7.2 Kinetic energy  (Page 2/4)

Because velocity is a relative quantity, you can see that the value of kinetic energy must depend on your frame of reference. You can generally choose a frame of reference that is suited to the purpose of your analysis and that simplifies your calculations. One such frame of reference is the one in which the observations of the system are made (likely an external frame). Another choice is a frame that is attached to, or moves with, the system (likely an internal frame). The equations for relative motion, discussed in Motion in Two and Three Dimensions , provide a link to calculating the kinetic energy of an object with respect to different frames of reference.

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 ${\overrightarrow{v}}_{\text{PT}}={\overrightarrow{v}}_{\text{PC}}+{\overrightarrow{v}}_{\text{CT}}.$ 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, $v_{\text{PT}}=v_{\text{CT}}\pm v_{\text{PC}}$ , as shown in [link] .

Solution

1. $K=\frac{1}{2}(75.0\text{kg})(1.50{\text{m/s})}^2=84.4\text{J}\text{.}$
2. $v_{\text{PT}}=(15.0\pm 1.50)\text{m/s}\text{.}$ 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.

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The kinetic energy of a particle is a single quantity, but the kinetic energy of a system of particles can sometimes be divided into various types, depending on the system and its motion. For example, if all the particles in a system have the same velocity, the system is undergoing translational motion and has translational kinetic energy. If an object is rotating, it could have rotational kinetic energy, or if it’s vibrating, it could have vibrational kinetic energy. The kinetic energy of a system, relative to an internal frame of reference, may be called internal kinetic energy. The kinetic energy associated with random molecular motion may be called thermal energy. These names will be used in later chapters of the book, when appropriate. Regardless of the name, every kind of kinetic energy is the same physical quantity, representing energy associated with motion.