Work (physics)

"Mechanical work" redirects here. For other uses of "Work" in physics, see Work (electrical) and Work (thermodynamics).

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.
Common symbols	W
SI unit	joule (J)
Derivations from other quantities	W = F · d W = τ θ

In physics, a force is said to do work when it acts on a body so that there is a displacement of the point of application, however small, in the direction of the force. Thus a force does work when there is movement under the action of the force.

The term work was introduced in 1826 by the French mathematician Gaspard-Gustave Coriolis 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-meter or joule (J).

The work done by a constant force of magnitude F on a point that moves a distance d in the direction of the force is the product,

$W=Fd.$

For example, if a force of 10 newton (F = 10 N) acts along 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.

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 metre. The dimensionally equivalent newton-metre (N·m) is sometimes used for work, but this can be confused with the units newton-metre of torque.

Non-SI units of work include the erg, the foot-pound, the foot-poundal, the kilowatt hour, the litre-atmosphere, and the horsepower-hour. Since work has the same physical dimension as heat it is sometimes also measured using units originally used for heat or energy content such as the therm, the BTU and Calorie.

Work and energy

Work is closely related to energy. The conservation of energy states that the change in total internal energy of a system equals the added heat minus the work performed by the system (see the first law of thermodynamics),

$\delta E=\delta Q-\delta W.$

Also, from Newton's second law for rigid bodies it can be shown that work on an object is equal to the change in kinetic energy of that object,

$W=\Delta KE$.

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 moving in a conservative force field is equal to minus the change of potential energy of the object,

$W=-\Delta PE.$

This shows that work is the energy associated with the action of a force, and so has the physical dimensions and units of energy.

The work energy principle discussed here is identical in the case of Electrical work.

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 motionsideways 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.

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 do work. It can change the direction of motion but never change the speed.

Mathematical calculation

For moving objects, the quantity of work/time enters calculations 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) with the velocity vector of the point of application. This scalar product of force and velocity is called 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.

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 δW that occurs over an instant of time δt is given by

$\delta W=\mathbf {F} \cdot \mathbf {v} \delta t,$

where the F.v is the power over the instant δt. The sum of these small amounts of work over the trajectory of the point yields the work,

$W=\int _{t_{1}}^{t_{2}}{\mathbf {F}}\cdot \mathbf {v} dt=\int _{t_{1}}^{t_{2}}{\mathbf {F}}\cdot {\tfrac {d\mathbf {x} }{dt}}dt=\int _{\mathbf {X} (t_{1})}^{\mathbf {X} (t_{2})}{\mathbf {F}}\cdot d{\mathbf {x}}.$

This integral is computed along the trajectory of the particle and is therefore said to be path dependent.

If the path of the particle is a straight line and the force F is constant and directed along this line, then this integral simplifies to

$W=\int _{C}{\mathbf {F}}\cdot d{\mathbf {x}}=Fd,$

where F is the magnitude of the force and d is the distance traveled by the point.

Gravity F=mg does work W=mgh along any descending path

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 F·dx = Fcosθdx, where θ is the angle between the force vector and the direction of movement, that is

$W=\int _{C}{\mathbf {F}}\cdot d{\mathbf {x}}=Fd\cos \theta .$

In the notable case of a force applied to a body always in 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 done by gravity on a planet in a circular orbit (this is an ideal, as all orbits are slightly ellipical), and no work is done on a body moving in a circle at constant speed while constrained by mechanical force, such as moving at constant speed in a frictionless ideal centrifuge.

Calculating the work as "force times straight path segment" can only be done in the simple circumstances described above. If the force is changing, 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 the force can be described by the scalar quantity called scalar tangential component ($\scriptstyle F\cos \theta$, where $\scriptstyle \theta$ 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.

A force of constant magnitude and perpendicular to the lever arm

Torque and rotation

A torque acting on a rotating body will also perform work. If the torque $\tau$ is constant and around the same axis as the rotation the work done for a rotation angle $\varphi$ is

$W=\tau \varphi$.

This can be understood by seeing the torque as a force of constant magnitude F applied perpendicularly to a lever arm r as in the figure to the right. The force will act through the length of the circular arc $s$, which is always parallel to the force, so the work done is $W=Fs$. Inserting the length of the arc $s=r\varphi$ and identifying the torque $\tau =Fr$ we get $W=Fr\varphi =\tau \varphi$, as stated above.

In the more general case of a variable torque but still in the direction of rotation, the work can be calculated with the integral

$W=\int _{\theta _{1}}^{\theta _{2}}\tau d\theta$,

where the rotation is from angle $\theta _{1}$ to $\theta _{2}$.

When the torque and rotation are not parallel only the component of the torque in the direction of the rotation does work

$W=\int _{\theta _{1}}^{\theta _{2}}\tau _{\parallel }d\theta =\int _{\theta _{1}}^{\theta _{2}}{\boldsymbol {\tau }}\cdot d{\boldsymbol {\theta }}$.

Work and potential energy

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:

$W=\int _{C}{\mathbf {F}}\cdot \mathrm {d} {\mathbf {x}}=\int _{t_{1}}^{t_{2}}{\mathbf {F}}\cdot {\mathbf {v}}dt,$

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,

${\frac {dW}{dt}}=P(t)=\mathbf {F} \cdot \mathbf {v} .$

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

$W=\int _{C}{\mathbf {F}}\cdot \mathrm {d} {\mathbf {x}}=\int _{\mathbf {x} (t_{1})}^{\mathbf {x} (t_{2})}{\mathbf {F}}\cdot \mathrm {d} {\mathbf {x}}=U(\mathbf {x} (t_{1}))-U(\mathbf {x} (t_{2})).$

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

${\frac {\partial W}{\partial \mathbf {x} }}=-{\frac {\partial U}{\partial \mathbf {x} }}=-{\big (}{\frac {\partial U}{\partial x}},{\frac {\partial U}{\partial y}},{\frac {\partial U}{\partial z}}{\big )}=\mathbf {F} ,$

and the force F is said to be "derivable from a potential."

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 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

$P(t)=-{\frac {\partial U}{\partial \mathbf {x} }}\cdot \mathbf {v} =\mathbf {F} \cdot \mathbf {v} .$

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, 0, W) on the center of mass of a body moving near the surface of the earth. The work of gravity on a body moving along a trajectory X(t) = (x(t), y(t), z(t)), such as the track of a roller coaster is calculated using its velocity, V=(v_x, v_y, v_z), to obtain

$W=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot \mathbf {V} dt=\int _{t_{1}}^{t_{2}}Wv_{z}dt=W\Delta z.$

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).

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=(v_x, v_y, v_z), to obtain

$W=\int _{0}^{t}\mathbf {F} \cdot \mathbf {V} dt=\int _{0}^{t}kxv_{x}dt={\frac {1}{2}}kx^{2}.$

For convenience, consider contact with the spring occurs at t=0, then the integral of the product of the distance x and the x-velocity, xv_x, is (1/2)x².

Principle of work and kinetic energy

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. This definition can be extended to rigid bodies by defining the work of the resultant torque and rotational kinetic energy.

The work W done by the resultant force on a particle equals the change in the particle's kinetic energy $E_{k}$,

$W=\Delta E_{k}={\tfrac {1}{2}}mv_{2}^{2}-{\tfrac {1}{2}}mv_{1}^{2}$,

where $v_{1}$ and $v_{2}$ are the speeds of the particle before and after the change and m is its mass.

Overview

The derivation of the work-energy principle begins with Newton's 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.

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.

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.

Derivation for a particle moving along a straight line

In the case the resultant force F is constant in both magnitude and direction, and parallel to the velocity of the particle, the particle is moving with constant acceleration a along a straight line. The relation between the net force and the acceleration is given by the equation F = ma (Newton's second law), and the particle displacement d can be expressed by the equation

$d={\frac {v_{2}^{2}-v_{1}^{2}}{2a}}$

which follows from $v_{2}^{2}=v_{1}^{2}+2ad$ (see Equations 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:

$W=Fd=mad=ma\left({\frac {v_{2}^{2}-v_{1}^{2}}{2a}}\right)={\frac {mv_{2}^{2}}{2}}-{\frac {mv_{1}^{2}}{2}}=\Delta {E_{k}}$

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:

$W=\int _{t_{1}}^{t_{2}}{\mathbf {F}}\cdot {\mathbf {v}}dt=\int _{t_{1}}^{t_{2}}F\,vdt=\int _{t_{1}}^{t_{2}}ma\,vdt=m\int _{t_{1}}^{t_{2}}v\,{dv \over dt}\,dt=m\int _{v_{1}}^{v_{2}}v\,dv={\tfrac {1}{2}}m(v_{2}^{2}-v_{1}^{2}).$

Derivation for a particle in constrained movement

In particle dynamics, a formula equating work applied to a system to its change in kinetic energy is obtained as a first integral of Newton'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 trajectory X(t) with a force F acting on it. Isolate the particle from its environment to expose constraint forces R, then Newton's Law takes the form

$\mathbf {F} +\mathbf {R} =m{\ddot {\mathbf {X} }},$

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 Newton's law with the velocity vector yields

$\mathbf {F} \cdot {\dot {\mathbf {X} }}=m{\ddot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }},$

because the constraint forces are perpendicular to the particle velocity. Integrate this equation along its trajectory from the point X(t₁) to the point X(t₂) to obtain

$\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot {\dot {\mathbf {X} }}dt=m\int _{t_{1}}^{t_{2}}{\ddot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}dt.$

The left side of this equation is the work of the applied force as it acts on the particle along the trajectory from time t₁ to time t₂. This can also be written as

$W=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot {\dot {\mathbf {X} }}dt=\int _{\mathbf {X} (t_{1})}^{\mathbf {X} (t_{2})}\mathbf {F} \cdot d\mathbf {X} .$

This integral is computed along the trajectory X(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 identity

${\frac {1}{2}}{\frac {d}{dt}}({\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }})={\ddot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }},$

(see product rule for derivation). Now it is integrated explicitly to obtain the change in kinetic energy,

$\Delta K=m\int _{t_{1}}^{t_{2}}{\ddot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}dt={\frac {m}{2}}\int _{t_{1}}^{t_{2}}{\frac {d}{dt}}({\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }})dt={\frac {m}{2}}{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}(t_{2})-{\frac {m}{2}}{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}(t_{1})={\frac {1}{2}}m\Delta \mathbf {v^{2}} ,$

where the kinetic energy of the particle is defined by the scalar quantity,

$K={\frac {m}{2}}{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}={\frac {1}{2}}m{\mathbf {v^{2}} }$

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

${\dot {\mathbf {X} }}=v\mathbf {T} ,\quad {\mbox{and}}\quad {\ddot {\mathbf {X} }}={\dot {v}}\mathbf {T} +v^{2}\kappa \mathbf {N} .$

where

$v=|{\dot {\mathbf {X} }}|={\sqrt {{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}}}.$

Then, the scalar product of velocity with acceleration in Newton's second law takes the form

$\Delta K=m\int _{t_{1}}^{t_{2}}{\dot {v}}vdt={\frac {m}{2}}\int _{t_{1}}^{t_{2}}{\frac {d}{dt}}v^{2}dt={\frac {m}{2}}v^{2}(t_{2})-{\frac {m}{2}}v^{2}(t_{1}),$.

where the kinetic energy of the particle is defined by the scalar quantity,

$K={\frac {m}{2}}v^{2}={\frac {m}{2}}{\dot {\mathbf {X} }}\cdot {\dot {\mathbf {X} }}.$

The result is the work-energy principle for particle dynamics,

$W=\Delta K.\!$

This derivation can be generalized to arbitrary rigid body systems.

Moving in a straight line (skid to a stop)

Consider the case of a vehicle moving along a straight horizontal trajectory under the action of a driving force and gravity that sum to F. The constraint forces between the vehicle and the road define R, and we have

$\mathbf {F} +\mathbf {R} =m{\ddot {\mathbf {X} }}.$

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=F_xv, where F_x is the component of F along the X-axis, so

$F_{x}v=m{\dot {v}}v.$

Integration of both sides yields

$\int _{t_{1}}^{t_{2}}F_{x}vdt={\frac {m}{2}}v^{2}(t_{2})-{\frac {m}{2}}v^{2}(t_{1}).$

If F_x is constant along the trajectory, then the integral of velocity is distance, so

$F_{x}(d(t_{2})-d(t_{1}))={\frac {m}{2}}v^{2}(t_{2})-{\frac {m}{2}}v^{2}(t_{1}).$

As an example consider a car skidding to a stop, where k is the coefficient of friction and W is the weight of the car. Then the force along the trajectory is F_x =-kW. The velocity v of the car can be determined from the length s of the skid using the work-energy principle,

$kWs={\frac {W}{2g}}v^{2},\quad {\mbox{or}}\quad v={\sqrt {2ksg}}.$

Notice that this formula uses the fact that the mass of the vehicle is m=W/g.

Coasting down a mountain road (gravity racing)

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

Let the trajectory of the vehicle following the road be X(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 gravity F=(0,0,W), while the force of the road on the vehicle is the constraint force R. Newton's second law yields,

$\mathbf {F} +\mathbf {R} =m{\ddot {\mathbf {X} }}.$

The scalar product of this equation with the velocity, V=(v_x, v_y,v_z), yields

$Wv_{z}=m{\dot {V}}V,$

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

$\int _{t_{1}}^{t_{2}}Wv_{z}dt={\frac {m}{2}}V^{2}(t_{2})-{\frac {m}{2}}V^{2}(t_{1}).$

The weight force W is constant along the trajectory and the integral of the vertical velocity is the vertical distance, therefore,

$W\Delta z={\frac {m}{2}}V^{2}.$

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 distance s in feet down a 6% grade to reach the velocity V is at least

$s={\frac {\Delta z}{0.06}}=8.3{\frac {V^{2}}{g}},\quad {\mbox{or}}\quad s=8.3{\frac {88^{2}}{32.2}}\approx 2000{\mbox{ft}}.$

This formula uses the fact that the weight of the vehicle is W=mg.

Work of forces acting on a rigid body

The work of forces acting at various points on a single rigid body can be calculated from the work of a resultant force and torque. To see this, let the forces F₁, F₂ ... F_n act on the points X₁, X₂ ... X_n in a rigid body.

The trajectories of X_i, 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 x_i i=1,...,n define these points in the moving rigid body's reference frame M, so that the trajectories traced in the fixed frame F are given by

$\mathbf {X} _{i}(t)=[A(t)]\mathbf {x} _{i}+\mathbf {d} (t)\quad i=1,\ldots ,n.$

The velocity of the points X_i along their trajectories are

$\mathbf {V} _{i}={\vec {\omega }}\times (\mathbf {X} _{i}-\mathbf {d} )+{\dot {\mathbf {d} }},$

where ω is the angular velocity vector obtained from the skew symmetric matrix

$[\Omega ]={\dot {A}}A^{T},$

known as the angular velocity matrix.

The small amount of work by the forces over the instant of time δt is given by

$\delta W=\mathbf {F} _{1}\cdot \mathbf {V} _{1}\delta t+\mathbf {F} _{2}\cdot \mathbf {V} _{2}\delta t+\ldots +\mathbf {F} _{n}\cdot \mathbf {V} _{n}\delta t$

or

$\delta W=\sum _{i=1}^{n}\mathbf {F} _{i}\cdot ({\vec {\omega }}\times (\mathbf {X} _{i}-\mathbf {d} )+{\dot {\mathbf {d} }})\delta t.$

This formula can be rewritten to obtain

$\delta W=(\sum _{i=1}^{n}\mathbf {F} _{i})\cdot {\dot {\mathbf {d} }}\delta t+(\sum _{i=1}^{n}(\mathbf {X} _{i}-\mathbf {d} )\times \mathbf {F} _{i})\cdot {\vec {\omega }}\delta t=\mathbf {F} \cdot {\dot {\mathbf {d} }}\delta t+\mathbf {T} \cdot {\vec {\omega }}\delta t,$

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.

Bibliography

• Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers (6th ed. ed.). Brooks/Cole. ISBN 0-534-40842-7. {{cite book}}: |edition= has extra text (help)CS1 maint: multiple names: authors list (link)
• 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 – a chapter from an online textbook
• Work-energy principle