Work of a force

See also: Work (homonymy)

The work of a force is the energy provided by this force when its point of application moves (the object undergoing the force moves or becomes deformed). If for example a car is pushed, the work of the push is the energy produced by this push. Work is expressed in joules ( J ), and is often noted W , initial of the English word Work which means work .

Definition

A constant force $\backslash \; vec\; \{F\}$ which applies to an object traversing a way rectilinear $\backslash \; vec\; \{U\}$ provides a energy, a work W

$W\; =\; \backslash \; vec\; \{F\}\; \backslash \; cdot\; \backslash \; vec\; \{U\}$

$\backslash \; overrightarrow\; \{F\}.\; \backslash \; overrightarrow\; \{U\}\; =\; F\; \backslash \; times\; U\; \backslash \; times\; cos\; (\backslash \; widehat\; \{fu\})$

It is noticed that only the component of $\backslash \; vec\; \{F\}$ which is parallel to $\backslash \; vec\; \{U\}$ works (property of the scalar Produit: the scalar of 2 orthogonal forces is null).

If the force changes during the way, or if the way is not rectilinear, one brings back oneself to one short duration dt during which the force can be presumedly constant and the way traversed $\backslash \; vec\; \{of\; the\}$ is regarded as rectilinear (tangent with the curve); this elementary work is noted $\backslash \; W$ delta and is worth:

$\backslash \; delta\; W\; =\; \backslash \; vec\; \{F\}\; \backslash \; cdot\; \backslash \; vec\; \{of\; the\}$.

One can then obtain the total work provided by the force $\backslash \; vec\; \{F\}$, by summoning work on the trajectory $\backslash \; mathcal\; \{C\}$ traversed by the point of application of $\backslash \; vec\; \{F\}$:

$W=\; \backslash \; int\_\; \{\backslash \; mathcal\; \{C\}\}\; \backslash \; vec\; \{F\}\; \backslash \; cdot\; \backslash \; vec\; \{of\; the\}$

If the trajectory is circular (for example if the point of application of a force is in rotation around an axis $(\backslash \; Delta)\; \backslash ,$), then the elementary work of the resulting moment is worth $\backslash \; delta\; W\; =\; \backslash \; vec\; \{M\}\; \backslash \; cdot\; \backslash \; vec\; \{D\; \backslash \; theta\}\; \backslash ,$, where $\backslash \; vec\; \{M\}$ is the Moment force compared to $(\backslash \; Delta)\; \backslash ,$ and $\backslash \; vec\; \{D\; \backslash \; theta\}$ the angle traversed by the solid during one short duration dt .

Particular cases

Let us consider a force $\backslash \; vec\; \{F\}$ constant applying to an object moving on a rectilinear trajectory (There are not other forces being exerted on the object). A certain number of particular cases make it possible to illustrate the concept of work of a force:

• If the force $\backslash \; vec\; \{F\}$ is parallel to displacement $\backslash \; vec\; \{U\}$ and directed in the same direction, work $W\; =\; \backslash \; vec\; \{F\}\; \backslash \; cdot\; \backslash \; vec\; \{U\}$ provided by the force is positive: according to the theorem of the kinetic energy, the force increased the kinetic energy of the system, this one thus moves more quickly. Such a force is sometimes called driving force .
• If         , angles being in degrees, then     $1\; >\; \backslash \; cos\; \backslash \; widehat\; \{(\backslash \; vec\; \{F\},\; \backslash \; vec\; \{U\})\}\; >\; 0$     the work provided by the force is positive.
  La force is driving .
• One can say more simply than if the force $\backslash \; vec\; \{F\}$ is driving, it supports displacement (speed increases)

• If the force $\backslash \; vec\; \{F\}$ is parallel to displacement $\backslash \; vec\; \{U\}$ but directed in the opposed direction, work $W\; =\; \backslash \; vec\; \{F\}\; \backslash \; cdot\; \backslash \; vec\; \{U\}$, provided by the force is negative: according to the theorem of the kinetic energy, the force decreased the kinetic energy of the system, this one thus moves more slowly. One calls sometimes such a force, a resistant force .

• If         , angles being in degrees, then     $0\; >\; \backslash \; cos\; \backslash \; widehat\; \{(\backslash \; vec\; \{F\},\; \backslash \; vec\; \{U\})\}\; >\; -1$     the work provided by the force is negative.
  La force is resistant .
• One can say more simply than if the force $\backslash \; vec\; \{F\}$ is resistant, it is opposed to displacement (speed decreases)

• If the force $\backslash \; vec\; \{F\}$ is perpendicular to displacement $\backslash \; vec\; \{U\}$, the work of the force is null $W\; =\; 0$: the force did not modify the kinetic energy of the system.

• One can say more simply than if the force $\backslash \; vec\; \{F\}$ is perpendicular to displacement, it does not modify displacement.

This last case should not let think that a force whose work is null does not have any effect on a system. Thus, in the case of a solid in uniform circular motion, the centripetal Force has a null work (the uniform circular motion is not modified). For as much, if one removes the centripetal force the solid ceases his circular motion and moves in rectilinear motion, in accordance with the 1^ère law of Newton.

Circular motion uniforme.
La centripetal force which creates the acceleration of the same name is perpendicular to the movement: its work is null.

The forces whose work is null do not modify the kinetic energy of the solid. In particular, they do not modify the standard of the vitesse  ; they can however modify the direction of it.

The case of the conservative forces: example of the weight

See also: conservative Force

The conservative forces are, by definition, of the forces whose work does not depend on the followed way. The Poids is an example.

Let us consider a body of Masse m moving of has towards B in a reference mark galiléen $\backslash \; left\; (O,\; \backslash \; vec\; \{X\},\; \backslash \; vec\; \{there\},\; \backslash \; vec\; \{Z\}\; \backslash \; right)$, the axis $\backslash \; vec\; \{Z\}$ being supposed vertical and being directed in the opposite direction of the Gravité: $\backslash \; vec\; \{G\}\; =-g\; \backslash \; vec\; \{Z\}$. In this case, the work of the weight is worth:

$W\; =\; \backslash \; vec\; \{P\}\; \backslash \; cdot\; \backslash \; vec\; \{AB\}\; =\; m\; \backslash \; vec\; \{G\}\; \backslash \; cdot\; \backslash \; vec\; \{AB\}\; =\; -\; Mg\; \backslash \; vec\; \{Z\}\; \backslash \; cdot\; \backslash \; vec\; \{AB\}$

If one notes $\backslash \; left\; (x\_A,\; y\_A,\; z\_A\; \backslash \; right)$ the Cartesian Coordonnées of the point has in this reference mark and $\backslash \; left\; (x\_B,\; y\_B,\; z\_B\; \backslash \; right)$ those of B then the coordinates of the vectors $\backslash \; vec\; \{P\}$ and $\backslash \; vec\; \{AB\}$ in the reference mark galiléen is the following ones:

$\backslash \; vec\; \{P\}\; =-mg\; \backslash \; vec\; \{Z\}$

$\backslash \; vec\; \{AB\}\; =\; \backslash \; left\; (x\_B-x\_A\; \backslash \; right)\; \backslash \; vec\; \{X\}\; +\; \backslash \; left\; (y\_B\; there\; \_A\; \backslash \; right)\; \backslash \; vec\; \{there\}\; +\; \backslash \; left\; (z\_B-z\_A\; \backslash \; right)\; \backslash \; vec\; \{Z\}$

and, by definition of the Produces scalar, the work of the weight is simplified in the following way:

$W=\; \backslash \; vec\; \{P\}\; \backslash \; cdot\; \backslash \; vec\; \{AB\}\; =\; Mg\; \backslash \; left\; (z\_A-z\_B\; \backslash \; right)$

The work of the weight of a body is thus independent of the way followed during its displacement, it depends only on the variation of altitude of the Center of gravity of this body.

Example

A person of mass 80 kg goes up upright on a 50 height centimetres chair. Which is the work carried out by the weight of this person?
$W\; =\; m\; \backslash \; G\; \backslash \; left\; (z\_A-z\_B\; \backslash \; right)$, is $W\; =\; 80\; \backslash \; times\; 10\; \backslash \; times\; (0\; -\; 0,5)\; =\; -\; 400\; \backslash \; J$
The weight is a resistant force in this case (It " oppose" with the displacement of the person).

See too

• Work and heat

Simple: Mechanical work

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