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Work (physics)

From Wikipedia, the free encyclopedia
Process of energy transfer to an object via force application through displacement

Work
Abaseballpitcher does positive work on the ball by applying a force to it over the distance it moves while in his grip.
Common symbols
W
SI unitjoule (J)
Other units
Foot-pound,Erg
InSI base units1kg⋅m2⋅s−2
Derivations from
other quantities
W =F ⋅s
W =τθ
DimensionML2T−2{\displaystyle {\mathsf {M}}{\mathsf {L}}^{2}{\mathsf {T}}^{-2}}
For otheruses of "Work" in physics, seeWork (electric field) andWork (thermodynamics).
Part of a series on
Classical mechanics
F=dpdt{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}}

In science,work is theenergy transferred to or from anobject via the application offorce along adisplacement. In its simplest form, for a constant force aligned with the direction of motion, the work equals theproduct of the force strength and the distance traveled. A force is said to dopositive work if it has a component in the direction of the displacement of thepoint of application. A force doesnegative work if it has a component opposite to the direction of the displacement at the point of application of the force.[1]

For example, when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is positive, and is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement). If the ball is thrown upwards, the work done by the gravitational force is negative, and is equal to the weight multiplied by the displacement in the upwards direction.

Both force and displacement arevectors. The work done is given by thedot product of the two vectors, where the result is ascalar. When the forceF is constant and the angleθ between the force and the displacements is also constant, then the work done is given by:W=F⋅s=Fscos⁡θ{\displaystyle W=\mathbf {F} \cdot \mathbf {s} =Fs\cos {\theta }}

If the force and/or displacement is variable, then work is given by theline integral:

W=∫F⋅ds=∫F⋅dsdtdt=∫F⋅vdt{\displaystyle {\begin{aligned}W&=\int \mathbf {F} \cdot d\mathbf {s} \\&=\int \mathbf {F} \cdot {\frac {d\mathbf {s} }{dt}}dt\\&=\int \mathbf {F} \cdot \mathbf {v} dt\end{aligned}}}

whereds{\displaystyle d\mathbf {s} } is the infinitesimal change in displacement vector,dt{\displaystyle dt} is the infinitesimal increment of time, andv{\displaystyle \mathbf {v} } represents the velocity vector. The first equation represents force as a function of the position and the second and third equations represent force as a function of time.

Work is ascalar quantity,[2] so it has only magnitude and no direction. Work transfers energy from one place to another, or one form to another. TheSI unit of work is thejoule (J), the same unit as for energy.

History

[edit]

Theancient Greek understanding of physics was limited to thestatics of simple machines (the balance of forces), and did not includedynamics or the concept of work. During theRenaissance the dynamics of theMechanical Powers, as thesimple machines were called, began to be studied from the standpoint of how far they could lift a load, in addition to the force they could apply, leading eventually to the new concept of mechanical work. The complete dynamic theory of simple machines was worked out by Italian scientistGalileo Galilei in 1600 inLe Meccaniche (On Mechanics), in which he showed the underlying mathematical similarity of the machines as force amplifiers.[3][4] He was the first to explain that simple machines do not create energy, only transform it.[3]

Early concepts of work

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Althoughwork was not formally used until 1826, similar concepts existed before then. Early names for the same concept includedmoment of activity, quantity of action, latent live force, dynamic effect, efficiency, and evenforce.[5] In 1637, the French philosopherRené Descartes wrote:[6]

Lifting 100 lb one foot twice over is the same as lifting 200 lb one foot, or 100 lb two feet.

— René Descartes, Letter to Huygens

In 1686, the German philosopherGottfried Leibniz wrote:[7]

The same force ["work" in modern terms] is necessary to raise body A of 1 pound (libra) to a height of 4 yards (ulnae), as is necessary to raise body B of 4 pounds to a height of 1 yard.

— Gottfried Leibniz, Brevis demonstratio

In 1759,John Smeaton described a quantity that he called "power" "to signify the exertion of strength, gravitation, impulse, or pressure, as to produce motion." Smeaton continues that this quantity can be calculated if "the weight raised is multiplied by the height to which it can be raised in a given time," making this definition remarkably similar toCoriolis's.[8]

Etymology and modern usage

[edit]

The termwork (ormechanical work), and the use of thework-energy principle in mechanics, was introduced in the late 1820s independently by French mathematicianGaspard-Gustave Coriolis and French Professor of Applied MechanicsJean-Victor Poncelet.[9][10][11] Both scientists were pursuing a view of mechanics suitable for studying the dynamics and power of machines, for examplesteam engines lifting buckets of water out of flooded ore mines. According to Rene Dugas, French engineer and historian, it is toSolomon of Caux "that we owe the termwork in the sense that it is used in mechanics now".[12] The concept ofvirtual work, and the use ofvariational methods in mechanics, preceded the introduction of "mechanical work" but was originally called "virtual moment". It was re-named once the terminology of Poncelet and Coriolis was adopted.[13][14]

Units

[edit]

TheSI unit of work is thejoule (J), named after English physicistJames Prescott Joule (1818–1889). According to theInternational Bureau of Weights and Measures it is defined as "the work done when the point of application of 1MKS unit of force [newton] moves a distance of 1 metre in the direction of the force."[15]

The dimensionally equivalentnewton-metre (N⋅m) is sometimes used as the measuring unit for work, but this can be confused with the measurement unit oftorque. Usage of N⋅m is discouraged by theSI authority, since it can lead to confusion as to whether the quantity expressed in newton-metres is a torque measurement, or a measurement of work.[16]

Another unit for work is thefoot-pound, which comes from the English system of measurement. As the unit name suggests, it is the product of pounds for the unit of force and feet for the unit of displacement. One joule is approximately equal to 0.7376 ft-lbs.[17][18]

Non-SI units of work include the newton-metre,erg, the foot-pound, thefoot-poundal, thekilowatt hour, thelitre-atmosphere, and thehorsepower-hour. Due to work having the samephysical dimension asheat, occasionally measurement units typically reserved for heat or energy content, such astherm,BTU andcalorie, are used as a measuring unit.

Work and energy

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The workW done by a constant force of magnitudeF on a point that moves a displacements in a straight line in the direction of the force is the productW=F⋅s{\displaystyle W=\mathbf {F} \cdot \mathbf {s} }

For example, if a force of 10 newtons (F = 10 N) acts along a point that travels 2 metres (s = 2 m), thenW =Fs = (10 N) (2 m) = 20 J. This is approximately the work done lifting a 1 kg object from ground level to over a person's head against the force of gravity.

The work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance.

Work is closely related toenergy. Energy shares the same unit of measurement with work (Joules) because the energy from the object doing work is transferred to the other objects it interacts with when work is being done.[18] The work–energy principle states that an increase in the kinetic energy of arigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force. Thus, if the net work is positive, then the particle's kinetic energy increases by the amount of the work. If the net work done is negative, then the particle's kinetic energy decreases by the amount of work.[19]

FromNewton's 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 energyEk corresponding to the linear velocity andangular velocity of that body,W=ΔEk.{\displaystyle W=\Delta E_{\text{k}}.}The work of forces generated by a potential function is known aspotential energy and the forces are said to beconservative. Therefore, work on an object that is merely displaced in a conservative forcefield, without change in velocity or rotation, is equal tominus the change of potential energyEp of the object,W=−ΔEp.{\displaystyle W=-\Delta E_{\text{p}}.}These formulas show that work is the energy associated with the action of a force, so work subsequently possesses thephysical dimensions, and units, of energy.The work/energy principles discussed here are identical to electric work/energy principles.

Constraint forces

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Constraint forces determine the object's displacement in the system, limiting it within a range. For example, in the case of aslope plus gravity, the object isstuck to the slope and, when attached to a taut string, it cannot move in an outwards direction to make the string any 'tauter'. It eliminates all displacements in that direction, that is, the velocity in the direction of the constraint is limited to 0, so that the constraint forces do not perform work on the system.

For amechanical system,[20] constraint forces eliminate movement in directions that characterize the constraint. Thus thevirtual work done by the forces of constraint is zero, a result which is only true if friction forces are excluded.[21]

Fixed, frictionless constraint forces do not perform work on the system,[22] as the angle between the motion and the constraint forces is always90°.[22] Examples of workless constraints are: rigid interconnections between particles, sliding motion on a frictionless surface, and rolling contact without slipping.[23]

For example, in a pulley system like theAtwood machine, the internal forces on the rope and at the supporting pulley do no work on the system. Therefore, work need only be computed for the gravitational forces acting on the bodies. Another example is thecentripetal force exertedinwards by a string on a ball in uniformcircular motionsideways constrains the ball to circular motion restricting its movement away from the centre of the circle. This force does zero work because it is perpendicular to the velocity of the ball.

Themagnetic force on a charged particle isF =qv ×B, whereq is the charge,v is the velocity of the particle, andB is themagnetic field. The result of across product is always perpendicular to both of the original vectors, soF ⊥v. Thedot product of two perpendicular vectors is always zero, so the workW =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

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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, orwatts) is thescalar 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 instantaneouspower. Just as velocities may be integrated over time to obtain a total distance, by thefundamental 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.[24]

Work is the result of a force on a point that follows a curveX, with a velocityv, at each instant. The small amount of workδW that occurs over an instant of timedt is calculated asδW=F⋅ds=F⋅vdt{\displaystyle \delta W=\mathbf {F} \cdot d\mathbf {s} =\mathbf {F} \cdot \mathbf {v} dt}where theF ⋅v is the power over the instantdt. The sum of these small amounts of work over the trajectory of the point yields the work,W=∫t1t2F⋅vdt=∫t1t2F⋅dsdtdt=∫CF⋅ds,{\displaystyle 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} ,}whereC is the trajectory fromx(t1) tox(t2). This integral is computed along the trajectory of the particle, and is therefore said to bepath dependent.

If the force is always directed along this line, and the magnitude of the force isF, then this integral simplifies toW=∫CFds{\displaystyle W=\int _{C}F\,ds}wheres is displacement along the line. IfF is constant, in addition to being directed along the line, then the integral simplifies further toW=∫CFds=F∫Cds=Fs{\displaystyle W=\int _{C}F\,ds=F\int _{C}ds=Fs}wheres 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 thedot productF ⋅ds =F cosθds, whereθ is the angle between the force vector and the direction of movement,[24] that isW=∫CF⋅ds=Fscos⁡θ.{\displaystyle W=\int _{C}\mathbf {F} \cdot d\mathbf {s} =Fs\cos \theta .}

When a force component is perpendicular to the displacement of the object (such as when a body moves in a circular path under acentral force), no work is done, since the cosine of 90° is zero.[19] 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

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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 pointvelocity 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 calledscalar tangential component (F cos(θ), whereθ is the angle between the force and the velocity). And then the most general definition of work can be formulated as follows:

Area under the curve gives work done by F(x).
Work done by a variable force is the line integral of its scalar tangential component along the path of its application point.

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 ofx is given byF(x), then the work done by the force along the x-axis fromx1 tox2 is:

W=limΔx→0∑x1x2F(x)Δx=∫x1x2F(x)dx.{\displaystyle 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} .}

Thus, the work done for a variable force can be expressed as adefinite integral of force over displacement.[25]

If the displacement as a variable of time is given by∆x(t), then work done by the variable force fromt1 tot2 is:

W=∫t1t2F(t)⋅v(t)dt=∫t1t2P(t)dt.{\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {F} (t)\cdot \mathbf {v} (t)dt=\int _{t_{1}}^{t_{2}}P(t)dt.}

Thus, the work done for a variable force can be expressed as a definite integral ofpower over time.

Torque and rotation

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Aforce 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 torqueT. The work of the torque is calculated asδW=T⋅ωdt,{\displaystyle \delta W=\mathbf {T} \cdot {\boldsymbol {\omega }}\,dt,}where theT ⋅ω is the power over the instantdt. The sum of these small amounts of work over the trajectory of the rigid body yields the work,W=∫t1t2T⋅ωdt.{\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {T} \cdot {\boldsymbol {\omega }}\,dt.}This integral is computed along the trajectory of the rigid body with an angular velocityω that varies with time, and is therefore said to bepath dependent.

If the angular velocity vector maintains a constant direction, then it takes the form,ω=ϕ˙S,{\displaystyle {\boldsymbol {\omega }}={\dot {\phi }}\mathbf {S} ,}whereϕ{\displaystyle \phi } is the angle of rotation about the constant unit vectorS. In this case, the work of the torque becomes,W=∫t1t2T⋅ωdt=∫t1t2T⋅Sdϕdtdt=∫CT⋅Sdϕ,{\displaystyle 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 ,}whereC is the trajectory fromϕ(t1){\displaystyle \phi (t_{1})} toϕ(t2){\displaystyle \phi (t_{2})}. This integral depends on the rotational trajectoryϕ(t){\displaystyle \phi (t)}, and is therefore path-dependent.

If the torqueτ{\displaystyle \tau } is aligned with the angular velocity vector so that,T=τS,{\displaystyle \mathbf {T} =\tau \mathbf {S} ,}and both the torque and angular velocity are constant, then the work takes the form,[2]W=∫t1t2τϕ˙dt=τ(ϕ2−ϕ1).{\displaystyle W=\int _{t_{1}}^{t_{2}}\tau {\dot {\phi }}\,dt=\tau (\phi _{2}-\phi _{1}).}

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 magnitudeF, being applied perpendicularly to a lever arm at a distancer{\displaystyle r}, as shown in the figure. This force will act through the distance along the circular arcl=s=rϕ{\displaystyle l=s=r\phi }, so the work done isW=Fs=Frϕ.{\displaystyle W=Fs=Fr\phi .}Introduce the torqueτ =Fr, to obtainW=Frϕ=τϕ,{\displaystyle W=Fr\phi =\tau \phi ,}as presented above.

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

Work and potential energy

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The scalar product of a forceF and the velocityv of its point of application defines thepower 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

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Therefore, thework done by a forceF on an object that travels along a curveC is given by theline integral:W=∫CF⋅dx=∫t1t2F⋅vdt,{\displaystyle W=\int _{C}\mathbf {F} \cdot d\mathbf {x} =\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot \mathbf {v} dt,}wheredx(t) defines the trajectoryC andv is the velocity along this trajectory.In general this integral requires that 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,dWdt=P(t)=F⋅v.{\displaystyle {\frac {dW}{dt}}=P(t)=\mathbf {F} \cdot \mathbf {v} .}

Path independence

[edit]

If the work for an applied force is independent of the path, then the work done by the force, by thegradient theorem, defines a potential function which is evaluated at the start and end of the trajectory of the point of application. This means that there is a potential functionU(x), that can be evaluated at the two pointsx(t1) andx(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 isW=∫CF⋅dx=∫x(t1)x(t2)F⋅dx=U(x(t1))−U(x(t2)).{\displaystyle W=\int _{C}\mathbf {F} \cdot d\mathbf {x} =\int _{\mathbf {x} (t_{1})}^{\mathbf {x} (t_{2})}\mathbf {F} \cdot d\mathbf {x} =U(\mathbf {x} (t_{1}))-U(\mathbf {x} (t_{2})).}

The functionU(x) is called thepotential energy associated with the applied force. The force derived from such a potential function is said to beconservative. Examples of forces that have potential energies are gravity and spring forces.

In this case, thegradient of work yields∇W=−∇U=−(∂U∂x,∂U∂y,∂U∂z)=F,{\displaystyle \nabla W=-\nabla U=-\left({\frac {\partial U}{\partial x}},{\frac {\partial U}{\partial y}},{\frac {\partial U}{\partial z}}\right)=\mathbf {F} ,}and the forceF is said to be "derivable from a potential."[26]

Because the potentialU defines a forceF at every pointx in space, the set of forces is called aforce field. The power applied to a body by a force field is obtained from the gradient of the work, or potential, in the direction of the velocityV of the body, that isP(t)=−∇U⋅v=F⋅v.{\displaystyle P(t)=-\nabla U\cdot \mathbf {v} =\mathbf {F} \cdot \mathbf {v} .}

Work by gravity

[edit]
GravityF =mg does workW =mgh along any descending path

In the absence of other forces, gravity results in a constant downward acceleration of every freely moving object. Near Earth's surface the acceleration due to gravity isg = 9.8 m⋅s−2 and the gravitational force on an object of massm isFg =mg. It is convenient to imagine this gravitational force concentrated at thecenter of mass of the object.

If an object with weightmg is displaced upwards or downwards a vertical distancey2 −y1, the workW done on the object is:W=Fg(y2−y1)=FgΔy=mgΔy{\displaystyle W=F_{g}(y_{2}-y_{1})=F_{g}\Delta y=mg\Delta y}whereFg is weight (pounds in imperial units, and newtons in SI units), and Δy is the change in heighty. Notice that the work done by gravity depends only on the vertical movement of the object. The presence of friction does not affect the work done on the object by its weight.

Gravity in 3D space

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The force of gravity exerted by a massM on another massm is given byF=−GMmr2r^=−GMmr3r,{\displaystyle \mathbf {F} =-{\frac {GMm}{r^{2}}}{\hat {\mathbf {r} }}=-{\frac {GMm}{r^{3}}}\mathbf {r} ,}wherer is the position vector fromM tom andr̂ is the unit vector in the direction ofr.

Let the massm move at the velocityv; then the work of gravity on this mass as it moves from positionr(t1) tor(t2) is given byW=−∫r(t1)r(t2)GMmr3r⋅dr=−∫t1t2GMmr3r⋅vdt.{\displaystyle W=-\int _{\mathbf {r} (t_{1})}^{\mathbf {r} (t_{2})}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot d\mathbf {r} =-\int _{t_{1}}^{t_{2}}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot \mathbf {v} \,dt.}Notice that the position and velocity of the massm are given byr=rer,v=drdt=r˙er+rθ˙et,{\displaystyle \mathbf {r} =r\mathbf {e} _{r},\qquad \mathbf {v} ={\frac {d\mathbf {r} }{dt}}={\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t},}whereer andet are the radial and tangential unit vectors directed relative to the vector fromM tom, and we use the fact thatder/dt=θ˙et.{\displaystyle d\mathbf {e} _{r}/dt={\dot {\theta }}\mathbf {e} _{t}.} Use this to simplify the formula for work of gravity to,W=−∫t1t2GmMr3(rer)⋅(r˙er+rθ˙et)dt=−∫t1t2GmMr3rr˙dt=GMmr(t2)−GMmr(t1).{\displaystyle W=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}(r\mathbf {e} _{r})\cdot \left({\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t}\right)dt=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}r{\dot {r}}dt={\frac {GMm}{r(t_{2})}}-{\frac {GMm}{r(t_{1})}}.}This calculation uses the fact thatddtr−1=−r−2r˙=−r˙r2.{\displaystyle {\frac {d}{dt}}r^{-1}=-r^{-2}{\dot {r}}=-{\frac {\dot {r}}{r^{2}}}.}The functionU=−GMmr,{\displaystyle U=-{\frac {GMm}{r}},}is the gravitational potential function, also known asgravitational potential energy. The negative sign follows the convention that work is gained from a loss of potential energy.

Work by a spring

[edit]
Forces in springs assembled in parallel

Consider a spring that exerts a horizontal forceF = (−kx, 0, 0) that is proportional to its deflection in thex direction independent of how a body moves. The work of this spring on a body moving along the space with the curveX(t) = (x(t),y(t),z(t)), is calculated using its velocity,v = (vx,vy,vz), to obtainW=∫0tF⋅vdt=−∫0tkxvxdt=−12kx2.{\displaystyle W=\int _{0}^{t}\mathbf {F} \cdot \mathbf {v} dt=-\int _{0}^{t}kxv_{x}dt=-{\frac {1}{2}}kx^{2}.}For convenience, consider contact with the spring occurs att = 0, then the integral of the product of the distancex and the x-velocity,xvxdt, over timet is⁠1/2⁠x2. The work is the product of the distance times the spring force, which is also dependent on distance; hence thex2 result.

Work by a gas

[edit]

The workW{\displaystyle W} done by a body of gas on its surroundings is:W=∫abPdV{\displaystyle W=\int _{a}^{b}P\,dV}whereP is pressure,V is volume, anda andb are initial and final volumes.

Work–energy principle

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The principle of work andkinetic energy (also known as thework–energy principle) states thatthe 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.[27] That is, the workW done by theresultant force on aparticle equals the change in the particle's kinetic energyEk{\displaystyle E_{\text{k}}},[2]W=ΔEk=12mv22−12mv12{\displaystyle W=\Delta E_{\text{k}}={\frac {1}{2}}mv_{2}^{2}-{\frac {1}{2}}mv_{1}^{2}}wherev1{\displaystyle v_{1}} andv2{\displaystyle v_{2}} are thespeeds of the particle before and after the work is done, andm is itsmass.

The derivation of thework–energy principle begins withNewton's second law of motion and the resultant force on a particle. Computation of the scalar product of the force with the velocity of the particle evaluates the instantaneous power added to the system.[28](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 acceleration with velocity. The fact that the work–energy principle eliminates the constraint forces underliesLagrangian mechanics.[29]

This section focuses on the work–energy principle as it applies to particle dynamics. In more general systems work can change thepotential energy of a mechanical device, the thermal energy in a thermal system, or theelectrical energy in an electrical device. Work transfers energy from one place to another or one form to another.

Derivation for a particle moving along a straight line

[edit]

In the case theresultant forceF is constant in both magnitude and direction, and parallel to the velocity of the particle, the particle is moving with constant accelerationa along a straight line.[30] The relation between the net force and the acceleration is given by the equationF =ma (Newton's second law), and the particledisplacements can be expressed by the equations=v22−v122a{\displaystyle s={\frac {v_{2}^{2}-v_{1}^{2}}{2a}}}which follows fromv22=v12+2as{\displaystyle v_{2}^{2}=v_{1}^{2}+2as} (seeEquations of motion).

The work of the net force is calculated as the product of its magnitude and the particle displacement. Substituting the above equations, one obtains:W=Fs=mas=mav22−v122a=12mv22−12mv12=ΔEk{\displaystyle W=Fs=mas=ma{\frac {v_{2}^{2}-v_{1}^{2}}{2a}}={\frac {1}{2}}mv_{2}^{2}-{\frac {1}{2}}mv_{1}^{2}=\Delta E_{\text{k}}}

Other derivation:W=Fs=mas=mv22−v122ss=12mv22−12mv12=ΔEk{\displaystyle W=Fs=mas=m{\frac {v_{2}^{2}-v_{1}^{2}}{2s}}s={\frac {1}{2}}mv_{2}^{2}-{\frac {1}{2}}mv_{1}^{2}=\Delta E_{\text{k}}}

In the general case of rectilinear motion, when the net forceF 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:W=∫t1t2F⋅vdt=∫t1t2Fvdt=∫t1t2mavdt=m∫t1t2vdvdtdt=m∫v1v2vdv=12m(v22−v12).{\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot \mathbf {v} dt=\int _{t_{1}}^{t_{2}}F\,v\,dt=\int _{t_{1}}^{t_{2}}ma\,v\,dt=m\int _{t_{1}}^{t_{2}}v\,{\frac {dv}{dt}}\,dt=m\int _{v_{1}}^{v_{2}}v\,dv={\tfrac {1}{2}}m\left(v_{2}^{2}-v_{1}^{2}\right).}

General derivation of the work–energy principle for a particle

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For any net force acting on a particle moving along any curvilinear path, it can be demonstrated that its work equals the change in the kinetic energy of the particle by a simple derivation analogous to the equation above. It is known asthe work–energy principle:W=∫t1t2F⋅vdt=m∫t1t2a⋅vdt=m2∫t1t2dv2dtdt=m2∫v12v22dv2=mv222−mv122=ΔEk{\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot \mathbf {v} dt=m\int _{t_{1}}^{t_{2}}\mathbf {a} \cdot \mathbf {v} dt={\frac {m}{2}}\int _{t_{1}}^{t_{2}}{\frac {dv^{2}}{dt}}\,dt={\frac {m}{2}}\int _{v_{1}^{2}}^{v_{2}^{2}}dv^{2}={\frac {mv_{2}^{2}}{2}}-{\frac {mv_{1}^{2}}{2}}=\Delta E_{\text{k}}}

The identitya⋅v=12dv2dt{\textstyle \mathbf {a} \cdot \mathbf {v} ={\frac {1}{2}}{\frac {dv^{2}}{dt}}} requires some algebra.From the identityv2=v⋅v{\textstyle v^{2}=\mathbf {v} \cdot \mathbf {v} } and definitiona=dvdt{\textstyle \mathbf {a} ={\frac {d\mathbf {v} }{dt}}}it followsdv2dt=d(v⋅v)dt=dvdt⋅v+v⋅dvdt=2dvdt⋅v=2a⋅v.{\displaystyle {\frac {dv^{2}}{dt}}={\frac {d(\mathbf {v} \cdot \mathbf {v} )}{dt}}={\frac {d\mathbf {v} }{dt}}\cdot \mathbf {v} +\mathbf {v} \cdot {\frac {d\mathbf {v} }{dt}}=2{\frac {d\mathbf {v} }{dt}}\cdot \mathbf {v} =2\mathbf {a} \cdot \mathbf {v} .}

The remaining part of the above derivation is just simple calculus, same as in the preceding rectilinear case.

Derivation for a particle in constrained movement

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In particle dynamics, a formula equating work applied to a system to its change in kinetic energy is obtained as a first integral ofNewton's second law of motion. It is useful to notice that the resultant force used in Newton's laws can be separated into forces that are applied to the particle and forces imposed by constraints on the movement of the particle. Remarkably, the work of a constraint force is zero, therefore only the work of the applied forces need be considered in the work–energy principle.

To see this, consider a particle P that follows the trajectoryX(t) with a forceF acting on it. Isolate the particle from its environment to expose constraint forcesR, then Newton's Law takes the formF+R=mX¨,{\displaystyle \mathbf {F} +\mathbf {R} =m{\ddot {\mathbf {X} }},}wherem is the mass of the particle.

Vector formulation

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Note that n dots above a vector indicates its nthtime derivative.Thescalar product of each side of Newton's law with the velocity vector yieldsF⋅X˙=mX¨⋅X˙,{\displaystyle \mathbf {F} \cdot {\dot {\mathbf {X} }}=m{\ddot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }},}because the constraint forces are perpendicular to the particle velocity. Integrate this equation along its trajectory from the pointX(t1) to the pointX(t2) to obtain∫t1t2F⋅X˙dt=m∫t1t2X¨⋅X˙dt.{\displaystyle \int _{t_{1}}^{t_{2}}\mathbf {F} \cdot {\dot {\mathbf {X} }}dt=m\int _{t_{1}}^{t_{2}}{\ddot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}dt.}

The left side of this equation is the work of the applied force as it acts on the particle along the trajectory from timet1 to timet2. This can also be written asW=∫t1t2F⋅X˙dt=∫X(t1)X(t2)F⋅dX.{\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot {\dot {\mathbf {X} }}dt=\int _{\mathbf {X} (t_{1})}^{\mathbf {X} (t_{2})}\mathbf {F} \cdot d\mathbf {X} .}This integral is computed along the trajectoryX(t) of the particle and is therefore path dependent.

The right side of the first integral of Newton's equations can be simplified using the following identity12ddt(X˙⋅X˙)=X¨⋅X˙,{\displaystyle {\frac {1}{2}}{\frac {d}{dt}}({\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }})={\ddot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }},}(seeproduct rule for derivation). Now it is integrated explicitly to obtain the change in kinetic energy,ΔK=m∫t1t2X¨⋅X˙dt=m2∫t1t2ddt(X˙⋅X˙)dt=m2X˙⋅X˙(t2)−m2X˙⋅X˙(t1)=12mΔv2,{\displaystyle \Delta K=m\int _{t_{1}}^{t_{2}}{\ddot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}dt={\frac {m}{2}}\int _{t_{1}}^{t_{2}}{\frac {d}{dt}}({\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }})dt={\frac {m}{2}}{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}(t_{2})-{\frac {m}{2}}{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}(t_{1})={\frac {1}{2}}m\Delta \mathbf {v} ^{2},}where the kinetic energy of the particle is defined by the scalar quantity,K=m2X˙⋅X˙=12mv2{\displaystyle K={\frac {m}{2}}{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}={\frac {1}{2}}m{\mathbf {v} ^{2}}}

Tangential and normal components

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It is useful to resolve the velocity and acceleration vectors into tangential and normal components along the trajectoryX(t), such thatX˙=vTandX¨=v˙T+v2κN,{\displaystyle {\dot {\mathbf {X} }}=v\mathbf {T} \quad {\text{and}}\quad {\ddot {\mathbf {X} }}={\dot {v}}\mathbf {T} +v^{2}\kappa \mathbf {N} ,}wherev=|X˙|=X˙⋅X˙.{\displaystyle v=|{\dot {\mathbf {X} }}|={\sqrt {{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}}}.}Then, thescalar product of velocity with acceleration in Newton's second law takes the formΔK=m∫t1t2v˙vdt=m2∫t1t2ddtv2dt=m2v2(t2)−m2v2(t1),{\displaystyle \Delta K=m\int _{t_{1}}^{t_{2}}{\dot {v}}v\,dt={\frac {m}{2}}\int _{t_{1}}^{t_{2}}{\frac {d}{dt}}v^{2}\,dt={\frac {m}{2}}v^{2}(t_{2})-{\frac {m}{2}}v^{2}(t_{1}),}where the kinetic energy of the particle is defined by the scalar quantity,K=m2v2=m2X˙⋅X˙.{\displaystyle K={\frac {m}{2}}v^{2}={\frac {m}{2}}{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}.}

The result is the work–energy principle for particle dynamics,W=ΔK.{\displaystyle W=\Delta K.}This derivation can be generalized to arbitrary rigid body systems.

Moving in a straight line (skid to a stop)

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Consider the case of a vehicle moving along a straight horizontal trajectory under the action of a driving force and gravity that sum toF. The constraint forces between the vehicle and the road defineR, and we haveF+R=mX¨.{\displaystyle \mathbf {F} +\mathbf {R} =m{\ddot {\mathbf {X} }}.}For convenience let the trajectory be along the X-axis, soX = (d, 0) and the velocity isV = (v, 0), thenR ⋅V = 0, andF ⋅V =Fxv, whereFx is the component ofF along the X-axis, soFxv=mv˙v.{\displaystyle F_{x}v=m{\dot {v}}v.}Integration of both sides yields∫t1t2Fxvdt=m2v2(t2)−m2v2(t1).{\displaystyle \int _{t_{1}}^{t_{2}}F_{x}vdt={\frac {m}{2}}v^{2}(t_{2})-{\frac {m}{2}}v^{2}(t_{1}).}IfFx is constant along the trajectory, then the integral of velocity is distance, soFx(d(t2)−d(t1))=m2v2(t2)−m2v2(t1).{\displaystyle F_{x}(d(t_{2})-d(t_{1}))={\frac {m}{2}}v^{2}(t_{2})-{\frac {m}{2}}v^{2}(t_{1}).}

As an example consider a car skidding to a stop, wherek is the coefficient of friction andw is the weight of the car. Then the force along the trajectory isFx = −kw. The velocityv of the car can be determined from the lengths of the skid using the work–energy principle,kws=w2gv2,orv=2ksg.{\displaystyle kws={\frac {w}{2g}}v^{2},\quad {\text{or}}\quad v={\sqrt {2ksg}}.}This formula uses the fact that the mass of the vehicle ism =w/g.

Lotus type 119Bgravity racer at Lotus 60th celebration
Gravity racing championship in Campos Novos, Santa Catarina, Brazil, 8 September 2010

Coasting down an inclined surface (gravity racing)

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Consider the case of a vehicle that starts at rest and coasts down an inclined surface (such as mountain road), the work–energy principle helps compute the minimum distance that the vehicle travels to reach a velocityV, of say 60 mph (88 fps). Rolling resistance and air drag will slow the vehicle down so the actual distance will be greater than if these forces are neglected.

Let the trajectory of the vehicle following the road beX(t) which is a curve in three-dimensional space. The force acting on the vehicle that pushes it down the road is the constant force of gravityF = (0, 0,w), while the force of the road on the vehicle is the constraint forceR. Newton's second law yields,F+R=mX¨.{\displaystyle \mathbf {F} +\mathbf {R} =m{\ddot {\mathbf {X} }}.}Thescalar product of this equation with the velocity,V = (vx,vy,vz), yieldswvz=mV˙V,{\displaystyle wv_{z}=m{\dot {V}}V,}whereV is the magnitude ofV. The constraint forces between the vehicle and the road cancel from this equation becauseR ⋅V = 0, which means they do no work.Integrate both sides to obtain∫t1t2wvzdt=m2V2(t2)−m2V2(t1).{\displaystyle \int _{t_{1}}^{t_{2}}wv_{z}dt={\frac {m}{2}}V^{2}(t_{2})-{\frac {m}{2}}V^{2}(t_{1}).}The weight forcew is constant along the trajectory and the integral of the vertical velocity is the vertical distance, therefore,wΔz=m2V2.{\displaystyle w\Delta z={\frac {m}{2}}V^{2}.}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 traveled—for angles this small the sin and tan functions are approximately equal. Therefore, the distances in feet down a 6% grade to reach the velocityV is at leasts=Δz0.06=8.3V2g,ors=8.388232.2≈2000ft.{\displaystyle s={\frac {\Delta z}{0.06}}=8.3{\frac {V^{2}}{g}},\quad {\text{or}}\quad s=8.3{\frac {88^{2}}{32.2}}\approx 2000\mathrm {ft} .}This formula uses the fact that the weight of the vehicle isw =mg.

Work of forces acting on a rigid body

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The work of forces acting at various points on a single rigid body can be calculated from the work of aresultant force and torque. To see this, let the forcesF1,F2, ...,Fn act on the pointsX1,X2, ...,Xn in a rigid body.

The trajectories ofXi,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 trajectoryd(t) of a reference point in the body. Let the coordinatesxii = 1, ...,n define these points in the moving rigid body'sreference frameM, so that the trajectories traced in the fixed frameF are given byXi(t)=[A(t)]xi+d(t)i=1,…,n.{\displaystyle \mathbf {X} _{i}(t)=[A(t)]\mathbf {x} _{i}+\mathbf {d} (t)\quad i=1,\ldots ,n.}

The velocity of the pointsXi along their trajectories areVi=ω×(Xi−d)+d˙,{\displaystyle \mathbf {V} _{i}={\boldsymbol {\omega }}\times (\mathbf {X} _{i}-\mathbf {d} )+{\dot {\mathbf {d} }},}whereω is the angular velocity vector obtained from the skew symmetric matrix[Ω]=A˙AT,{\displaystyle [\Omega ]={\dot {A}}A^{\mathsf {T}},}known as the angular velocity matrix.

The small amount of work by the forces over the small displacementsδri can be determined by approximating the displacement byδr =vδt soδW=F1⋅V1δt+F2⋅V2δt+…+Fn⋅Vnδt{\displaystyle \delta W=\mathbf {F} _{1}\cdot \mathbf {V} _{1}\delta t+\mathbf {F} _{2}\cdot \mathbf {V} _{2}\delta t+\ldots +\mathbf {F} _{n}\cdot \mathbf {V} _{n}\delta t}orδW=∑i=1nFi⋅(ω×(Xi−d)+d˙)δt.{\displaystyle \delta W=\sum _{i=1}^{n}\mathbf {F} _{i}\cdot ({\boldsymbol {\omega }}\times (\mathbf {X} _{i}-\mathbf {d} )+{\dot {\mathbf {d} }})\delta t.}

This formula can be rewritten to obtainδW=(∑i=1nFi)⋅d˙δt+(∑i=1n(Xi−d)×Fi)⋅ωδt=(F⋅d˙+T⋅ω)δt,{\displaystyle \delta W=\left(\sum _{i=1}^{n}\mathbf {F} _{i}\right)\cdot {\dot {\mathbf {d} }}\delta t+\left(\sum _{i=1}^{n}\left(\mathbf {X} _{i}-\mathbf {d} \right)\times \mathbf {F} _{i}\right)\cdot {\boldsymbol {\omega }}\delta t=\left(\mathbf {F} \cdot {\dot {\mathbf {d} }}+\mathbf {T} \cdot {\boldsymbol {\omega }}\right)\delta t,}whereF andT are theresultant force and torque applied at the reference pointd of the moving frameM in the rigid body.

References

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  1. ^NCERT (2020)."Physics Book"(PDF).ncert.nic.in. Retrieved24 November 2021.
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Bibliography

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External links

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Linear/translational quantitiesAngular/rotational quantities
Dimensions1LL2Dimensions1θθ2
Ttime:t
s		absement:A
m s		Ttime:t
s
1distance:d,position:r,s,x,displacement
m		area:A
m2		1angle:θ,angular displacement:θ
rad		solid angle:Ω
rad2, sr
T−1frequency:f
s−1,Hz		speed:v,velocity:v
m s−1		kinematic viscosity:ν,
specific angular momentum: h
m2 s−1		T−1frequency:f,rotational speed:n,rotational velocity:n
s−1,Hz		angular speed:ω,angular velocity:ω
rad s−1
T−2acceleration:a
m s−2		T−2rotational acceleration
s−2		angular acceleration:α
rad s−2
T−3jerk:j
m s−3		T−3angular jerk:ζ
rad s−3
Mmass:m
kg		weighted position:M ⟨x⟩ = ∑mxmoment of inertia: I
kg m2		ML
MT−1Mass flow rate:m˙{\displaystyle {\dot {m}}}
kg s−1		momentum:p,impulse:J
kg m s−1,N s		action:𝒮,actergy:ℵ
kg m2 s−1,J s		MLT−1angular momentum:L,angular impulse:ΔL
kg m rad s−1
MT−2force:F,weight:Fg
kg m s−2,N		energy:E,work:W,Lagrangian:L
kg m2 s−2,J		MLT−2torque:τ,moment:M
kg m rad s−2,N m
MT−3yank:Y
kg m s−3, N s−1		power:P
kg m2 s−3, W		MLT−3rotatum:P
kg m rad s−3, N m s−1
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