Example

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 $$ 4.2\,×\,{10}^{23}\,\text{J} $$ 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 (Figure). 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. $$K=\frac{1}{2}(80\,\text{kg})(10\,{\text{m/s})}^{2}=4.0\,\text{kJ}\text{.}$$
2. $$m=2K\text{/}{v}^{2}=2(4.2\,×\,{10}^{23}\text{J})\text{/}{(22\,\text{km/s})}^{2}=1.7\,×\,{10}^{15}\,\text{kg}\text{.}$$
3. $$K=\frac{1}{2}(1.68\,×\,{10}^{-27}\,\text{kg}){(2.2\,\text{km/s})}^{2}=4.1\,×\,{10}^{-21}\,\text{J}\text{.}$$

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, $$ 1\,\text{megaton}=4.18\,×\,{10}^{15}\,\text{J}\text{.} $$ 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, $$ 1\,\text{eV}=1.6\,×\,{10}^{-19}\,\text{J}\text{.} $$ The thermal neutron in part (c) has a kinetic energy of about one fortieth of an electron-volt.

Example

Kinetic Energy Relative to Different Frames

A 75.0-kg person walks down the central aisle of a subway car at a speed of 1.50 m/s relative to the car, whereas the train is moving at 15.0 m/s relative to the tracks. (a) What is the person’s kinetic energy relative to the car? (b) What is the person’s kinetic energy relative to the tracks? (c) What is the person’s kinetic energy relative to a frame moving with the person?

Strategy

Since speeds are given, we can use $$ \frac{1}{2}m{v}^{2} $$ to calculate the person’s kinetic energy. However, in part (a), the person’s speed is relative to the subway car (as given); in part (b), it is relative to the tracks; and in part (c), it is zero. If we denote the car frame by C, the track frame by T, and the person by P, the relative velocities in part (b) are related by $$ {\overset{\to }{v}}_{\text{PT}}={\overset{\to }{v}}_{\text{PC}}+{\overset{\to }{v}}_{\text{CT}}. $$ We can assume that the central aisle and the tracks lie along the same line, but the direction the person is walking relative to the car isn’t specified, so we will give an answer for each possibility, $$ {v}_{\text{PT}}={v}_{\text{CT}}±{v}_{\text{PC}}$$, as shown in (Figure).

Figure 7.10 The possible motions of a person walking in a train are (a) toward the front of the car and (b) toward the back of the car.

Solution

1. $$K=\frac{1}{2}(75.0\,\text{kg})(1.50\,{\text{m/s})}^{2}=84.4\,\text{J}\text{.}$$
2. $${v}_{\text{PT}}=(15.0±1.50)\,\text{m/s}\text{.} $$ Therefore, the two possible values for kinetic energy relative to the car are
   $$K=\frac{1}{2}(75.0\,\text{kg})(13.5\,{\text{m/s})}^{2}=6.83\,\text{kJ}$$

   and

   $$K=\frac{1}{2}(75.0\,\text{kg})(16.5\,{\text{m/s})}^{2}=10.2\,\text{kJ}\text{.}$$
3. In a frame where $$ {v}_{\text{P}}=0,K=0 $$ as well.

Significance

You can see that the kinetic energy of an object can have very different values, depending on the frame of reference. However, the kinetic energy of an object can never be negative, since it is the product of the mass and the square of the speed, both of which are always positive or zero.

Example

Special Names for Kinetic Energy

(a) A player lobs a mid-court pass with a 624-g basketball, which covers 15 m in 2 s. What is the basketball’s horizontal translational kinetic energy while in flight? (b) An average molecule of air, in the basketball in part (a), has a mass of 29 u, and an average speed of 500 m/s, relative to the basketball. There are about $$ 3\,×\,{10}^{23} $$ molecules inside it, moving in random directions, when the ball is properly inflated. What is the average translational kinetic energy of the random motion of all the molecules inside, relative to the basketball? (c) How fast would the basketball have to travel relative to the court, as in part (a), so as to have a kinetic energy equal to the amount in part (b)?

Strategy

In part (a), first find the horizontal speed of the basketball and then use the definition of kinetic energy in terms of mass and speed, $$ K=\frac{1}{2}m{v}^{2}$$. Then in part (b), convert unified units to kilograms and then use $$ K=\frac{1}{2}m{v}^{2} $$ to get the average translational kinetic energy of one molecule, relative to the basketball. Then multiply by the number of molecules to get the total result. Finally, in part (c), we can substitute the amount of kinetic energy in part (b), and the mass of the basketball in part (a), into the definition $$ K=\frac{1}{2}m{v}^{2}$$, and solve for v.

Solution

1. The horizontal speed is (15 m)/(2 s), so the horizontal kinetic energy of the basketball is
   $$\frac{1}{2}(0.624\,\text{kg}){(7.5\,\text{m/s})}^{2}=17.6\,\text{J}\text{.}$$
2. The average translational kinetic energy of a molecule is
   $$\frac{1}{2}(29\,\text{u})(1.66\,×\,{10}^{-27}\,\text{kg/u}){(500\,\text{m/s})}^{2}=6.02\,×\,{10}^{-21}\,\text{J,}$$

   and the total kinetic energy of all the molecules is

   $$(3\,×\,{10}^{23})(6.02\,×\,{10}^{-21}\,\text{J})=1.80\,\text{kJ}\text{.}$$
3. $$v=\sqrt{2(1.8\,\text{kJ})\text{/}(0.624\,\text{kg})}=76.0\,\text{m/s}\text{.}$$

Significance

In part (a), this kind of kinetic energy can be called the horizontal kinetic energy of an object (the basketball), relative to its surroundings (the court). If the basketball were spinning, all parts of it would have not just the average speed, but it would also have rotational kinetic energy. Part (b) reminds us that this kind of kinetic energy can be called internal or thermal kinetic energy. Notice that this energy is about a hundred times the energy in part (a). How to make use of thermal energy will be the subject of the chapters on thermodynamics. In part (c), since the energy in part (b) is about 100 times that in part (a), the speed should be about 10 times as big, which it is (76 compared to 7.5 m/s).