Work (physics)

Although work was not formally used until 1826, similar concepts existed before then. Early names for the same concept included moment of activity, quantity of action, latent live force, dynamic effect, efficiency, and even force.[5] In 1637, the French philosopher René 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 philosopher Gottfried 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 to Coriolis's.[8]

According to the 1957 physics textbook by Max Jammer,[9] the term work was introduced in 1826 by the French mathematician Gaspard-Gustave Coriolis[10] 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. According to Rene Dugas, French engineer and historian, it is to Solomon of Caux "that we owe the term work in the sense that it is used in mechanics now".[11]

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 (F cos(θ), where θ 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.

If the force varies (e.g. compressing a spring) we need to use calculus to find the work done. If the force is given by F(x) (a function of x) then the work done by the force along the x-axis from a to b is:

$${\displaystyle W=\int _{a}^{b}\mathbf {F(s)} \cdot d\mathbf {s} }$$

A force 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 torque T. The work of the torque is calculated as

$${\displaystyle \delta W=\mathbf {T} \cdot {\boldsymbol {\omega }}\,dt,}$$

where the T ⋅ ω is the power over the instant dt. The sum of these small amounts of work over the trajectory of the rigid body yields the work,

$${\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {T} \cdot {\boldsymbol {\omega }}\,dt.}$$

This integral is computed along the trajectory of the rigid body with an angular velocity ω 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,

$${\displaystyle {\boldsymbol {\omega }}={\dot {\phi }}\mathbf {S} ,}$$

where ${\displaystyle \phi }$ is the angle of rotation about the constant unit vector S. In this case, the work of the torque becomes,

$${\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {T} \cdot {\boldsymbol {\omega }}\,dt=\int _{t_{1}}^{t_{2}}\mathbf {T} \cdot \mathbf {S} {\frac {d\phi }{dt}}dt=\int _{C}\mathbf {T} \cdot \mathbf {S} \,d\phi ,}$$

where C is the trajectory from ${\displaystyle \phi (t_{1})}$ to ${\displaystyle \phi (t_{2})}$. This integral depends on the rotational trajectory ${\displaystyle \phi (t)}$, and is therefore path-dependent.

If the torque ${\displaystyle \tau }$ is aligned with the angular velocity vector so that,

and both the torque and angular velocity are constant, then the work takes the form,[2]

$${\displaystyle W=\int _{t_{1}}^{t_{2}}\tau {\dot {\phi }}\,dt=\tau (\phi _{2}-\phi _{1}).}$$

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 magnitude F, being applied perpendicularly to a lever arm at a distance ${\displaystyle r}$, as shown in the figure. This force will act through the distance along the circular arc ${\displaystyle l=s=r\phi }$, so the work done is

$${\displaystyle W=Fs=Fr\phi .}$$

Introduce the torque τ = Fr, to obtain

$${\displaystyle W=Fr\phi =\tau \phi ,}$$

as presented above.

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

Therefore, the work done by a force F on an object that travels along a curve C is given by the line integral:

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

where dx(t) defines the trajectory C and v 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,

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

If the work for an applied force is independent of the path, then the work done by the force, by the gradient 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 function U(x), 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

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

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

In this case, the gradient of work yields

and the force F is said to be "derivable from a potential."[20]

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

Gravity F = mg does work W = 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 is g = 9.8 m⋅s⁻² and the gravitational force on an object of mass m is F_g = mg. It is convenient to imagine this gravitational force concentrated at the center of mass of the object.

If an object with weight mg is displaced upwards or downwards a vertical distance y₂ − y₁, the work W done on the object is:

$${\displaystyle W=F_{g}(y_{2}-y_{1})=F_{g}\Delta y=mg\Delta y}$$

where F_g is weight (pounds in imperial units, and newtons in SI units), and Δy is the change in height y. 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.

The force of gravity exerted by a mass M on another mass m is given by

$${\displaystyle \mathbf {F} =-{\frac {GMm}{r^{2}}}{\hat {\mathbf {r} }}=-{\frac {GMm}{r^{3}}}\mathbf {r} ,}$$

where r is the position vector from M to m and r̂ is the unit vector in the direction of r.

Let the mass m move at the velocity v; then the work of gravity on this mass as it moves from position r(t₁) to r(t₂) is given by

$${\displaystyle W=-\int _{\mathbf {r} (t_{1})}^{\mathbf {r} (t_{2})}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot d\mathbf {r} =-\int _{t_{1}}^{t_{2}}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot \mathbf {v} \,dt.}$$

Notice that the position and velocity of the mass m are given by

$${\displaystyle \mathbf {r} =r\mathbf {e} _{r},\qquad \mathbf {v} ={\frac {d\mathbf {r} }{dt}}={\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t},}$$

where e_r and e_t are the radial and tangential unit vectors directed relative to the vector from M to m, and we use the fact that ${\displaystyle d\mathbf {e} _{r}/dt={\dot {\theta }}\mathbf {e} _{t}.}$ Use this to simplify the formula for work of gravity to,

$${\displaystyle W=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}(r\mathbf {e} _{r})\cdot \left({\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t}\right)dt=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}r{\dot {r}}dt={\frac {GMm}{r(t_{2})}}-{\frac {GMm}{r(t_{1})}}.}$$

This calculation uses the fact that

$${\displaystyle {\frac {d}{dt}}r^{-1}=-r^{-2}{\dot {r}}=-{\frac {\dot {r}}{r^{2}}}.}$$

The function

$${\displaystyle U=-{\frac {GMm}{r}},}$$

is the gravitational potential function, also known as gravitational potential energy. The negative sign follows the convention that work is gained from a loss of potential energy.

Forces in springs assembled in parallel

Consider a spring that exerts a horizontal force F = (−kx, 0, 0) that is proportional to its deflection in the x direction independent of how a body moves. The work of this spring on a body moving along the space with the curve X(t) = (x(t), y(t), z(t)), is calculated using its velocity, v = (v_x, v_y, v_z), to obtain

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

For convenience, consider contact with the spring occurs at t = 0, then the integral of the product of the distance x and the x-velocity, xv_xdt, over time t is 1/2x². The work is the product of the distance times the spring force, which is also dependent on distance; hence the x² result.

The work ${\displaystyle W}$ done by a body of gas on its surroundings is:

$${\displaystyle W=\int _{a}^{b}P\,dV}$$

where P is pressure, V is volume, and a and b are initial and final volumes.

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.[24] The relation between the net force and the acceleration is given by the equation F = ma (Newton's second law), and the particle displacement s can be expressed by the equation

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

which follows from ${\displaystyle v_{2}^{2}=v_{1}^{2}+2as}$ (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:

$${\displaystyle W=Fs=mas=ma{\frac {v_{2}^{2}-v_{1}^{2}}{2a}}={\frac {mv_{2}^{2}}{2}}-{\frac {mv_{1}^{2}}{2}}=\Delta E_{\text{k}}}$$

Other derivation:

$${\displaystyle W=Fs=mas=m{\frac {v_{2}^{2}-v_{1}^{2}}{2s}}s={\frac {1}{2}}mv_{2}^{2}-{\frac {1}{2}}mv_{1}^{2}=\Delta E_{\text{k}}}$$

In the general case of rectilinear motion, when the net 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:

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

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 as the work–energy principle:

$${\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot \mathbf {v} dt=m\int _{t_{1}}^{t_{2}}\mathbf {a} \cdot \mathbf {v} dt={\frac {m}{2}}\int _{t_{1}}^{t_{2}}{\frac {dv^{2}}{dt}}\,dt={\frac {m}{2}}\int _{v_{1}^{2}}^{v_{2}^{2}}dv^{2}={\frac {mv_{2}^{2}}{2}}-{\frac {mv_{1}^{2}}{2}}=\Delta E_{\text{k}}}$$

The identity ${\textstyle \mathbf {a} \cdot \mathbf {v} ={\frac {1}{2}}{\frac {dv^{2}}{dt}}}$ requires some algebra. From the identity ${\textstyle v^{2}=\mathbf {v} \cdot \mathbf {v} }$ and definition ${\textstyle \mathbf {a} ={\frac {d\mathbf {v} }{dt}}}$ it follows

$${\displaystyle {\frac {dv^{2}}{dt}}={\frac {d(\mathbf {v} \cdot \mathbf {v} )}{dt}}={\frac {d\mathbf {v} }{dt}}\cdot \mathbf {v} +\mathbf {v} \cdot {\frac {d\mathbf {v} }{dt}}=2{\frac {d\mathbf {v} }{dt}}\cdot \mathbf {v} =2\mathbf {a} \cdot \mathbf {v} .}$$

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

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

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

where m is the mass of the particle.

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

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

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

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

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

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

This integral is computed along the 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

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

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

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

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

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

It is useful to resolve the velocity and acceleration vectors into tangential and normal components along the trajectory X(t), such that

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

where

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

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

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

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

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

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

$${\displaystyle W=\Delta K.}$$

This derivation can be generalized to arbitrary rigid body systems.

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

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

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

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

Integration of both sides yields

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

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

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

As an example consider a car skidding to a stop, 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,

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

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

Lotus type 119B gravity racer at Lotus 60th celebration

Gravity racing championship in Campos Novos, Santa Catarina, Brazil, 8 September 2010

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 velocity V, 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 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,

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

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

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

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

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

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

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

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

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

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