Kinetic Energy

### Kinetic energy is the energy of motion

Kinetic energy is the easiest form of energy to think about. If something is moving, it has kinetic energy. If it's moving faster, it has more kinetic energy. And if two things are moving at the same velocity but one has more mass, it also has more kinetic energy.

Kinetic energy (E or Ek or KE) is proportional to mass (m) and proportional to the square of the velocity (v) of an object. The constant of proportionality turns out to be ½ in this case:

$$KE = \frac{1}{2} mv^2$$

The units of KE are Joules (J), 1 J = 1 Kg·m·s-2. Kinetic energy is not a vector. In squaring the velocity vector, v, we lose it's directional part, so KE is a scalar quantity – no direction.

In this section, we'll discuss the kinetic energy of linear motion, or motion in a plane, which can be curved. We'll reserve for another section talking about the kinetic energy of rotation or circular motion.

#### Examples of kinetic energy

Linear motion

Curved or 2-D motion

Rotation of a body about an axis

KE of electrons and atoms – quantum-mechanical KE

We'll develop the relationship between KE and momentum (p), and we'll look at the conservation of kinetic energy in mechanical processes ... here we go.

#### Kinetic energy

An object has kinetic energy (KE) if it is moving. KE is the energy of motion.

An astute student of mine once asked, "If kinetic energy is   $\frac{1}{2} mv^2$, what happened to the other half?"

### Derivation of the formula

We arrive at the formula for kinetic energy by considering the amount of work that a moving object can do. Think of something like a hammer driving a nail. The hammer has kinetic energy because it has mass and it is moving. It strikes the nail with some initial velocity, vi, and eventually comes to rest (vf = 0). During the motion, work is done on the nail. That work is:

$$w = F \cdot d$$

Newton's 2nd law tells us that force is mass times acceleration, so we make the substitution F = ma:

$$w = (ma) \cdot d$$

Now let's turn to the definition of acceleration. In this case, the initial velocity is finite, and the final velocity is zero,

$$a = \frac{v_f - v_i}{t}$$

So we can reduce that definition to one involving only the initial velocity of the hammer. Here I've just made the acceleration positive to make things simpler:

$$a = \frac{v}{t}$$

Now the average velocity is distance divided by time:

$$\bar{v} = \frac{d}{t}$$

The numerical average of two velocities (one of which is zero) gives us

$$\frac{v}{2} = \frac{d}{t}$$

Solving for time gives us this expression

$$t = \frac{2d}{v}$$

which we can plug into the acceleration formula like this

$$a = \frac{v}{\frac{2d}{v}}$$

Division is multiplication by the reciprocal of the denominator, so that gives us a new acceleration formula:

$$a = \frac{v^2}{2d}$$

which we can plug into our original work expression,

$$w = (ma)\cdot d$$

Cancelling the distances (d/d = 1),

$$w = m \left( \frac{v^2}{2d} \right) \cdot d$$

gives us the total work done:

$$w = \frac{1}{2} mv^2$$

Now that work, must equal the amount of kinetic energy (KE) lost, so that gives us the familiar formula for KE:

$$\bf KE = \frac{1}{2} mv^2$$

### Units of kinetic energy

The units of kinetic energy (all kinds of energy, actually) can be determined by looking at the formula:

$$KE = \frac{1}{2} mv^2 \; \longleftarrow \frac{Kg\cdot m^2}{s^2}$$

This collection of units is given the name Joules (symbol J), after James Prescott Joule (1818-1889), an English mathematician & physicist who was an important figure in early thermodynamics.

$$1 \, \frac{Kg\cdot m^2}{s^2} = 1 \; Joule \;(J)$$

The Joule is related to the Newton (N), the unit of force. Recall that a Newton is

$$1 \; N = 1 \; \frac{Kg\cdot m}{s^2}$$

So a Joule is a Newton-meter (Newtons of force multiplied by meters of distance over which that force is exerted:

#### 1 Joule = 1 Newton-meter

$$1 \; J = 1 \; N\cdot m$$

Joules and Newtons and their sub-units (the basic units inside them) are two important units that you should commit to memory.

### KE is quadratic in velocity and linear in mass

Kinetic energy depends quadratically (to the second power) on velocity, and linearly on the mass.

You can think of it this way: Say an object like a car is moving toward you. What factors will increase the likelihood of injury, and by how much?

If we double the mass, the KE will double. That's shown in the graph below. The black curve is KE vs. velocity for a 1 Kg object. If we double the mass to 2 Kg (magenta curve), the energy is doubled. If we further double the mass to 4 Kg, the green curve is obtained. Values of KE for a velocity of 6 m/s are shown. You can see that doubling from 1 Kg to 2 Kg doubles the KE from 18 J to 36 J. Further doubling of the mass from 2 Kg to 4 Kg again doubles the KE from 36 J to 72 J. This is a linear relationship between KE and mass.

Now look at the graph in a different way. Consider only the 2 Kg curve. Notice that it's not linear; as we increase the velocity, the KE increases more rapidly. In fact, a doubling of velocity leads to a four-fold increase in KE, because the velocity is squared in the equation.

Consider these cases, in which v is some number (which we just call v), v is twice that value, and v is three times that value.

The resulting kinetic energies rise by factors of 4 and 9 respectively, the squares of the velocity increase.

Things like this are very important when we consider things like being hit by a car: If velocity is doubled, the kinetic energy rises by a factor of four. This translates to braking distance. The braking distance at 30 mi./h is quadrupled in the same car moving at 60 mi./h.

### Algebraic rearrangements of the KE formula

You should practice all of the algebraic rearrangements of the kinetic energy formula so that you can use it to solve for KE, mass or velocity if you have the other two bits of information. Here are the results:

### Energy is conserved

#### Energy can be converted from one form to another

Energy exists in many forms in nature. We measure energy by the ability to do work – to exert a force over a distance.

Kinetic energy can be transformed into potential energy – the energy of position. For example, when a roller coaster rolls down a hill, its kinetic energy increases due to the force of gravity working on it. As the car travels back up the next hill, it slows down, therefore it loses KE. That

In climbing the hill the KE of the car is used to do work against the gravitational force. It is converted into gravitational potential energy, the energy of position. On the downhill trip, that PE is converted back into KE.

Other forms of energy are also similarly conserved. For example, heat flows from hotter objects to cooler ones, but the total amount of heat (which is actually the kinetic energy of the small motions of atoms and molecules) remains the same.

Chemical energy is the energy "stored" in chemical bonds. When, in a reaction, the difference between the bond energies of the reactants and products (products minus reactants) is negative, we say that the reaction is exothermic, and it gives off heat. Sometimes reactions require heat energy from the surroundings in order to proceed; these are endothermic reactions.

In the process depicted below, the chemical energy stored in a liquid mixture is converted to kinetic energy as the liquid converts to a gas and pushes outward, moving a piston in the container upward.

### Practice problems

 1 To hit the ball farther, a baseball player may either increase bat weight or increase the bat speed (or both, of course, but there's a trade-off). Which will create more kinetic energy increase at the end of a swinging bat: Increasing the bat weight from 32 oz. to 34 oz. (1 oz = 28.3495 g) Increasing the bat head speed from 95 mi/h to 100 mi/h (1 mi/h = 0.44704 m/s) Solution First some unit conversions: \begin{align} 95 \frac{mi}{h} \left( \frac{0.4407 \, m/s}{1 \, mi/h} \right) &= 42.47 \frac{m}{s} \; \; and \\ 100 \frac{mi}{h} &= 44.07 \frac{m}{s} \\ \\ 32 \frac{mi}{h} \left( \frac{28.3495 \, g}{1 \, oz} \right) &= .907 \, Kg \; \; and \\ 34 \frac{mi}{h} &= 0.964 \, Kg \end{align} Light bat, low speed: $$KE = \frac{1}{2} 0.907 \,Kg \cdot \left( 42.47 \frac{m}{s} \right)^2 = 818 \, J$$ Heavy bat, low speed: $$KE = \frac{1}{2} 0.964 \,Kg \cdot \left( 42.47 \frac{m}{s} \right)^2 = 870 \, J$$ Light bat, high speed: $$KE = \frac{1}{2} 0.907 \,Kg \cdot \left( 44.07 \frac{m}{s} \right)^2 = 880 \, J$$ A 6.25% increase in bat weight doesn't make as much difference as a 5% increase in speed. 2 Compare the kinetic energy of a car weighing 3500 lbs., the weight of an average passenger car, and traveling at 20 mi/h (8.94 m/s), with the weight of a 150 lb. cyclist riding a 20 lb. bike at the same speed. (1 lb. = 2.2046 Kg on Earth) Solution First some unit conversions: $$3500 \, lbs \left( \frac{2.2046 \, Kg}{1 \, lb} \right) = 7716 \, Kg$$ $$170 \, lbs \left( \frac{2.2046 \, Kg}{1 \, lb} \right) = 375 \, Kg$$ By the way, why do I say "on Earth" when giving the unit conversion? The kinetic energies: \begin{align} KE_{car} &= \frac{1}{2} mv^2 = \frac{1}{2}(7716 \, Kg)\left(8.94 \frac{m}{s}\right)^2 \\ &= 308 \, KJ \\ \\ KE_{car} &= \frac{1}{2} mv^2 = \frac{1}{2}(375 \, Kg)\left(8.94 \frac{m}{s}\right)^2 \\ &= 14.9 \, KJ \end{align} The bike has less than 5% of the KE of the car, not to mention that the cyclist isn't surrounded by a cage of metal, plastic and airbags. Be careful out there! 3 A photon of light has no mass (m = 0). Calculate the momentum of a photon moving at the speed of light, 2.99792458 × 108 m·s-1. Solution Because a photon has no mass and   $KE = \frac{1}{2} mv^2$,   a photon has zero KE. However, a photon actually does have momentum, but that's a surprising result from quantum mechanics, a later subject. Nothing to worry about right now, but interesting, right?

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