Editing Work (physics) (section)

==Mathematical calculation==
For moving objects, the quantity of work/time (power) is integrated along the trajectory of the point of application of the force. Thus, at any instant, the rate of the work done by a force (measured in joules/second, or [[watt]]s) 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 known 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, 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>

Work is the result of a force on a point that follows a curve {{math|'''X'''}}, with a velocity {{math|'''v'''}}, at each instant.  The small amount of work {{math|''δW''}} that occurs over an instant of time {{math|''dt''}} is calculated as
<math display="block"> \delta W = \mathbf{F} \cdot d\mathbf{s} = \mathbf{F} \cdot \mathbf{v}dt </math>
where the {{math|'''F''' ⋅ '''v'''}} is the power over the instant {{math|''dt''}}.  The sum of these small amounts of work over the trajectory of the point yields the work,
<math display="block"> W = \int_{t_1}^{t_2}\mathbf{F} \cdot \mathbf{v} \, dt = \int_{t_1}^{t_2}\mathbf{F} \cdot \tfrac{d\mathbf{s}}{dt} \, dt =\int_C \mathbf{F} \cdot d\mathbf{s},</math>
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''.

If the force is always directed along this line, and the magnitude of the force is {{math|''F''}}, then this integral simplifies to
<math display="block"> W = \int_C F\,ds</math>
where {{mvar|s}} is displacement along the line. If {{math|'''F'''}} is constant, in addition to being directed along the line, then the integral simplifies further to
<math display="block"> W = \int_C F\,ds = F\int_C ds = Fs</math>
where ''s'' is the displacement of the point along the line.

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]] {{math|1='''F''' ⋅ ''d'''''s''' = ''F'' cos ''θ'' ''ds''}}, where {{mvar|θ}} is the angle between the force vector and the direction of movement,<ref name="Resnick"/> that is
<math display="block">W = \int_C \mathbf{F} \cdot d\mathbf{s} = Fs\cos\theta.</math>

When a force component is perpendicular to the displacement of the object (such as when a body moves in a circular path under a [[central force]]), no work is done, since the cosine of 90° is zero.<ref name="walker" /> 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 frictionless ideal centrifuge.

=== Work done by a variable force ===

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|''F'' cos(''θ'')}}, where {{mvar|θ}} is the angle between the force and the velocity). And then the most general definition of work can be formulated as follows:
[[File:Force-distance-diagram.svg|thumb|[[Area under the curve]] gives work done by F(x).]]
{{block indent | em = 1.5 | text =
''Work done by a variable force is the line integral of its scalar tangential component along the path of its application point.''{{pb}}
If the force varies (e.g. compressing a spring) we need to use calculus to find the work done.  If the force as a variable of {{mvar|'''x'''}} is given by {{math|'''F'''('''x''')}}, then the work done by the force along the x-axis from {{mvar|'''x'''<sub>1</sub>}} to {{mvar|'''x'''<sub>2</sub>}} is:
<math display="block">W = \lim_{\Delta\mathbf{x} \to 0}\sum_{x_1}^{x_2}\mathbf{F(x)}\Delta\mathbf{x} = \int_{x_1}^{x_2}\mathbf{F(x)}d\mathbf{x}.</math>}}

Thus, the work done for a variable force can be expressed as a [[definite integral]] of force over displacement.<ref>{{Cite web |title=MindTap - Cengage Learning |url=https://ng.cengage.com/static/nb/ui/evo/index.html?snapshotId=1529049&id=677758950&eISBN=9780357049105 |access-date=2023-10-16 |website=ng.cengage.com}}</ref>

If the displacement as a variable of time is given by {{mvar|∆'''x'''(t)}}, then work done by the variable force from {{mvar|t<sub>1</sub>}} to {{mvar|t<sub>2</sub>}} is:
:<math display="block">W = \int_{t_1}^{t_2}\mathbf{F}(t)\cdot \mathbf{v}(t)dt = \int_{t_1}^{t_2}P(t)dt.</math>

Thus, the work done for a variable force can be expressed as a definite integral of [[Power (physics)|power]] over time.

===Torque and rotation===
A [[Couple (mechanics)|force couple]] results from equal and opposite forces, acting on two different points of a rigid body.  The sum (resultant) of these forces may cancel, but their effect on the body is the couple or torque '''T'''.  The work of the torque is calculated as
<math display="block"> \delta W = \mathbf{T} \cdot \boldsymbol{\omega} \, dt,</math>
where the {{math|'''T''' ⋅ '''''ω'''''}} is the power over the instant {{math|''dt''}}.  The sum of these small amounts of work over the trajectory of the rigid body yields the work,
<math display="block"> W = \int_{t_1}^{t_2} \mathbf{T} \cdot \boldsymbol{\omega} \, dt.</math>
This integral is computed along the trajectory of the rigid body with an angular velocity {{math|'''''ω'''''}} that varies with time, and is therefore said to be ''path dependent''.

If the angular velocity vector maintains a constant direction, then it takes the form,
<math display="block"> \boldsymbol{\omega} = \dot{\phi}\mathbf{S},</math>
where <math>\phi</math> is the angle of rotation about the constant unit vector {{math|'''S'''}}.  In this case, the work of the torque becomes,
<math display="block">W = \int_{t_1}^{t_2} \mathbf{T} \cdot \boldsymbol{\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>
where {{math|''C''}} is the trajectory from <math>\phi (t_{1})</math> to <math>\phi (t_{2})</math>.  This integral depends on the rotational trajectory <math>\phi (t)</math>, and is therefore path-dependent.

If the torque <math>\tau</math> is aligned with the angular velocity vector so that,
<math display="block" qid=Q48103> \mathbf{T} = \tau \mathbf{S},</math>
and both the torque and angular velocity are constant, then the work takes the form,<ref name="Young">{{cite book | author1 = Hugh D. Young |author2 = Roger A. Freedman | name-list-style=amp | title = University Physics | edition = 12th | publisher = Addison-Wesley | year = 2008| isbn = 978-0-321-50130-1| page = 329}}</ref>
<math display="block">W = \int_{t_1}^{t_2} \tau \dot{\phi} \, dt = \tau(\phi_2 - \phi_1).</math>
[[File:Work on lever arm.png|250px|thumb|right|alt=Work on lever arm|A force of constant magnitude and perpendicular to the lever arm]]

This result can be understood more simply by considering the torque as arising from a force of constant magnitude {{math|''F''}}, being applied perpendicularly to a lever arm at a distance <math>r</math>, as shown in the figure.  This force will act through the distance along the circular arc <math>l=s=r\phi</math>, so the work done is
<math display="block"> W = F s = F r \phi .</math>
Introduce the torque {{math|1=''τ'' = ''Fr''}}, to obtain
<math display="block"> W = F r \phi = \tau \phi ,</math>
as presented above.

Notice that only the component of torque in the direction of the angular velocity vector contributes to the work.