# 7.2 Kinetic energy

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By the end of this section, you will be able to:
• Calculate the kinetic energy of a particle given its mass and its velocity or momentum
• Evaluate the kinetic energy of a body, relative to different frames of reference

It’s plausible to suppose that the greater the velocity of a body, the greater effect it could have on other bodies. This does not depend on the direction of the velocity, only its magnitude. At the end of the seventeenth century, a quantity was introduced into mechanics to explain collisions between two perfectly elastic bodies, in which one body makes a head-on collision with an identical body at rest. The first body stops, and the second body moves off with the initial velocity of the first body. (If you have ever played billiards or croquet, or seen a model of Newton’s Cradle, you have observed this type of collision.) The idea behind this quantity was related to the forces acting on a body and was referred to as “the energy of motion.” Later on, during the eighteenth century, the name kinetic energy    was given to energy of motion.

With this history in mind, we can now state the classical definition of kinetic energy. Note that when we say “classical,” we mean non-relativistic, that is, at speeds much less that the speed of light. At speeds comparable to the speed of light, the special theory of relativity requires a different expression for the kinetic energy of a particle, as discussed in Relativity .

Since objects (or systems) of interest vary in complexity, we first define the kinetic energy of a particle with mass m .

## Kinetic energy

The kinetic energy of a particle is one-half the product of the particle’s mass m and the square of its speed v :

$K=\frac{1}{2}m{v}^{2}.$

We then extend this definition to any system of particles by adding up the kinetic energies of all the constituent particles:

$K=\sum \frac{1}{2}m{v}^{2}.$

Note that just as we can express Newton’s second law in terms of either the rate of change of momentum or mass times the rate of change of velocity, so the kinetic energy of a particle can be expressed in terms of its mass and momentum $\left(\stackrel{\to }{p}=m\stackrel{\to }{v}\right),$ instead of its mass and velocity. Since $v=p\text{/}m$ , we see that

$K=\frac{1}{2}m{\left(\frac{p}{m}\right)}^{2}=\frac{{p}^{2}}{2m}$

also expresses the kinetic energy of a single particle. Sometimes, this expression is more convenient to use than [link] .

The units of kinetic energy are mass times the square of speed, or $\text{kg}·{\text{m}}^{2}{\text{/s}}^{2}$ . But the units of force are mass times acceleration, $\text{kg}·{\text{m/s}}^{2}$ , so the units of kinetic energy are also the units of force times distance, which are the units of work, or joules. You will see in the next section that work and kinetic energy have the same units, because they are different forms of the same, more general, physical property.

## Kinetic energy of an object

(a) What is the kinetic energy of an 80-kg athlete, running at 10 m/s? (b) The Chicxulub crater in Yucatan, one of the largest existing impact craters on Earth, is thought to have been created by an asteroid, traveling at
22 km/s and releasing $4.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}\phantom{\rule{0.2em}{0ex}}\text{J}$ of kinetic energy upon impact. What was its mass? (c) In nuclear reactors, thermal neutrons, traveling at about 2.2 km/s, play an important role. What is the kinetic energy of such a particle?

## Strategy

To answer these questions, you can use the definition of kinetic energy in [link] . You also have to look up the mass of a neutron.

## Solution

Don’t forget to convert km into m to do these calculations, although, to save space, we omitted showing these conversions.

1. $K=\frac{1}{2}\left(80\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(10\phantom{\rule{0.2em}{0ex}}{\text{m/s}\right)}^{2}=4.0\phantom{\rule{0.2em}{0ex}}\text{kJ}\text{.}$
2. $m=2K\text{/}{v}^{2}=2\left(4.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}\text{J}\right)\text{/}{\left(22\phantom{\rule{0.2em}{0ex}}\text{km/s}\right)}^{2}=1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{15}\phantom{\rule{0.2em}{0ex}}\text{kg}\text{.}$
3. $K=\frac{1}{2}\left(1.68\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(2.2\phantom{\rule{0.2em}{0ex}}\text{km/s}\right)}^{2}=4.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-21}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$

## Significance

In this example, we used the way mass and speed are related to kinetic energy, and we encountered a very wide range of values for the kinetic energies. Different units are commonly used for such very large and very small values. The energy of the impactor in part (b) can be compared to the explosive yield of TNT and nuclear explosions, $1\phantom{\rule{0.2em}{0ex}}\text{megaton}=4.18\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{15}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$ The Chicxulub asteroid’s kinetic energy was about a hundred million megatons. At the other extreme, the energy of subatomic particle is expressed in electron-volts, $1\phantom{\rule{0.2em}{0ex}}\text{eV}=1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$ The thermal neutron in part (c) has a kinetic energy of about one fortieth of an electron-volt.

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