Opening the book…
- Explain work as a transfer of energy and net work as the work done by the net force.
- Explain and apply the work-energy theorem.
Work Transfers Energy
What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if a lawn mower is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on a briefcase by a person carrying it up stairs is stored in the briefcase-Earth system and can be recovered at any time. In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work.
In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion.
Net Work and the Work-Energy Theorem
We know from the study of Newton’s laws that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion.
Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all external forces—that is, net work is the work done by the net external force Fnet over a distance, d, and is given by the equation
\[W_{\mathrm{net}}=F_{\mathrm{net}}d. \nonumber \]
By using Newton’s second law, and doing some algebra, we can reach an interesting conclusion. Substituting F=ma from Newton’s second law gives
\[W_{\mathrm{net}}=mad. \nonumber \]
After doing more algebraic substitution, we eventually derive the work-energy theorem, which tells us that the net work on a system depends on the mass and velocity of an object.
The net work on a system equals the change in the quantity \(\frac{1}{2} m v^{2}\).
\[W_{\mathrm{net}}=\frac{1}{2} m v^{2}-\frac{1}{2} m v_{0}^{2} \nonumber\]
The quantity \(\frac{1}{2} m v^{2}\) in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass \(m\) moving at a speed \(v\). In equation form, the translational kinetic energy,
\[\mathrm{KE}=\frac{1}{2} m v^{2}, \nonumber \]
is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together. The work-energy theorem can be compactly written as,
\[W_{\text {net }}=\Delta \mathrm{KE} \nonumber \]
where \(\triangle \mathrm{KE}\) is understood to mean "change in KE."We are aware that it takes energy to get an object, like a car, up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 40 m/s has four times the kinetic energy it has at 20 m/s, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy.
Suppose a 30.0-kg package on a roller belt conveyor system is moving at 0.500 m/s. What is its kinetic energy?
Strategy
Because the mass m and speed v are given, the kinetic energy can be calculated from its definition as given in the equation \(\mathrm{KE}=\frac{1}{2} m v^{2}\).
Solution
The kinetic energy is given by
\[\mathrm{KE}=\frac{1}{2} m v^{2}. \nonumber\]
Entering known values gives
\[\mathrm{KE}=0.5(30.0 \mathrm{~kg})(0.500 \mathrm{~m} / \mathrm{s})^{2}, \nonumber\]
which yields
\[\mathrm{KE}=3.75 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2}=3.75 \mathrm{~J}. \nonumber\]
Discussion
Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves.
Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics and dynamics alone.
Section Summary
- The net work \(W_{\text {net }}\) is the work done by the net force acting on an object.
- Work done on an object transfers energy to the object.
- The translational kinetic energy of an object of mass m moving at speed v is \(\mathrm{KE}=\frac{1}{2} m v^{2}\).
- The work-energy theorem states that the net work \(W_{\text {net }}\) on a system changes its kinetic energy, \(W_{\text {net }}=\Delta \mathrm{KE}=\frac{1}{2} m v^{2}-\frac{1}{2} m v_{0}^{2}\).
Glossary
- net work
- work done by the net force, or vector sum of all the forces, acting on an object
- work-energy theorem
- the result, based on Newton’s laws, that the net work done on an object is equal to its change in kinetic energy
- kinetic energy
- the energy an object has by reason of its motion, equal to \(\frac{1}{2} m v^{2}\) for the translational (i.e., non-rotational) motion of an object of mass \(m\) moving at speed \(v\)
This page titled 4.3.2: Net Force, Work, and Kinetic Energy was last modified on Fri, 06 Jun 2025 16:03:54 GMT and is shared under a mixed 4.0 license and was authored, remixed, and/or curated by OpenStax.
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