Work (physics)

In physics, work is the energy transferred to a body when it is moved along a path by a force. This definition is similar to one that we use in everyday speech; that is, to proceed along a path through an activity, as "the politician worked her way through the crowd," "the professor worked the problem on the chalkboard," or "the host was working on his second martini when the guests arrived" (one hopes that the path in the latter case was a short one).

The terms force, work, and energy have well-defined meanings in physics, and they have their counterparts in everyday life. For example, force is that which produces or prevents motion, or imposes a change of velocity on something (a police force can do any of those, depending on the orders given). When a force acts to move an object in physics, we say that work is done on the object by the force (the police officers are not exactly taking it easy when they push back a crowd of fans trying to rush the latest pop singing star: they are doing work). And energy in physics, as in everyday life, is the capacity of a physical system to perform work. When you are "too pooped to pop," you have no energy.

Force is the agent of change, and work is a measure of the change.

In the physical world, work done on a body is accomplished not only by a displacement of the body as a whole from one place to another (like crowd control) but also, for example, by compressing a gas, rotating a shaft, and even directing small particles along a magnetic field.

Definition of mechanical work

PD Image
Constant force F along path s = r₁ − r₀.

The mechanical work W is defined as force times path length: When a constant force F acts on a body along a straight path and the body is moved over a length |s| along this path, then

$W=\mathbf {F} \cdot \mathbf {s} =|\mathbf {F} |\,|\mathbf {s} |\,\cos \alpha \,.$

Here s is vector of magnitude |s| along the path. The dot product between the two vectors F and s is the product of their magnitudes |F| and |s| times the cosine of the angle, α, between the vectors.

When the force is directed along the path (F and s parallel), α = 0, cosα = 1, and we have simply W = |F| |s|. The positive work W is converted into an increase of kinetic energy T,

W = ΔT ≡ T₁ − T₀ > 0.

In the next section this relation will be proved and shown to be a consequence of Newton's second law: F = m a.

When a force is perpendicular to the path, α = 90°, cosα = 0, the force performs no work. Examples of such a situation are the centrifugal force on a mass in uniform circular motion and the gravitational force acting on a satellite in a circular orbit. If a body is in uniform (i.e., has constant speed) straight motion and the only force acting on it is perpendicular to its path, then the body will persist in its uniform straight motion (Newton's first law).

When a force F is anti-parallel to the path (α = 180°), it performs negative work W < 0, since cosα = −1, and |F| and |s| are both greater than zero. The negative work done by the force is converted into a decrease of the kinetic energy T,

W = ΔT < 0

(ΔT < 0 because T decreases, i.e., T₁ < T₀).

For example, think of an old-fashioned type of cannon that shoots a cannon ball of mass m straight up. The gun powder explosion gives the ball initial kinetic energy T₀ = ½ m v₀². The gravitational attraction of the earth performs negative work (force downward, motion of the cannon ball upward), until at the highest point the speed v₁ and the kinetic energy T₁ = ½ m v₁² of the ball are zero. The amount of work performed is W = ΔT = 0−T₀ = −T₀. At the highest point the motion of the cannon ball reverts direction, it starts falling to earth, and from there on the gravitational attraction performs positive work (direction of motion and force are parallel). The kinetic energy increases again until it achieves its original value T₀ = ½ m v₀² at the point where the cannon ball arrives again at the cannon. The total work done by the gravitation is zero, the work done on the cannon ball going up cancels the work done on the ball going down. (In this example we ignore friction by the air).

When the the force is conservative (non-dissipative), the work is independent of path, and when furthermore the path is a closed curve, the total work is zero (one way the work is positive and the other way the work is equally large in absolute value, but negative).

As the gravitational field is conservative, we just saw an example of a mass (a cannon ball) making a closed path in a conservative force field. The potential energy U of a mass m close to the surface of the earth is equal to U = mgh, where g is the gravitational acceleration and h is the height. When the cannon is positioned at height h = 0, the cannon ball starts with potential energy zero, U₀ = 0. The work W done by the gravitational field on the cannon ball going upward has two effects: it decreases its kinetic energy—as just discussed—and it increases its potential energy, U₁ > U₀, so that

W = −ΔU ≡ −(U₁ − U₀) = U₀ − U₁ < 0.

At the highest point, h₁, the potential energy is maximum U₁ = mgh₁ (equal to minus the work performed) and the kinetic energy is zero (this is the point where the motion of the cannon ball changes direction, for a small amount of time its speed is zero). The work done by a conservative force converts a decrease in kinetic energy into an increase of potential energy and conversely:

W = ΔT = −ΔU, so that ΔT + ΔU = 0.

When the cannon ball dropping down is at height h = 0 again, its kinetic energy is maximum (equal to the energy imparted to it by the exploding gun powder) and its potential energy is zero again.

Mathematical formulation

When the force is not constant along the path, or the path is not straight, it is possible to compute the work by infinitesimal calculus. One chooses N + 1 consecutive points on the path

$\mathbf {r} _{0},\mathbf {r} _{1},\ldots ,\mathbf {r} _{N}\quad {\hbox{and defines}}\quad \Delta \mathbf {s} _{i}\equiv \mathbf {r} _{i+1}-\mathbf {r} _{i}.$

This divides the path in N pieces Δs_i, which are small enough to assume that the force is constant, F(r_i), on the i-th piece. By their definition as the differences of two vectors, the pieces are straight. The (approximate) total work is obtained by summing the work done along the individual small pieces,

$W\approx \sum _{i=0}^{N-1}W_{i}=\sum _{i=0}^{N-1}\mathbf {F} (\mathbf {r} _{i})\Delta \mathbf {s} _{i}\,.$

Obviously, when the total path is straight (as in the figure above) and the force is constant over the path, two points give an exact result, i.e., N = 1 suffices.

To improve the approximation one chooses the sampling points r_i closer and closer, which makes the pieces Δs_i smaller and smaller, so that their lengths go to zero. The limit is the path integral

$W=\int _{\mathbf {r} _{0}}^{\mathbf {r} _{1}}\mathbf {F} (\mathbf {r} )\,\mathrm {d} \mathbf {s} \,,$

where we wrote the upper limit (endpoint of the curve) now as r₁ (when the number of intervals N goes to infinity, writing the upper limit as r_N becomes meaningless).

To show that the work is converted into an increase in kinetic energy, we write for one mass m

${\dot {T}}\equiv {\frac {dT}{dt}}={\frac {d({\tfrac {1}{2}}mv^{2})}{dt}}={\tfrac {1}{2}}m{\frac {d({\dot {\mathbf {r} }}\cdot {\dot {\mathbf {r} }})}{dt}}=m{\dot {\mathbf {r} }}\cdot {\ddot {\mathbf {r} }}={\dot {\mathbf {r} }}\cdot \mathbf {F} ,$

where we used F = m a (Newton's second law) and that the acceleration a is the second derivative of r with respect to time. Integrate

$\int _{t_{0}}^{t_{1}}{\frac {dT}{dt}}\mathrm {d} t=T_{1}-T_{0}=\int _{t_{0}}^{t_{1}}\mathbf {F} \cdot {\dot {\mathbf {r} }}\;\mathrm {d} t=\int _{\mathbf {r} _{0}}^{\mathbf {r} _{1}}\mathbf {F} \cdot \mathrm {d} \mathbf {s} =W$

where we used that Δs is tangent to the path, i.e.,

$\Delta \mathbf {s} =\mathbf {r} (t+\Delta t)-\mathbf {r} (t)\quad \Longrightarrow \quad d\mathbf {s} =\lim _{\Delta t\rightarrow 0}{\frac {\mathbf {r} (t+\Delta t)-\mathbf {r} (t)}{\Delta t}}\Delta t={\dot {\mathbf {r} }}\,dt,$

Note that from the usual definition of the time derivative of a vector function used here, follows dr ≡ ds and that accordingly the limits in the path integral are r₀ ≡ r(t₀) and r₁ ≡ r(t₁). The following important result is obtained that connects the work and change in kinetic energy over the path,

$T_{1}-T_{0}\equiv T(t_{1})-T(t_{0})=W.\,$

To show that work by a conservative force is converted into decrease in potential energy, we use the relation between force and potential, valid for conservative fields

$\mathbf {F} =-{\boldsymbol {\nabla }}U=-\left({\frac {\partial U}{\partial x}},{\frac {\partial U}{\partial y}},{\frac {\partial U}{\partial z}}\right)$

and by the chain rule

${\frac {dU}{dt}}={\frac {\partial U}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial U}{\partial y}}{\frac {dy}{dt}}+{\frac {\partial U}{\partial z}}{\frac {dz}{dt}}\;\Longrightarrow \;dU={\frac {\partial U}{\partial x}}dx+{\frac {\partial U}{\partial y}}dy+{\frac {\partial U}{\partial z}}dz={\boldsymbol {\nabla }}U\cdot d\mathbf {s}$

Integrate

$\int _{\mathbf {r} _{0}}^{\mathbf {r} _{1}}\mathrm {d} U=U_{1}-U_{0}=\int _{\mathbf {r} _{0}}^{\mathbf {r} _{1}}{\boldsymbol {\nabla }}U\cdot \mathrm {d} \mathbf {s} =-\int _{\mathbf {r} _{0}}^{\mathbf {r} _{1}}\mathbf {F} \cdot \mathrm {d} \mathbf {s} =-W.$

In summary,

$U_{1}-U_{0}\equiv U(\mathbf {r} _{1})-U(\mathbf {r} _{0})=-W.\,$

In total, the conservation of mechanical energy follows

$T_{1}+U_{1}=T_{0}+U_{0}\;\Longrightarrow \;E_{1}=E_{0}\quad {\hbox{with}}\quad E\equiv T+U.$

Example of mechanical work

• Work to lift a mass m in the gravitational field of the earth. Close to the surface of the earth, the attractive force is constant and equal to the gravitational acceleration g. The work to lift the mass to a height h is,

$W(h)=F_{g}\,h=m\,g\,h\,.$

When h is positive, and the mass is at rest before and after the lifting (no kinetic energy), the work is completely converted into potential energy.

Numeric examples:

$m=1\,\mathrm {kg} ,g=9.81\,\mathrm {m} /\mathrm {s^{2}} ,F_{g}=m\,g=9.81\,\mathrm {N} ,h=40\,\mathrm {cm} ,W(h)=9.81\,\mathrm {N} \cdot 0.4\,\mathrm {m} \approx 4\,\mathrm {J} \,.$

Here N is the unit newton and J is the unit joule.

History

The following is en excerpt of Ref.

(To be continued)