Conservation of Mechanical Energy
Conservation of Mechanical Energy
Let us now consider what form the work-energy theorem takes when only conservative forces are involved. This will lead us to the conservation of energy principle. The work-energy theorem states that the net work done by all forces acting on a system equals its change in kinetic energy. In equation form, this is
If only conservative forces act, then
where \({W}_{c}\) is the total work done by all conservative forces. Thus,
Now, if the conservative force, such as the gravitational force or a spring force, does work, the system loses potential energy. That is, \({W}_{\text{c}}=-\text{Δ}\text{PE}\). Therefore,
or
This equation means that the total kinetic and potential energy is constant for any process involving only conservative forces. That is,
where i and f denote initial and final values. This equation is a form of the work-energy theorem for conservative forces; it is known as the conservation of mechanical energy principle. Remember that this applies to the extent that all the forces are conservative, so that friction is negligible. The total kinetic plus potential energy of a system is defined to be its mechanical energy, \((\text{KE}+\text{PE})\). In a system that experiences only conservative forces, there is a potential energy associated with each force, and the energy only changes form between \(\text{KE}\) and the various types of \(\text{PE}\), with the total energy remaining constant.
Example: Using Conservation of Mechanical Energy to Calculate the Speed of a Toy Car
A 0.100-kg toy car is propelled by a compressed spring, as shown in the figure below. The car follows a track that rises 0.180 m above the starting point. The spring is compressed 4.00 cm and has a force constant of 250.0 N/m. Assuming work done by friction to be negligible, find (a) how fast the car is going before it starts up the slope and (b) how fast it is going at the top of the slope.
A toy car is pushed by a compressed spring and coasts up a slope. Assuming negligible friction, the potential energy in the spring is first completely converted to kinetic energy, and then to a combination of kinetic and gravitational potential energy as the car rises. The details of the path are unimportant because all forces are conservative—the car would have the same final speed if it took the alternate path shown.
Strategy
The spring force and the gravitational force are conservative forces, so conservation of mechanical energy can be used. Thus,
\({\text{KE}}_{\text{i}}+{\text{PE}}_{\text{i}}={\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}\)
or
\(\cfrac{1}{2}{{\text{mv}}_{i}}^{2}+{\text{mgh}}_{i}+\cfrac{1}{2}{{\text{kx}}_{i}}^{2}=\cfrac{1}{2}{{\text{mv}}_{f}}^{2}+{\text{mgh}}_{f}+\cfrac{1}{2}{{\text{kx}}_{f}}^{2},\)
where \(h\) is the height (vertical position) and \(x\) is the compression of the spring. This general statement looks complex but becomes much simpler when we start considering specific situations. First, we must identify the initial and final conditions in a problem; then, we enter them into the last equation to solve for an unknown.
Solution for (a)
This part of the problem is limited to conditions just before the car is released and just after it leaves the spring. Take the initial height to be zero, so that both \({h}_{\text{i}}\) and \({h}_{\text{f}}\) are zero. Furthermore, the initial speed \({v}_{\text{i}}\) is zero and the final compression of the spring \({x}_{\text{f}}\) is zero, and so several terms in the conservation of mechanical energy equation are zero and it simplifies to
\(\cfrac{1}{2}{{\text{kx}}_{i}}^{2}=\cfrac{1}{2}{{\text{mv}}_{f}}^{2}\text{.}\)
In other words, the initial potential energy in the spring is converted completely to kinetic energy in the absence of friction. Solving for the final speed and entering known values yields
\(\begin{array}{lll}{v}_{f}& =& \sqrt{\frac{k}{m}}{x}_{i}\\ & =& \sqrt{\frac{\text{250}\text{.0 N/m}}{\text{0.100 kg}}}(\text{0.0400 m})\\ & =& \text{2.00 m/s.}\end{array}\)
Solution for (b)
One method of finding the speed at the top of the slope is to consider conditions just before the car is released and just after it reaches the top of the slope, completely ignoring everything in between. Doing the same type of analysis to find which terms are zero, the conservation of mechanical energy becomes
\(\cfrac{1}{2}{\text{kx}}_{\text{i}}^{ 2}=\cfrac{1}{ 2}{\text{mv}}_{\text{f}}^{\text{ 2}}+{\text{mgh}}_{\text{f}}\text{.}\)
This form of the equation means that the spring’s initial potential energy is converted partly to gravitational potential energy and partly to kinetic energy. The final speed at the top of the slope will be less than at the bottom. Solving for \({v}_{\text{f}}\) and substituting known values gives
\(\begin{array}{lll}{v}_{f}& =& \sqrt{\frac{{{\text{kx}}_{i}}^{2}}{m}-2{\text{gh}}_{f}}\\ & =& \sqrt{(\frac{\text{250.0 N/m}}{\text{0.100 kg}})(\text{0.0400 m}{)}^{2}-2(\text{9.80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2})(\text{0.180 m})}\\ & =& \text{0.687 m/s.}\end{array}\)
Discussion
Another way to solve this problem is to realize that the car’s kinetic energy before it goes up the slope is converted partly to potential energy—that is, to take the final conditions in part (a) to be the initial conditions in part (b).
Note that, for conservative forces, we do not directly calculate the work they do; rather, we consider their effects through their corresponding potential energies, just as we did in the example above. Note also that we do not consider details of the path taken—only the starting and ending points are important (as long as the path is not impossible). This assumption is usually a tremendous simplification, because the path may be complicated and forces may vary along the way.
This lesson is part of:
Work, Energy and Energy Resources