Monday, 19 September 2011

Work

In physics, work is the amount of energy transferred to a body by a force acting through a distance in the direction of the force. Like energy, it is a scalar quantity, with SI units of joules. The term work was first coined in 1826 by the French mathematician Gaspard-Gustave Coriolis.[1][2]
According to the work-energy theorem, if one or more external forces act upon a rigid object, causing its kinetic energy to change from Ek1 to Ek2, then the mechanical work (W) done by the net force is equal to the change in kinetic energy. For translational motion, the theorem can be specified as:[3]
W = \Delta E_k = E_{k_2} - E_{k_1} = \tfrac12 m (v_2^2 - v_1^2) \,\!
where m is the mass of the object and v is the object's velocity.
If a force F that is constant with respect to time acts on an object while the object is displaced in a straight line along the length and direction of a vector d, the mechanical work done by the force on the object is the dot product of the vectors F and d:[4]
W = \bold{F} \cdot \bold{d} = F d \cos\theta
If the force and the displacement are parallel and in the same direction (θ = 0), the mechanical work is positive. If the force and the displacement are parallel but in opposite directions (i.e. antiparallel, θ = 180⁰), the mechanical work is negative. If a force F is applied at an angle θ, only the component of the force in the same direction as the displacement (Fcosθ) does work. Thus, if the force acts perpendicular to the displacement (θ = 90⁰ or 270⁰), zero work is done by the force.

Units

The SI unit of work is the joule (J), which is defined as the work done by a force of one newton acting over a distance of one meter. This definition is based on Sadi Carnot's 1824 definition of work as "weight liftednewton-metertorque to distinguish its units from work or energy. through a height", which is based on the fact that early steam engines were principally used to lift buckets of water, through a gravitational height, out of flooded ore mines. The dimensionally equivalent (N·m) is sometimes used instead; however, it is also sometimes reserved for
Non-SI units of work include the erg, the foot-pound, the foot-poundal, and the liter-atmosphere.
Heat conduction is not considered to be a form of work, since the energy is transferred into atomic vibration rather than a macroscopic displacement.

[edit] Zero work

A baseball pitcher does positive work on the ball by transferring energy into it.
Work can be zero even when there is a force. The centripetal force in a uniform circular motion, for example, does zero work since the kinetic energy of the moving object doesn't change. This is because the force is always perpendicular to the motion of the object; only the component of a force parallel to the velocity vector of an object can do work on that object. Likewise when a book sits on a table, the table does no work on the book despite exerting a force equivalent to mg upwards, because no energy is transferred into or out of the book.
If the table were moving upward at a constant velocity, then it will be doing work on the book, since the force of the table on the book will be acting through a distance. However, the force of gravity will be doing equal and opposite work on the book, and the net rate of work done on the book will still be zero, as evidenced by the fact that its kinetic energy remains constant throughout the process. A current that generates a magnetic field can also produce a magnetic force where a charged particle exerts a force on a magnetic field, but the magnetic force can do no work because the charge velocity is perpendicular to the magnetic field and in order for a force or an object to perform work, the force has to be in the same direction as the distance that it moves.

Mathematical calculation

Force and displacement

Force and displacement are both vector quantities and they are combined using the dot product to evaluate the mechanical work, a scalar quantity:
W = \bold{F} \cdot \bold{d} = F d \cos\theta        (1)
where \textstyle\theta is the angle between the force and the displacement vector.
Gravity F=mg does work W=mgh along any descending path
In order for this formula to be valid for translational motion, 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 \scriptstyle\bold{d}. 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 \scriptstyle d \cos\theta = h , which gives the familiar result \scriptstyle mgh .
More generally, if the force causes (or affects) rotation of the object, 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 \scriptstyle \mathrm{d}\bold{x} can equally be interpreted as differential displacement magnitude or differential length of the path of the application point.
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.
The general definition of mechanical work is given by the following line integral:
W_C = \int_{C} \bold{F} \cdot \mathrm{d}\bold{x}       (2)
where:
\textstyle _C is the path or curve traversed by the object;
\bold F is the force vector; and
\bold x is the position vector.
The expression \delta W = \bold{F} \cdot \mathrm{d}\bold{x} is an inexact differential which means that the calculation of \textstyle{ W_C} is path-dependent and cannot be differentiated to give \bold{F} \cdot \mathrm{d}\bold{x}.
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.
The possibility of a nonzero force doing zero work illustrates the difference between work and a related quantity, 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.
A force of constant magnitude and perpendicular to the lever arm

Torque and rotation

Work done by a torque can be calculated in a similar manner, as is easily seen when a force of constant magnitude is applied perpendicularly to a lever arm. After extraction of this constant value, the integral in the equation (2) gives the path length of the application point, i.e. the circular arc \ s , and the work done is \ W=Fs .
However, the arc length can be calculated from the angle of rotation \varphi\; (expressed in radians) as \ s= r \varphi\; , and the ensuing product \ Fr \; is equal to the torque \tau\; applied to the lever arm. Therefore, a constant torque does work as follows:
W= \tau \varphi\

Frame of reference

The work done by a force acting on an object depends on the choice of reference frame because displacements and velocities are dependent on the reference frame in which the observations are being made.[5]
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.[6]

References

  1. ^ Jammer, Max (1957). Concepts of Force. Dover Publications, Inc.. ISBN 0-486-40689-X. 
  2. ^ Sur une nouvelle dénomination et sur une nouvelle unité à introduire dans la dynamique, Académie des sciences, August 1826
  3. ^ Tipler (1991), page 138.
  4. ^ a b Resnick, Robert and Halliday, David (1966), Physics, Section 7-2 (Vol I and II, Combined edition), Wiley International Edition, Library of Congress Catalog Card No. 66-11527
  5. ^ 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
  6. ^ Resnick, Robert and Halliday, David, Physics, Section 4-6

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