Work (Physics) - Wik
Work (Physics) - Wik
Work (Physics) - Wik
http://en.wikipedia.org/wiki/Mechanical_work
Work (physics)
From Wikipedia, the free encyclopedia
(Redirected from Mechanical work) In physics, a force is said to do work when it acts on a body, and there is a displacement of the point of application in the direction of the force. The force does not need to cause the displacement. For example, when you lift a suitcase from the floor, there are two forces that do work: the normal force by your hand and the gravitational force. The term work was introduced in 1826 by the French mathematician Gaspard-Gustave Coriolis[1][2] as "weight lifted through a height", which is based on the use of early steam engines to lift buckets of water out of flooded ore mines. The SI unit of work is the newton-metre or joule (J). The work done by a constant force of magnitude F on a point that moves a displacement (NOT distance) d in the direction of the force is the product,
Work
A baseball pitcher does positive work on the ball by applying a force to it over the distance it moves while in his grip.
For example, if a force of 10 newton (F = 10 N) acts along a point that travels 2 metres (d = 2 m), then it does the work W = (10 N)(2 m) = 20 N m = 20 J. This is approximately the work done lifting a 1 kg weight from ground to over a persons head against the force of gravity. Notice that the work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance.
Contents
1 Units 2 Work and energy 3 Constraint forces 4 Mathematical calculation 4.1 Torque and rotation 5 Work and potential energy 5.1 Path dependence 5.2 Path independence 5.3 Work by gravity 5.4 Work by a spring 5.5 Work by a gas 6 Work-energy principle 6.1 Overview 6.2 Derivation for a particle moving along a straight line 6.3 General derivation of the work-energy theorem for a particle 6.4 Derivation for a particle in constrained movement 6.4.1 Vector formulation 6.4.2 Tangential and normal components 6.5 Moving in a straight line (skid to a stop) 6.6 Coasting down a mountain road (gravity racing) 7 Work of forces acting on a rigid body
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Units
The SI unit of work is the joule (J), which is defined as the work expended by a force of one newton through a distance of one metre. The dimensionally equivalent newton-metre (Nm) is sometimes used as the measuring unit for work, but this can be confused with the unit newton-metre, which is the measurement unit of torque. Usage of Nm is discouraged by the SI authority, since it can lead to confusion as to whether the quantity expressed in newton metres is a torque measurement, or a measurement of energy.[3] Non-SI units of work include the erg, the foot-pound, the foot-poundal, the kilowatt hour, the litre-atmosphere, and the horsepower-hour. Due to work having the same physical dimension as heat, occasionally measurement units typically reserved for heat or energy content, such as therm, BTU and Calorie, are utilized as a measuring unit.
indicates that heat (Q) and work (W) are inexact differentials.
From Newtons second law, it can be shown that work on a free (no fields), rigid (no internal degrees of freedom) body, is equal to the change in kinetic energy of the velocity and rotation of that body,
The work of forces generated by a potential function is known as potential energy and the forces are said to be conservative. Therefore work on an object that is merely displaced in a conservative force field, without change in velocity or rotation, is equal to minus the change of potential energy of the object,
These formulas demonstrate that work is the energy associated with the action of a force, so work subsequently possesses the physical dimensions, and units, of energy. The work/energy principles discussed here are identical to Electric work/energy principles.
Constraint forces
Constraint forces determine the movement of components in a system, constraining the object within a boundary (in the case of a slope plus gravity, the object is stuck to the slope, when attached to a taut string it cannot move in an outwards direction to make the string any tauter). They eliminate all movement in the direction of the constraint, thus constraint forces do not perform work on the system, as the velocity of that object is constrained to be 0 parallel to this force, due to this force. For example, the centripetal force exerted inwards by a string on a ball in uniform circular motion sideways constrains the ball to circular motion restricting its movement away from the center of the circle. This force does zero work because it is perpendicular to the velocity of the ball.
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Another example is a book on a table. If external forces are applied to the book so that it slides on the table, then the force exerted by the table constrains the book from moving downwards. The force exerted by the table supports the book and is perpendicular to its movement which means that this constraint force does not perform work. The magnetic force on a charged particle is F = qv B, where q is the charge, v is the velocity of the particle, and B is the magnetic field. The result of a cross product is always perpendicular to both of the original vectors, so F v. The dot product of two perpendicular vectors is always zero, so the work W = F v = 0, and the magnetic force does not do work. It can change the direction of motion but never change the speed.
Mathematical calculation
For moving objects, the quantity of work/time is calculated as distance/time, or velocity. Thus, at any instant, the rate of the work done by a force (measured in joules/second, or watts) 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 classified as instantaneous 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.[4] Work is the result of a force on a point that moves through a distance. As the point moves, it follows a curve X, with a velocity v, at each instant. The small amount of work W that occurs over an instant of time dt is calculated as
where the F.v is the power over the instant dt. The sum of these small amounts of work over the trajectory of the point yields the work,
where C is the trajectory from x(t1) to x(t2). 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 F, then this integral simplifies to
where s is distance along the line. If F is constant, in addition to being directed along the line, then the integral simplifies further to
where d is the distance travelled by 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 Fdx = Fcosdx, where is the angle between the force vector and the direction of movement, [4] that is
In the notable case of a force applied to a body always at an angle of 90 degrees from the velocity vector (as when a body moves in a circle under a central force), no work is done at all, since the cosine of 90 degrees is zero. Thus, no work can be
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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 friction-less ideal centrifuge. 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 ( , where is the angle between the force and the velocity). And then the most general definition of work can be formulated as follows: Work of a force is the line integral of its scalar tangential component along the path of its application point.
where the T. is the power over the instant t. The sum of these small amounts of work over the trajectory of the rigid body yields the work,
This integral is computed along the trajectory of the rigid body with an angular velocity 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,
where is the angle of rotation about the constant unit vector S. In this case, the work of the torque becomes,
where C is the trajectory from (t1) to (t2). This integral depends on the rotational trajectory (t), and is therefore path-dependent. If the torque T is aligned with the angular velocity vector so that,
and both the torque and angular velocity are constant, then the work takes the form, [5]
This result can be understood more simply by considering the torque as arising from a force of constant magnitude F, being applied perpendicularly to a lever arm at a distance r, as shown in the figure. This force will act through the distance along the circular arc s=r, so the work done is
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as presented above. Notice that only the component of torque in the direction of the angular velocity vector contributes to the work.
The scalar product of a force F and the velocity v of its point of application defines the power input to a system at an instant of time. Integration of this power over the trajectory of the point of application, C=x(t), defines the work input to the system by the force.
Path dependence
Therefore, the work done by a force F on an object that travels along a curve C is given by the line integral:
where x(t) defines the trajectory C and v is the velocity along this trajectory. In general this integral requires the path along which the velocity is defined, so the evaluation of work is said to be path dependent. The time derivative of the integral for work yields the instantaneous power,
Path independence
If the work for an applied force is independent of the path, then the work done by the force is evaluated at the start and end of the trajectory of the point of application. This means that there is a function U (x), called a "potential," that can be evaluated at the two points x(t1) and x(t2) to obtain the work over any trajectory between these two points. It is tradition to define this function with a negative sign so that positive work is a reduction in the potential, that is
The function U(x) is called the potential energy associated with the applied force. Examples of forces that have potential energies are gravity and spring forces. In this case, the partial derivative of work yields
and the force F is said to be "derivable from a potential."[6] Because the potential U defines a force F at every point x in space, the set of forces is called a force field. The power
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applied to a body by a force field is obtained from the gradient of the work, or potential, in the direction of the velocity V of the body, that is
Examples of work that can be computed from potential functions are gravity and spring forces.
Work by gravity
Gravity exerts a constant downward force F=(0, mg, 0) on the center of mass of a body moving near the surface of the earth.
where the integral of the vertical component of velocity is the vertical distance. Notice that the work of gravity depends only on the vertical movement of the curve X(t). Where w is weight (pounds in imperial units, and mass*gravity in SI units), and y is the change in height.
Gravity F=mg does work W=mgh along any descending path
Work by a spring
A horizontal spring exerts a force F=(kx, 0, 0) that is proportional to its deflection in the x direction. The work of this spring on a body moving along the space curve X(t) = (x(t), y(t), z(t)), is calculated using its velocity, v=(vx, vy, vz), to obtain
For convenience, consider contact with the spring occurs at t=0, then the integral of the product of the distance x and the x-velocity, xvx, is (1/2)x2.
Work by a gas
Where P is pressure, V is volume, and a and b are initial and final volumes.
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.[7] The work W done by the resultant force on a particle equals the change in the particles kinetic energy , where and are the speeds of the particle before and after the change and m is its mass. ,[5]
If a resultant torque acts on a rigid body, the definition can be extended to equate the work done by the resultant torque and the change in rotational kinetic energy of the rigid body.
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Overview
The derivation of the work-energy principle begins with Newtons second law and the resultant force on a particle which includes forces applied to the particle and constraint forces imposed on its movement. Computation of the scalar product of the forces with the velocity of the particle evaluates the instantaneous power added to the system.[8] 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 velocity and acceleration. The fact the work-energy principle eliminates the constraint forces underlies Lagrangian mechanics.[9] This section focusses 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 heat 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.
The work of the net force is calculated as the product of its magnitude and the particle displacement. Substituting the above equations, one obtains:
In the general case of rectilinear motion, when the net force 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:
The identity
and definition
it follows
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The remaining part of the above derivation is just simple calculus, same as in the preceding rectilinear case.
where 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 Newtons law with the velocity vector yields
because the constraint forces are perpendicular to the particle velocity. Integrate this equation along its trajectory from the point X(t1) to the point X(t2) to obtain
The left side of this equation is the work of the applied force as it acts on the particle along the trajectory from time t 1 to time t2. This can also be written as
This integral is computed along the trajectory X(t) of the particle and is therefore path dependent. The right side of the first integral of Newtons equations can be simplified using the following identity
(see product rule for derivation). Now it is integrated explicitly to obtain the change in kinetic energy,
where the kinetic energy of the particle is defined by the scalar quantity,
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Tangential and normal components It is useful to resolve the velocity and acceleration vectors into tangential and normal components along the trajectory X(t), such that
where
Then, the scalar product of velocity with acceleration in Newtons second law takes the form . where the kinetic energy of the particle is defined by the scalar quantity,
For convenience let the trajectory be along the X-axis, so X=(d,0) and the velocity is V=(v, 0), then R.V=0, and F.V=Fxv, where Fx is the component of F along the X-axis, so
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 Fx =-kW. The velocity v of the car can be determined from the length s of the skid using the work-energy principle,
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Notice that this formula uses the fact that the mass of the vehicle is m=W/g.
The scalar product of this equation with the velocity, V=(vx, vy,vz), yields
where V is the magnitude of V. The constraint forces between the vehicle and the road cancel from this equation because R.V=0, which means they do no work. Integrate both sides to obtain
Gravity racing championship in Campos Novos, Santa Catarina, Brazil, 8 September 2010.
The weight force W is constant along the trajectory and the integral of the vertical velocity is the vertical distance, therefore,
Recall that V(t1)=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 traveledfor angles this small the sin and tan functions are approximately equal. Therefore, the distance s in feet down a 6% grade to reach the velocity V is at least
This formula uses the fact that the weight of the vehicle is W=mg.
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The trajectories of Xi, i=1,...,n are defined by the movement of the rigid body. This movement is given by the set of rotations [A(t)] and the trajectory d(t) of a reference point in the body. Let the coordinates xi i=1,...,n define these points in the moving rigid bodys reference frame M, so that the trajectories traced in the fixed frame F are given by
where is the angular velocity vector obtained from the skew symmetric matrix
known as the angular velocity matrix. The small amount of work by the forces over the small displacements ri can be determined by approximating the displacement by r=vt so
or
where F and T are the resultant force and torque applied at the reference point d of the moving frame M in the rigid body.
References
1. ^ Jammer, Max (1957). Concepts of Force (http://books.google.com/?id=CZtEBcmOe6gC&printsec=frontcover#PPA167,M1). Dover Publications, Inc. p. 167; footnote 14. ISBN 0-486-40689-X. 2. ^ Coriolis, Gustave. (1829). Calculation of the Effect of Machines, or Considerations on the Use of Engines and their Evaluation (Du Calcul de leffet des Machines, ou Considrations sur lemploi des Moteurs et sur Leur Evaluation). Paris: Carilian-Goeury, Libraire. 3. ^ http://www.bipm.org/en/si/si_brochure/chapter2/2-2/2-2-2.html 4. ^ a b 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 5. ^ a b Hugh D. Young and Roger A. Freedman (2008). University Physics (12th ed.). Addison-Wesley. p. 329. ISBN 978-0-321-50130-1. 6. ^ J. R. Taylor, Classical Mechanics, (http://books.google.com/books?id=P1kCtNr-pJsC&pg=PA117&lpg=PA117#v=onepage& q&f=false) University Science Books, 2005. 7. ^ Andrew Pytel, Jaan Kiusalaas (2010). Engineering Mechanics: Dynamics SI Version, Volume 2 (3rd ed.). Cengage Learning,. p. 654. ISBN 9780495295631. 8. ^ B. Paul, Kinematics and Dynamics of Planar Machinery, (http://books.google.com/books?ei=JGMfULH-KK2r2AWIzICYAw& id=3UdSAAAAMAAJ&dq=kinematics+and+dynamics+of+planar+machinery&q=instantaneous+power#search_anchor) Prentice-Hall, 1979. 9. ^ E. T. Whittaker, A treatise on the analytical dynamics of particles and rigid bodies, (http://books.google.com /books?id=OSckAAAAMAAJ&printsec=frontcover&dq=analytical+dynamics&source=bl&ots=AGqvlTnrw7&sig=x7DfbqV97-
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Bibliography
Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers (6th ed. ed.). Brooks/Cole. ISBN 0-534-40842-7. Tipler, Paul (1991). Physics for Scientists and Engineers: Mechanics (3rd ed., extended version ed.). W. H. Freeman. ISBN 0-87901-432-6.
External links
Work (http://www.lightandmatter.com/html_books/2cl/ch03/ch03.html) a chapter from an online textbook Work-energy principle (http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html) Retrieved from "http://en.wikipedia.org/w/index.php?title=Work_(physics)&oldid=582919069" Categories: Energy (physics) Physical quantities Mechanical engineering Mechanics Force Length This page was last modified on 23 November 2013 at 05:43. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization.
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