Editing Work (physics) (section)

==Work–energy principle==
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 | author1=Andrew Pytel | author2=Jaan Kiusalaas | title = Engineering Mechanics: Dynamics – SI Version, Volume 2| edition = 3rd | publisher = Cengage Learning| year = 2010| isbn = 9780495295631| page = 654}}</ref>  That is, the work ''W'' done by the [[resultant force]] on a [[particle]] equals the change in the particle's kinetic energy <math>E_\text{k}</math>,<ref name="Young"/>
<math display="block"> W = \Delta E_\text{k} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2</math>
where <math>v_1</math> and <math>v_2</math> are the [[speed]]s of the particle before and after the work is done, and {{mvar|m}} is its [[mass]].

The derivation of the ''work–energy principle'' begins with [[Newton's second law of motion]] and the resultant force on a particle.  Computation of the scalar product of the force with the velocity of the particle evaluates the instantaneous power added to the system.<ref>{{cite book |last1=Paul |first1=Burton |title=Kinematics and Dynamics of Planar Machinery |year=1979 |publisher=Prentice-Hall |isbn=978-0-13-516062-6 |url=https://books.google.com/books?id=3UdSAAAAMAAJ&q=instantaneous+power |language=en}}</ref>
(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 acceleration with velocity.  The fact that the work–energy principle eliminates the constraint forces underlies [[Lagrangian mechanics]].<ref>{{cite book|last=Whittaker|first=E. T.|author-link=E. T. Whittaker|title=A treatise on the analytical dynamics of particles and rigid bodies |publisher=Cambridge University Press |url=https://archive.org/details/atreatiseonanal00whitgoog|language=en|year=1904}}</ref>

This section focuses 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 thermal 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.

===Derivation for a particle moving along a straight line===
In the case the [[resultant force]] {{math|'''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>{{cite web| url=http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html|title=Work–energy principle| website=www.wwu.edu| access-date=2012-08-06|archive-url=https://web.archive.org/web/20120530075449/http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html|archive-date=2012-05-30|url-status=dead}}</ref> The relation between the net force and the acceleration is given by the equation {{math|1=''F'' = ''ma''}} ([[Newton's second law]]), and the particle [[displacement (vector)|displacement]] {{mvar|s}} can be expressed by the equation
<math display="block">s = \frac{v_2^2 - v_1^2}{2a}</math>
which follows from <math>v_2^2 = v_1^2 + 2as</math> (see [[Equations of motion]]).

The work of the net force is calculated as the product of its magnitude and the particle displacement. Substituting the above equations, one obtains:
<math display="block"> W = Fs = mas = ma\frac{v_2^2-v_1^2}{2a} = \frac{1}{2} mv_2^2- \frac{1}{2} mv_1^2 = \Delta E_\text{k}</math>

Other derivation:
<math display="block"> W = Fs = mas = m\frac{v_2^2 - v_1^2}{2s}s = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 = \Delta E_\text{k}</math>

In the general case of rectilinear motion, when the net force {{math|'''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:
<math display="block"> 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 \,\frac{dv}{dt}\,dt = m \int_{v_1}^{v_2} v\,dv = \tfrac12 m \left(v_2^2 - v_1^2\right) .</math>

===General derivation of the work–energy principle for a particle===
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. It is known as '''the work–energy principle''':
<math display="block"> 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_\text{k} </math>

The identity <math display="inline">\mathbf{a} \cdot \mathbf{v} = \frac{1}{2} \frac{d v^2}{dt}</math> requires some algebra.
From the identity <math display="inline">v^2 = \mathbf{v} \cdot \mathbf{v}</math> and definition <math display="inline">\mathbf{a} = \frac{d \mathbf{v}}{dt} </math>
it follows
<math display="block"> \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>

The remaining part of the above derivation is just simple calculus, same as in the preceding rectilinear case.

===Derivation for a particle in constrained movement===
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.

To see this, consider a particle P that follows the trajectory {{math|'''X'''(''t'')}} with a force {{math|'''F'''}} acting on it.  Isolate the particle from its environment to expose constraint forces {{math|'''R'''}}, then Newton's Law takes the form
<math display="block"> \mathbf{F} + \mathbf{R} = m \ddot{\mathbf{X}}, </math>
where {{mvar|m}} is the mass of the particle.

====Vector formulation====
Note that n dots above a vector indicates its nth [[time derivative]].
The [[scalar product]] of each side of Newton's law with the velocity vector yields
<math display="block"> \mathbf{F}\cdot\dot{\mathbf{X}} = m\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}},</math>
because the constraint forces are perpendicular to the particle velocity.  Integrate this equation along its trajectory from the point {{math|'''X'''(''t''<sub>1</sub>)}} to the point {{math|'''X'''(''t''<sub>2</sub>)}} to obtain
<math display="block"> \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>

The left side of this equation is the work of the applied force as it acts on the particle along the trajectory from time {{math|''t''<sub>1</sub>}} to time {{math|''t''<sub>2</sub>}}.  This can also be written as
<math display="block"> 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>
This integral is computed along the trajectory {{math|'''X'''(''t'')}} of the particle and is therefore path dependent.

The right side of the first integral of Newton's equations can be simplified using the following identity
<math display="block"> \frac{1}{2}\frac{d}{dt}(\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}) = \ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}, </math>
(see [[product rule]] for derivation). Now it is integrated explicitly to obtain the change in kinetic energy,
<math display="block">\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>
where the kinetic energy of the particle is defined by the scalar quantity,
<math display="block"> K = \frac{m}{2} \dot{\mathbf{X}} \cdot \dot{\mathbf{X}} =\frac{1}{2} m {\mathbf{v}^2}</math>

====Tangential and normal components====
It is useful to resolve the velocity and acceleration vectors into tangential and normal components along the trajectory {{math|'''X'''(''t'')}}, such that
<math display="block"> \dot{\mathbf{X}}=v \mathbf{T}\quad\text{and}\quad \ddot{\mathbf{X}}=\dot{v}\mathbf{T} + v^2\kappa \mathbf{N},</math>
where
<math display="block"> v=|\dot{\mathbf{X}}|=\sqrt{\dot{\mathbf{X}}\cdot\dot{\mathbf{X}}}.</math>
Then, the [[scalar product]] of velocity with acceleration in Newton's second law takes the form
<math display="block"> \Delta K = m\int_{t_1}^{t_2}\dot{v}v \, dt = \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>
where the kinetic energy of the particle is defined by the scalar quantity,
<math display="block"> K = \frac{m}{2} v^2 = \frac{m}{2} \dot{\mathbf{X}} \cdot \dot{\mathbf{X}}. </math>

The result is the work–energy principle for particle dynamics,
<math display="block"> W = \Delta K. </math>
This derivation can be generalized to arbitrary rigid body systems.

===Moving in a straight line (skid to a stop)===
Consider the case of a vehicle moving along a straight horizontal trajectory under the action of a driving force and gravity that sum to {{math|'''F'''}}.  The constraint forces between the vehicle and the road define {{math|'''R'''}}, and we have
<math display="block"> \mathbf{F} + \mathbf{R} = m\ddot{\mathbf{X}}. </math>
For convenience let the trajectory be along the X-axis, so {{math|1='''X''' = (''d'', 0)}} and the velocity is {{math|1='''V''' = (''v'', 0)}}, then {{math|1='''R''' ⋅ '''V''' = 0}}, and {{math|1='''F''' ⋅ '''V''' = ''F''<sub>x</sub>''v''}}, where ''F''<sub>x</sub> is the component of '''F''' along the X-axis, so
<math display="block"> F_x v = m\dot{v}v.</math>
Integration of both sides yields
<math display="block"> \int_{t_1}^{t_2}F_x v dt = \frac{m}{2} v^2(t_2) - \frac{m}{2} v^2(t_1). </math>
If {{math|''F''<sub>x</sub>}} is constant along the trajectory, then the integral of velocity is distance, so
<math display="block"> F_x (d(t_2)-d(t_1)) = \frac{m}{2} v^2(t_2) - \frac{m}{2} v^2(t_1). </math>

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 {{math|1=''F''<sub>x</sub> = −''kw''}}.  The velocity ''v'' of the car can be determined from the length {{mvar|s}} of the skid using the work–energy principle,
<math display="block">kws = \frac{w}{2g} v^2,\quad\text{or}\quad v = \sqrt{2ksg}.</math>
This formula uses the fact that the mass of the vehicle is {{math|1=''m'' = ''w''/''g''}}.

[[File:LotusType119B.jpg|thumb|Lotus type 119B [[gravity racer]] at Lotus 60th celebration]]
[[File:Campeonato de carrinhos de rolimã em Campos Novos SC.jpg|thumb|Gravity racing championship in Campos Novos, Santa Catarina, Brazil, 8 September 2010]]

===Coasting down an inclined surface (gravity racing)===
Consider the case of a vehicle that starts at rest and coasts down an inclined surface (such as mountain road), the work–energy principle helps compute the minimum distance that the vehicle travels to reach a velocity {{math|''V''}}, of say 60&nbsp;mph (88 fps).  Rolling resistance and air drag will slow the vehicle down so the actual distance will be greater than if these forces are neglected.

Let the trajectory of the vehicle following the road be {{math|'''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 {{math|1='''F''' = (0, 0, ''w'')}}, while the force of the road on the vehicle is the constraint force {{math|'''R'''}}.  Newton's second law yields,
<math display="block"> \mathbf{F} + \mathbf{R} = m \ddot{\mathbf{X}}. </math>
The [[scalar product]] of this equation with the velocity, {{math|1='''V''' = (''v''<sub>x</sub>, ''v''<sub>y</sub>, ''v''<sub>z</sub>)}}, yields
<math display="block"> w v_z = m\dot{V}V,</math>
where {{math|''V''}} is the magnitude of {{math|'''V'''}}.  The constraint forces between the vehicle and the road cancel from this equation because {{math|1='''R''' ⋅ '''V''' = 0}}, which means they do no work.
Integrate both sides to obtain
<math display="block"> \int_{t_1}^{t_2} w v_z dt = \frac{m}{2} V^2(t_2) - \frac{m}{2} V^2 (t_1). </math>
The weight force ''w'' is constant along the trajectory and the integral of the vertical velocity is the vertical distance, therefore,
<math display="block"> w \Delta z = \frac{m}{2}V^2. </math>
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.

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 {{mvar|s}} in feet down a 6% grade to reach the velocity {{mvar|V}} is at least
<math display="block"> s = \frac{\Delta z}{0.06}= 8.3\frac{V^2}{g},\quad\text{or}\quad s=8.3\frac{88^2}{32.2}\approx 2000\mathrm{ft}.</math>
This formula uses the fact that the weight of the vehicle is {{math|1=''w'' = ''mg''}}.