v_1^2}{2a}\right) = \frac{mv_2^2}{2} - \frac{mv_1^2}{2} = \Delta {E_k}</math>
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| In the general case of rectilinear motion, when the net force '''F''' is not constant in magnitude, but is constant in direction, and parallel to the velocity of the particle, the work must be integrated along the path of the particle:
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| :<math>W = \int_{t_1}^{t_2} \mathbf{F}\cdot \mathbf{v}dt = \int_{t_1}^{t_2} F \,v dt = \int_{t_1}^{t_2} ma \,v dt = m \int_{t_1}^{t_2} v \,{dv \over dt}\,dt = m \int_{v_1}^{v_2} v\,dv = \tfrac12 m (v_2^2 - v_1^2) .</math>
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| ===General derivation of the work-energy theorem for a particle===
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| For any net force acting on a particle moving along any curvilinear path, it can be demonstrated that its work equals the change in the kinetic energy of the particle by a simple derivation analogous to the equation above. Some authors call this result ''work-energy principle'', but it is more widely known as '''the work-energy theorem''':
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| :<math>W = \int_{t_1}^{t_2} \mathbf{F}\cdot \mathbf{v}dt = m \int_{t_1}^{t_2} \mathbf{a} \cdot \mathbf{v}dt = \frac{m}{2} \int_{t_1}^{t_2} \frac{d v^2}{dt}\,dt = \frac{m}{2} \int_{v^2_1}^{v^2_2} d v^2 = \frac{mv_2^2}{2} - \frac{mv_1^2}{2} = \Delta {E_k} </math>
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| The identity <math>\textstyle \mathbf{a} \cdot \mathbf{v} = \frac{1}{2} \frac{d v^2}{dt}</math> requires some algebra.
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| From the identity <math>\textstyle v^2 = \mathbf{v} \cdot \mathbf{v}</math> and definition <math>\textstyle \mathbf{a} = \frac{d \mathbf{v}}{dt} </math>
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| it follows
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| :<math> \frac{d v^2}{dt} = \frac{d (\mathbf{v} \cdot \mathbf{v})}{dt} = \frac{d \mathbf{v}}{dt} \cdot \mathbf{v} + \mathbf{v} \cdot \frac{d \mathbf{v}}{dt} = 2 \frac{d \mathbf{v}}{dt} \cdot \mathbf{v} = 2 \mathbf{a} \cdot \mathbf{v}</math>.
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| The remaining part of the above derivation is just simple calculus, same as in the preceding rectilinear case.
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| ===Derivation for a particle in constrained movement===
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| In particle dynamics, a formula equating work applied to a system to its change in kinetic energy is obtained as a first integral of [[Newton's laws of motion|Newton's second law of motion]]. It is useful to notice that the resultant force used in Newton's laws can be separated into forces that are applied to the particle and forces imposed by constraints on the movement of the particle. Remarkably, the work of a constraint force is zero, therefore only the work of the applied forces need be considered in the work-energy principle.
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| To see this, consider a particle P that follows the trajectory '''X'''(t) with a force '''F''' acting on it. Isolate the particle from its environment to expose constraint forces '''R''', then Newton's Law takes the form
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| :<math> \mathbf{F} + \mathbf{R} =m\ddot{\mathbf{X}}, </math>
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| where ''m'' is the mass of the particle.
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| ====Vector formulation====
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| Note that n dots above a vector indicates its nth [[time derivative]].
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| The [[scalar product]] of each side of Newton's law with the velocity vector yields
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| :<math> \mathbf{F}\cdot\dot{\mathbf{X}} = m\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}},</math>
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| because the constraint forces are perpendicular to the particle velocity. Integrate this equation along its trajectory from the point '''X'''(t<sub>1</sub>) to the point '''X'''(t<sub>2</sub>) to obtain
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| :<math> \int_{t_1}^{t_2} \mathbf{F}\cdot\dot{\mathbf{X}} dt = m\int_{t_1}^{t_2}\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}dt. </math>
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| The left side of this equation is the work of the applied force as it acts on the particle along the trajectory from time t<sub>1</sub> to time t<sub>2</sub>. This can also be written as
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| :<math> W = \int_{t_1}^{t_2} \mathbf{F}\cdot\dot{\mathbf{X}} dt = \int_{\mathbf{X}(t_1)}^{\mathbf{X}(t_2)} \mathbf{F}\cdot d\mathbf{X}. </math>
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| This integral is computed along the trajectory '''X'''(t) of the particle and is therefore path dependent.
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| The right side of the first integral of Newton's equations can be simplified using the following identity
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| :<math> \frac{1}{2}\frac{d}{dt}(\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}) = \ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}, </math>
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| (see [[product rule]] for derivation). Now it is integrated explicitly to obtain the change in kinetic energy,
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| :<math>\Delta K = m\int_{t_1}^{t_2}\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}dt = \frac{m}{2}\int_{t_1}^{t_2}\frac{d}{dt}(\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}) dt = \frac{m}{2}\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}(t_2) - \frac{m}{2}\dot{\mathbf{X}}\cdot \dot{\mathbf{X}} (t_1) = \frac{1}{2}m \Delta \mathbf{v^{2}} , </math>
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| where the kinetic energy of the particle is defined by the scalar quantity,
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| :<math> K = \frac{m}{2}\dot{\mathbf{X}}\cdot \dot{\mathbf{X}} =\frac{1}{2}m{\mathbf{v^{2}}}</math>
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| ====Tangential and normal components====
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| It is useful to resolve the velocity and acceleration vectors into tangential and normal components along the trajectory '''X'''(t), such that
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| :<math> \dot{\mathbf{X}}=v \mathbf{T},\quad\mbox{and}\quad \ddot{\mathbf{X}}=\dot{v}\mathbf{T} + v^2\kappa \mathbf{N}. </math>
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| where
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| :<math> v=|\dot{\mathbf{X}}|=\sqrt{\dot{\mathbf{X}}\cdot\dot{\mathbf{X}}}.</math>
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| Then, the [[scalar product]] of velocity with acceleration in Newton's second law takes the form
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| :<math> \Delta K = m\int_{t_1}^{t_2}\dot{v}vdt = \frac{m}{2}\int_{t_1}^{t_2}\frac{d}{dt}v^2 dt = \frac{m}{2}v^2(t_2) - \frac{m}{2}v^2(t_1), </math>.
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| where the kinetic energy of the particle is defined by the scalar quantity,
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| :<math> K = \frac{m}{2} v^2 = \frac{m}{2} \dot{\mathbf{X}}\cdot\dot{\mathbf{X}}. </math>
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| The result is the work-energy principle for particle dynamics,
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| :<math> W = \Delta K. \!</math>
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| This derivation can be generalized to arbitrary rigid body systems.
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| ===Moving in a straight line (skid to a stop)===
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| Consider the case of a vehicle moving along a straight horizontal trajectory under the action of a driving force and gravity that sum to '''F'''. The constraint forces between the vehicle and the road define '''R''', and we have
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| :<math> \mathbf{F} + \mathbf{R} =m\ddot{\mathbf{X}}. </math>
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| For convenience let the trajectory be along the X-axis, so '''X'''=(d,0) and the velocity is '''V'''=(v, 0), then '''R.V'''=0, and '''F.V'''=F<sub>x</sub>v, where F<sub>x</sub> is the component of '''F''' along the X-axis, so
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| :<math> F_x v = m\dot{v}v.</math>
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| Integration of both sides yields
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| :<math> \int_{t_1}^{t_2}F_x v dt = \frac{m}{2}v^2(t_2) - \frac{m}{2}v^2(t_1). </math>
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| If F<sub>x</sub> is constant along the trajectory, then the integral of velocity is distance, so
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| :<math> F_x (d(t_2)-d(t_1)) = \frac{m}{2}v^2(t_2) - \frac{m}{2}v^2(t_1). </math>
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| As an example consider a car skidding to a stop, where ''k'' is the coefficient of friction and ''W'' is the weight of the car. Then the force along the trajectory is F<sub>x</sub> =-kW. The velocity ''v'' of the car can be determined from the length ''s'' of the skid using the work-energy principle,
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| :<math>kWs = \frac{W}{2g} v^2,\quad\mbox{or}\quad v = \sqrt{2ksg}.</math>
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| Notice that this formula uses the fact that the mass of the vehicle is ''m''=''W/g''.
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| [[File:LotusType119B.jpg|thumb|Lotus type 119B gravity racer at Lotus 60th celebration.]]
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| [[File:Campeonato de carrinhos de rolimã em Campos Novos SC.jpg|thumb|Gravity racing championship in Campos Novos, Santa Catarina, Brazil, 8 September 2010.]]
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| ===Coasting down a mountain road (gravity racing)===
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| Consider the case of a vehicle that starts at rest and coasts down a mountain road, the work-energy principle helps compute the minimum distance that the vehicle travels to reach a velocity ''V'', of say 60 mph (88 fps). Rolling resistance and air drag will slow the vehicle down so the actual distance will be less than if these forces are neglected.
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| Let the trajectory of the vehicle following the road be '''X'''(t) which is a curve in three-dimensional space. The force acting on the vehicle that pushes it down the road is the constant force of gravity '''F'''=(0,0,''W''), while the force of the road on the vehicle is the constraint force '''R'''. Newton's second law yields,
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| :<math> \mathbf{F} + \mathbf{R} =m\ddot{\mathbf{X}}. </math>
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| The [[scalar product]] of this equation with the velocity, '''V'''=(v<sub>x</sub>, v<sub>y</sub>,v<sub>z</sub>), yields
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| :<math> W v_z = m\dot{V}V,</math>
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| where ''V'' is the magnitude of '''V'''. The constraint forces between the vehicle and the road cancel from this equation because '''R'''.'''V'''=0, which means they do no work.
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| Integrate both sides to obtain
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| :<math> \int_{t_1}^{t_2}W v_z dt = \frac{m}{2}V^2(t_2) - \frac{m}{2}V^2(t_1). </math>
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| The weight force ''W'' is constant along the trajectory and the integral of the vertical velocity is the vertical distance, therefore,
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| :<math> W \Delta z = \frac{m}{2}V^2. </math>
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| Recall that V(t<sub>1</sub>)=0. Notice that this result does not depend on the shape of the road followed by the vehicle.
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| In order to determine the distance along the road assume the downgrade is 6%, which is a steep road. This means the altitude decreases 6 feet for every 100 feet traveled—for angles this small the sin and tan functions are approximately equal. Therefore, the distance ''s'' in feet down a 6% grade to reach the velocity ''V'' is at least
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| :<math> s=\frac{\Delta z}{0.06}= 8.3\frac{V^2}{g},\quad\mbox{or}\quad s=8.3\frac{88^2}{32.2}\approx 2000\mbox{ft}.</math>
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| This formula uses the fact that the weight of the vehicle is ''W''=''mg''.
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| <!--This discussion of reference frame seems unnecessary ==Frame of reference==
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| The work done by a force acting on an object depends on the choice of [[frame of reference|reference frame]] because [[Displacement (vector)|displacements]] and velocities are dependent on the reference frame in which the observations are being made.<ref name="Resnick">Resnick, Robert and Halliday, David (1966), ''Physics'', Section 1-3 (Vol I and II, Combined edition), Wiley International Edition, Library of Congress Catalog Card No. 66-11527</ref>
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| The change in kinetic energy also depends on the choice of reference frame because kinetic energy is a function of velocity. However, regardless of the choice of reference frame, the work energy theorem remains valid and the work done on the object is equal to the change in kinetic energy.<ref name="Resnick"/>
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| -->
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| ==Work of forces acting on a rigid body==
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| The work of forces acting at various points on a single rigid body can be calculated from the work of a [[resultant force|resultant force and torque]]. To see this, let the forces '''F'''<sub>1</sub>, '''F'''<sub>2</sub> ... '''F'''<sub>n</sub> act on the points '''X'''<sub>1</sub>, '''X'''<sub>2</sub> ... '''X'''<sub>n</sub> in a rigid body.
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| The trajectories of '''X'''<sub>i</sub>, i=1,...,n are defined by the movement of the rigid body. This movement is given by the set of rotations [A(t)] and the trajectory '''d'''(t) of a reference point in the body. Let the coordinates '''x'''<sub>i</sub> i=1,...,n define these points in the moving rigid body's [[Cartesian coordinate system|reference frame]] ''M'', so that the trajectories traced in the fixed frame ''F'' are given by
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| :<math> \mathbf{X}_i(t)= [A(t)]\mathbf{x}_i + \mathbf{d}(t)\quad i=1,\ldots, n. </math>
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| The velocity of the points '''X'''<sub>i</sub> along their trajectories are
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| :<math>\mathbf{V}_i = \vec{\omega}\times(\mathbf{X}_i-\mathbf{d}) + \dot{\mathbf{d}},</math>
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| where '''ω''' is the angular velocity vector obtained from the skew symmetric matrix
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| :<math> [\Omega] = \dot{A}A^T,</math>
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| known as the angular velocity matrix.
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| The small amount of work by the forces over the small displacements δ'''r'''<sub>i</sub> can be determined by approximating the displacement by δ'''r'''='''v'''δt so
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| :<math> \delta W = \mathbf{F}_1\cdot\mathbf{V}_1\delta t+\mathbf{F}_2\cdot\mathbf{V}_2\delta t + \ldots + \mathbf{F}_n\cdot\mathbf{V}_n\delta t</math>
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| or
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| :<math> \delta W = \sum_{i=1}^n \mathbf{F}_i\cdot (\vec{\omega}\times(\mathbf{X}_i-\mathbf{d}) + \dot{\mathbf{d}})\delta t. </math>
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| This formula can be rewritten to obtain
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| :<math> \delta W = (\sum_{i=1}^n \mathbf{F}_i)\cdot\dot{\mathbf{d}}\delta t + (\sum_{i=1}^n (\mathbf{X}_i-\mathbf{d})\times\mathbf{F}_i) \cdot \vec{\omega}\delta t = (\mathbf{F}\cdot\dot{\mathbf{d}}+ \mathbf{T}\cdot \vec{\omega})\delta t, </math>
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| where '''F''' and '''T''' are the [[resultant force|resultant force and torque]] applied at the reference point '''d''' of the moving frame ''M'' in the rigid body.
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| ==References==
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| {{reflist}}
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| ==Bibliography==
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| * {{cite book | author=Serway, Raymond A.; Jewett, John W. | title=Physics for Scientists and Engineers | edition=6th ed. | publisher=Brooks/Cole | year=2004 | isbn=0-534-40842-7}}
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| * {{cite book | author=Tipler, Paul | title=Physics for Scientists and Engineers: Mechanics| edition=3rd ed., extended version | publisher=W. H. Freeman | year=1991 | isbn=0-87901-432-6}}
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| ==External links==
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| * [http://www.lightandmatter.com/html_books/2cl/ch03/ch03.html Work] – a chapter from an online textbook
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| * [http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html Work-energy principle]
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| {{DEFAULTSORT:Work (Physics)}}
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| [[Category:Energy (physics)]]
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| [[Category:Physical quantities]]
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| [[Category:Mechanical engineering]]
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| [[Category:Mechanics]]
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| [[Category:Force]]
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| [[Category:Length]]
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