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11.4 Kinetic energy by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.

vCalc Companion Formulas | |

vCalc Formulary | 11.4 Kinetic energy |

`KE=1/2mv^2` | Kinetic Energy |

The technical term for the energy associated with motion is kinetic energy, from the Greek word for motion. (The root is the same as the root of the word “cinema” for a motion picture, and in French the term for kinetic energy is “énergie cinétique.”) To find how much kinetic energy is possessed by a given moving object, we must convert all its kinetic energy into heat energy, which we have chosen as the standard reference type of energy. We could do this, for example, by firing projectiles into a tank of water and measuring the increase in temperature of the water as a function of the projectile's mass and velocity. Consider the following data from a series of three such experiments:

m (kg) | v (m/s) | energy (J) | |

1.00 | 1.00 | 0.50 | |

1.00 | 2.00 | 2.00 | |

2.00 | 1.00 | 1.00 |

Discussion question B

Comparing the first experiment with the second, we see that doubling the object's velocity doesn't just double its energy, it quadruples it. If we compare the first and third lines, however, we find that doubling the mass only doubles the energy. This suggests that kinetic energy is proportional to mass and to the square of velocity, `KE?mv^2`, and further experiments of this type would indeed establish such a general rule. The proportionality factor equals 0.5 because of the design of the metric system, so the kinetic energy of a moving object is given by

The metric system is based on the meter, kilogram, and second, with other units being derived from those. Comparing the units on the left and right sides of the equation shows that the joule can be reexpressed in terms of the basic units as `kg?m^2"/"s^2`.

Students are often mystified by the occurrence of the factor of 1/2, but it is less obscure than it looks. The metric system was designed so that some of the equations relating to energy would come out looking simple, at the expense of some others, which had to have inconvenient conversion factors in front. If we were using the old British Engineering System of units in this course, then we'd have the British Thermal Unit (BTU) as our unit of energy. In that system, the equation you'd learn for kinetic energy would have an inconvenient proportionality constant, `KE=(1.29×10^(?3))mv^2`, with `KE` measured in units of BTUs, `v` measured in feet per second, and so on. At the expense of this inconvenient equation for kinetic energy, the designers of the British Engineering System got a simple rule for calculating the energy required to heat water: one BTU per degree Fahrenheit per pound. The inventor of kinetic energy, Thomas Young, actually defined it as **KE=mv ^{2}**, which meant that all his other equations had to be different from ours by a factor of two. All these systems of units work just fine as long as they are not combined with one another in an inconsistent way.

? Comet Shoemaker-Levy, which struck the planet Jupiter in 1994, had a mass of roughly `4×10^13 kg`, and was moving at a speed of 60 km/s. Compare the kinetic energy released in the impact to the total energy in the world's nuclear arsenals, which is `2×10^19 J`. Assume for the sake of simplicity that Jupiter was at rest.

? Since we assume Jupiter was at rest, we can imagine that the comet stopped completely on impact, and 100% of its kinetic energy was converted to heat and sound. We first convert the speed to mks units, `v=6×10^4 m"/"s`, and then plug in to the equation to find that the comet's kinetic energy was roughly `7×10^22 J`, or about 3000 times the energy in the world's nuclear arsenals.

Is there any way to derive the equation `KE=1/2mv^2`. mathematically from first principles? No, it is purely empirical. The factor of 1/2 in front is definitely not derivable, since it is different in different systems of units. The proportionality to `v^2` is not even quite correct; experiments have shown deviations from the `v^2` rule at high speeds, an effect that is related to Einstein's theory of relativity. Only the proportionality to `m` is inevitable. The whole energy concept is based on the idea that we add up energy contributions from all the objects within a system. Based on this philosophy, it is logically necessary that a 2-kg object moving at 1 m/s have the same kinetic energy as two 1-kg objects moving side-by-side at the same speed.

Although I mentioned Einstein's theory of relativity above, it's more relevant right now to consider how conservation of energy relates to the simpler Galilean idea, which we've already studied, that motion is relative. Galileo's Aristotelian enemies (and it is no exaggeration to call them enemies!) would probably have objected to conservation of energy. After all, the Galilean idea that an object in motion will continue in motion indefinitely in the absence of a force is not so different from the idea that an object's kinetic energy stays the same unless there is a mechanism like frictional heating for converting that energy into some other form.

More subtly, however, it's not immediately obvious that what we've learned so far about energy is strictly mathematically consistent with the principle that motion is relative. Suppose we verify that a certain process, say the collision of two pool balls, conserves energy as measured in a certain frame of reference: the sum of the balls' kinetic energies before the collision is equal to their sum after the collision. (In reality we'd need to add in other forms of energy, like heat and sound, that are liberated by the collision, but let's keep it simple.) But what if we were to measure everything in a frame of reference that was in a different state of motion? A particular pool ball might have less kinetic energy in this new frame; for example, if the new frame of reference was moving right along with it, its kinetic energy in that frame would be zero. On the other hand, some other balls might have a greater kinetic energy in the new frame. It's not immediately obvious that the total energy before the collision will still equal the total energy after the collision. After all, the equation for kinetic energy is fairly complicated, since it involves the square of the velocity, so it would be surprising if everything still worked out in the new frame of reference. It *does* still work out. Homework problem 13 in this chapter gives a simple numerical example, and the general proof is taken up in problem 15 on p. 382 (with the solution given in the back of the book).

? Suppose that, like Young or Einstein, you were trying out different equations for kinetic energy to see if they agreed with the experimental data. Based on the meaning of positive and negative signs of velocity, why would you suspect that a proportionality to *mv* would be less likely than `mv^2`?

? The figure shows a pendulum that is released at A and caught by a peg as it passes through the vertical, B. To what height will the bob rise on the right?

11.4 Kinetic energy by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.