Work
Recall our thought experiment regarding some heavy lifting
Consider two weights: a 500 barbell versus a dinky little 2 hand weight. Also, consider two cases: lifting each weight to a height of 0.5 m or lifting them fully above the head to about 2 m. Which case does the most work / requires the most energy? Which case does the least work?
First, consider the weight, or the force (\(F=mg\)) that must be exerted to lift a barbell of mass \(m\), where \(g=\qty{9.81}{\meter\per\second\squared}\). Is it more work to lift a heavy 500 barbell or a dinky 2 hand weight, all else being equal?
Next, consider the distance we must lift the weight. Does it take more work to lift the weight 0.5 m or 2 m?
So the amount of work or energy it takes to do these tasks seems to depend on force and distance over which the force is exerted. To be precise, let’s combine those into an equation \[\begin{equation} W = \text{force}\cdot\text{distance} = \vec{F}\cdot\Delta\vec{x} \label{eq:workdefn} \end{equation}\] where \(W\) is work or energy, \(\vec{F}\) is force, and \(\Delta\vec{x}\) is the distance. This combines the idea that if we exert bigger forces (as in lifting a bigger weight) we are doing more work; and that as we go longer distances we also do more work.
The units of \(W\) are N m, or alternatively, joules (J), where \(\qty{1}{\joule}=\qty{1}{\newton\meter}\). The quantity \(W\) is a scalar. Sometimes, rather than the letter \(W\), you may see the letters \(Q\), \(U\), or \(E\) used to denote energy of various forms or accounting categories.
How to compute work / energy
There are various ways to compute energy but let’s start with a simple one to warm up. Xuan the Guide Dog Puppy (mass 65 , 29.4 kg) smells a half-eaten dropped hamburger and starts pulling his handler towards it. Xuan exerts a force of 294 N and pulls his handler a distance of 1 m. How much work did Xuan do?
Solution
The equation you need to use at this point is : \[\begin{equation} \text{work}\ W = \vec{F}\cdot\Delta\vec{x}. \label{eq:worksoln1} \end{equation}\]
Next, identify from the problem statement the various known quantities. We know Xuan is exerting \(F=\qty{294}{\newton}\) of force, and that the distance traveled is \(\Delta x=\qty{1}{\meter}\). For now we assume the two are in line (along his leash). Substitution into gives \[\begin{align} W &= F\cdot \Delta x \\ &= (\qty{294}{\newton})\cdot(\qty{1}{\meter}) \\ &= \qty{294}{\joule}. \end{align}\] The mechanical work Xuan does in pulling 1 m is 294 J.
Defining work
Work is the amount of energy transferred into or out of a system by a force exerted on that system over a distance.
Conservative versus non-conservative forces
The work done by a conservative force exerted on a system is path-independent and only depends on the initial and final configurations of that system. The work done by a conservative force on a system – or the change in the potential energy of the system – will be zero if the system returns to its initial configuration.
Potential energies are associated only with conservative forces.
In contrast to a conservative force, the work done by a nonconservative force is path-dependent. The most common nonconservative forces are friction and air resistance.
Work
Work is a scalar quantity that may be positive, negative, or zero. The work done on an object by a variable force is calculated using \[\begin{equation} W = \int_a^b \vec{F}(\vec{r}) \cdot d\vec{r}, \end{equation}\] where the integral is taken over the path from point \(a\) to point \(b\). Work is measured in joules (J), same as for other energy-type quantities.
Note the use of the dot product in computing work. As a result of the dot product, only the component of the force exerted on a system that is parallel to the displacement of the point of application of the force will change the system’s total energy. If the component of the force exerted on a system that is parallel to the displacement is constant, the work done on the system by the force is given by the derived equation \[\begin{equation} W = F d \cos{\theta} \end{equation}\]
Work-energy theorem
The work–energy theorem states that the change in an object’s kinetic energy is equal to the sum of the work (net work) being done by all forces exerted on the object. \[\begin{equation} \Delta KE = \Sigma W_i = \Sigma F_{||,i} d_i. \end{equation}\] An external force may change the configuration of a system. The component of the external force parallel to the displacement times the displacement of the point of application of the force gives the change in kinetic energy of the system.
If the system’s center of mass and the point of application of the force move the same distance when a force is exerted on a system, then the system may be modeled as an object, and only the system’s kinetic energy can change.
Energy dissipated by friction
The energy dissipated by friction is typically equated to the force of friction times the length of the path over which the force is exerted. \[\begin{equation} \Delta E_{mech} = F_{fr} d \cos{\theta} \end{equation}\]
Work on a graph of force versus displacement
Work is also equal to the area under the curve of a graph of \(F_{||}\) as a function of displacement.
See also
- list here