Work (physics): Difference between revisions

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imported>Paul Wormer
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== Definition ==
== Definition ==
As stated, work is force times path length: When  a ''constant'' force '''F''' acts on a body along a ''straight'' path and the body is moved over a length |'''s'''|  along this path, then
As stated, mechanical  work ''W'' is force times path length: When  a ''constant'' force '''F''' acts on a body along a ''straight'' path and the body is moved over a length |'''s'''|  along this path, then
:<math>
:<math>
W=\mathbf F \cdot \mathbf s = |\mathbf F| \, |\mathbf s|  \, \cos\alpha\,.
W=\mathbf F \cdot \mathbf s = |\mathbf F| \, |\mathbf s|  \, \cos\alpha\,.
</math>
</math>
Here '''s''' is  [[vector]] of magnitude |'''s'''| in direction of the path. The [[inner product]] between the two vectors  '''F''' and '''s'''  is the product of their magnitudes |'''F'''| and |'''s'''| times the [[cosine]] of the angle, &alpha;, between the vectors.  
Here '''s''' is  [[vector]] of magnitude |'''s'''| along the path. The [[inner product]] between the two vectors  '''F''' and '''s'''  is the product of their magnitudes |'''F'''| and |'''s'''| times the [[cosine]] of the angle, &alpha;, between the vectors.  


When the force is directed along the path, &alpha; = 0 and  the work is simply ''W'' = |'''F'''| |'''s'''|.
When the force is directed along the path ('''F''' and '''s''' parallel), &alpha; = 0, cos&alpha; = 1, and  we have simply ''W'' = |'''F'''| |'''s'''|.


When a force is perpendicular to the path, &alpha; = 90°, cos(&alpha;) = 0, and the force performs no work. Examples of such a situation are the [[centrifugal force]] on a mass in uniform circular motion and the gravitational force acting on a satellite in a circular orbit. If a body is in uniform (i.e., with constant speed)  straight motion  and the only force acting on it is perpendicular to its path, then the body will persist in its uniform straight motion ([[Newton's first law]]).  
When a force is perpendicular to the path, &alpha; = 90°, cos&alpha; = 0, the force performs no work. Examples of such a situation are the [[centrifugal force]] on a mass in uniform circular motion and the gravitational force acting on a satellite in a circular orbit. If a body is in uniform (i.e., has constant speed)  straight motion  and the only force acting on it is perpendicular to its path, then the body will persist in its uniform straight motion ([[Newton's first law]]).  


When a force '''F''' is antiparallel to the path, it performs negative work (and usually another force than '''F''' will move the body and perform positive work).
When a force '''F''' is anti-parallel to the path, it performs negative work, cos&alpha; = &minus;1, |'''F'''| and |'''s'''| both greater than zero. Work being a form of energy, it is conserved;  the negative work done by the force is converted into a decrease of the [[kinetic energy]] ''T'': &nbsp; ''W'' = &Delta;''T'' (both sides of the equation are negative, ''W'' because path and force are anti-parallel, &Delta;''T'' &equiv; ''T''<sub>1</sub> &minus; ''T''<sub>0</sub> because ''T'' ''decreases'', ''T''<sub>1</sub> < ''T''<sub>0</sub> ).
 
For example, think of an old-fashioned type of cannon that shoots a cannon ball of mass ''m''  straight up. The gun powder explosion gives the ball initial kinetic energy ''T''<sub>0</sub> = &frac12; ''m v''<sub>0</sub><sup>2</sup>. The gravitational attraction of the earth performs negative work (force downward, motion of the cannon ball upward), until at the highest point the speed and the kinetic energy of the ball are zero. The amount of work performed is &Delta;''T'' &equiv; 0&minus;''T''<sub>0</sub> = &minus;''T''<sub>0</sub>. At the highest point the motion of the cannon ball reverts direction,  it starts falling  to earth, and from there on the gravitational attraction performs positive work (direction of motion and force are parallel). The kinetic energy increases again until it achieves its original value ''T''<sub>0</sub> = &frac12; ''m v''<sub>0</sub><sup>2</sup> at the point where the cannon ball arrives again at the cannon. The total work done by the gravitation is zero, the work done on the cannon ball going up cancels the work done on the ball going down. (In this example we ignore friction by the air).
 
When the the force is [[conservative force|conservative]] (non-dissipative), the work is independent of path, and when furthermore the path is a closed curve, the total work is zero (one way the work is positive and the other way the work is equally large in absolute value, but negative).
 
As the gravitational field is conservative, we just saw an example  of a mass making a closed path in a conservative force field. The potential energy of a mass ''m'' close to the surface of the earth is equal to ''mgh'', where ''g'' is the [[gravitational acceleration]] and ''h'' is the height.
Assuming that the cannon is positioned  on height ''h'' = 0, the cannon ball starts with potential energy zero. The work done by the gravitational field on the cannon ball going upward has two effects: it decreases its kinetic energy&mdash;as just discussed&mdash;and it increases its potential energy. At the highest point ''h''<sub>max</sub> the potential energy is maximum ''mgh''<sub>max</sub> (equal to minus the work performed)  and the kinetic energy is zero (this is the point where the motion of the cannon ball changes direction, for a small amount of time its speed is zero). The work done by a conservative force converts a decrease in kinetic energy into an increase of potential energy and conversely. When the cannon ball dropping down is at height ''h'' = 0 again, its kinetic energy is maximum (equal to the energy imparted to it by the gun powder) and its potential energy is zero again.


Work being a form of energy, it is conserved;  the work done by the force is converted into [[kinetic energy]], if the speed of the body increases, or into [[potential energy]], or a mixture of both.


When the force is not constant along the path, or the path is not straight, it is possible to compute the work by [[infinitesimal calculus]]. One divides  the path in ''N''  pieces &Delta;'''s'''<sub>''i''</sub>, which are small enough to assume that the force is constant and the piece is straight. The (approximate) total work is obtained by summing the work done along the individual small pieces,
When the force is not constant along the path, or the path is not straight, it is possible to compute the work by [[infinitesimal calculus]]. One divides  the path in ''N''  pieces &Delta;'''s'''<sub>''i''</sub>, which are small enough to assume that the force is constant and the piece is straight. The (approximate) total work is obtained by summing the work done along the individual small pieces,
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</math>
</math>
where '''s'''<sub>1</sub> is the start and '''s'''<sub>2</sub> the end point of the path.
where '''s'''<sub>1</sub> is the start and '''s'''<sub>2</sub> the end point of the path.
When the the force is [[conservative force|conservative]], the work is independent of path, and when furthermore the path is a closed curve, the total work is zero (one way the work is positive and the other way the work is equally large in absolute value, but negative).


== Example of mechanical work ==
== Example of mechanical work ==

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In physics, work is the energy that is transferred to a body when it is moved along a path by a force. When the force is conservative (non-dissipative) the work is independent of the path. Work done on a body is accomplished not only by a displacement of the body as a whole from one place to another but also, for example, by compressing a gas, by rotating a shaft, and even by directing small magnetic particles within a body along an external magnetic field.

The use of the terms "work", "energy", "force" have a well-defined, quantitative, meaning in physics, which differs somewhat from their more qualitative meaning in daily life. For instance, in physics work can be negative.

Definition

As stated, mechanical work W is force times path length: When a constant force F acts on a body along a straight path and the body is moved over a length |s| along this path, then

Here s is vector of magnitude |s| along the path. The inner product between the two vectors F and s is the product of their magnitudes |F| and |s| times the cosine of the angle, α, between the vectors.

When the force is directed along the path (F and s parallel), α = 0, cosα = 1, and we have simply W = |F| |s|.

When a force is perpendicular to the path, α = 90°, cosα = 0, the force performs no work. Examples of such a situation are the centrifugal force on a mass in uniform circular motion and the gravitational force acting on a satellite in a circular orbit. If a body is in uniform (i.e., has constant speed) straight motion and the only force acting on it is perpendicular to its path, then the body will persist in its uniform straight motion (Newton's first law).

When a force F is anti-parallel to the path, it performs negative work, cosα = −1, |F| and |s| both greater than zero. Work being a form of energy, it is conserved; the negative work done by the force is converted into a decrease of the kinetic energy T:   W = ΔT (both sides of the equation are negative, W because path and force are anti-parallel, ΔTT1T0 because T decreases, T1 < T0 ).

For example, think of an old-fashioned type of cannon that shoots a cannon ball of mass m straight up. The gun powder explosion gives the ball initial kinetic energy T0 = ½ m v02. The gravitational attraction of the earth performs negative work (force downward, motion of the cannon ball upward), until at the highest point the speed and the kinetic energy of the ball are zero. The amount of work performed is ΔT ≡ 0−T0 = −T0. At the highest point the motion of the cannon ball reverts direction, it starts falling to earth, and from there on the gravitational attraction performs positive work (direction of motion and force are parallel). The kinetic energy increases again until it achieves its original value T0 = ½ m v02 at the point where the cannon ball arrives again at the cannon. The total work done by the gravitation is zero, the work done on the cannon ball going up cancels the work done on the ball going down. (In this example we ignore friction by the air).

When the the force is conservative (non-dissipative), the work is independent of path, and when furthermore the path is a closed curve, the total work is zero (one way the work is positive and the other way the work is equally large in absolute value, but negative).

As the gravitational field is conservative, we just saw an example of a mass making a closed path in a conservative force field. The potential energy of a mass m close to the surface of the earth is equal to mgh, where g is the gravitational acceleration and h is the height. Assuming that the cannon is positioned on height h = 0, the cannon ball starts with potential energy zero. The work done by the gravitational field on the cannon ball going upward has two effects: it decreases its kinetic energy—as just discussed—and it increases its potential energy. At the highest point hmax the potential energy is maximum mghmax (equal to minus the work performed) and the kinetic energy is zero (this is the point where the motion of the cannon ball changes direction, for a small amount of time its speed is zero). The work done by a conservative force converts a decrease in kinetic energy into an increase of potential energy and conversely. When the cannon ball dropping down is at height h = 0 again, its kinetic energy is maximum (equal to the energy imparted to it by the gun powder) and its potential energy is zero again.


When the force is not constant along the path, or the path is not straight, it is possible to compute the work by infinitesimal calculus. One divides the path in N pieces Δsi, which are small enough to assume that the force is constant and the piece is straight. The (approximate) total work is obtained by summing the work done along the individual small pieces,

To improve the approximation one makes the pieces Δs smaller and smaller, so that their lengths go to zero. The limit is the path integral

where s1 is the start and s2 the end point of the path.

Example of mechanical work

When h is positive, and the mass is at rest before and after the lifting (no kinetic energy), the work is completely converted into potential energy.
Numeric examples:
Here N is the unit newton and J is the unit joule.