We're almost through with our whirlwind introduction to physics. Armed with what we've talked about so far (including today's lecture), we'll be able to start considering sound from a physics perspective. In fact, in our next meeting we'll begin talking about motion, and the forces that cause it, that leads to the production of a pure tone in sound. Before we get there, we have one more very important preliminary. That is the introduction of an abstract concept called energy. Energy appears in many forms: motion, light, sound, etc. Energy is not something we can measure directly; instead, we observe consequences when there is a change in energy from one form to another. Perhaps the best way to begin is to give the punchline before getting into details?
Got it? I suspect not; so lets go through things a little bit more slowly. We'll talk about
- what's work?
- what do we mean by kinetic and potential energy?
- why is any of this useful?
Let's consider all the forces acting on a box that we push on, causing it to accelerate across the floor. We'll assume that the floor is very smooth, so that we can ignore friction. We'll represent the forces acting by arrows, and also show the acceleration of the box as an arrow.
What forces act on the box?
- "W" is the weight of the box. It is the force of gravity on the box, and equal to Mg, where M is the mass of the box and g is the acceleration due to gravity.
- "N" is a force on the box arising from the contact with the floor.
- "F" is the constant force we apply to the box.
The acceleration of the box is represented by the arrow labeled "a". The box speeds up in the direction that we push on it. Two of the forces acting, "W" and "N" do not cause the box to either slow down nor speed up since they are perpendicular (make a 90 degree angle) to the motion. These two forces balance each other.
In the above situation, the force "F" does work on the box, because it tends to cause the box to speed up. The forces "N" and "W" do no work on the box. If the box moves by a distance d in the direction of its acceleration vector, the work done on the box by the constant force "F" is
Work done = Fd
Forces that don't act to slow down or speed up an object do no work. Forces that act to slow an object down do negative work and forces that act to speed an object up do positive work. We say that the total work done on the box (equal to the work done by the force "F") changes the kinetic energy of the box. We should briefly say something about the units of work (and energy). Since the SI units of force are Newtons, and the units of distance are meters, the units of work are N.m We introduce a new unit of work (and energy)
1 Joule (J) = 1 N.m
Another concept we'll find useful is the rate at which work is done; this is called power. Power is the rate of change of energy with time. The SI units of power are Joules/second. Once again, we introduce another new unit for power
1 Watt (W) = 1 Joule/second
We need to define what we mean by kinetic energy (KE); as the name implies it has to do with motion. As with all other forms of energy, we can't actually measure the kinetic energy of an object. Instead, we can measure consequences of the change in kinetic energy of an object. Such a change results from total work done on an object! For an object (say our box from the above example) with mass M, moving with speed v,
kinetic energy = 1/2 Mv2
We can now write a compact mathematical equation expressing our statement that a change in kinetic energy results from the total work done on an object.
total Work = Fnetd = delta_KE = KEfinal - KEinitial
The total work is the work done by the net (or total) force acting on some object over a distance d. It's not too surprising that an object can slow down or speed up (change it's kinetic energy) when acted upon by a net force since Fnet = Ma. All we're really doing is finding a different way of expressing Newton's 2nd law of motion. Why does this help?
Before we answer that, we need to introduce another form of energy that's important when analyzing motion: potential energy. If we ignore the effects of air resistance, if we threw a 1-kg ball straight upwards with an initial speed of vi = 20 m/s, it will go up to a maximum height h at which point it is instantaneously at rest, vf = 0. The only force acting on the ball is its weight (Mg). From the information given, we can find the total work done on the ball by gravity and the change in kinetic energy of the ball:
total Work = Force times distance = -(Mg) (h) = 0 - 1/2 Mv2 = -200 J
What happens from that point on? We know that the ball falls back down to our hand. When it reaches our hand, it has exactly the same speed as it did when we threw it upwards, now just going in the other direction! Gravity has now done positive work on the ball, causing it to speed up on its downward trip. Somehow, the work done by gravity on the ball on its upward trip has been recovered! Whenever we have forces that do recoverable work, we associate a potential energy with them. In the case of gravity, when an object of mass M is raised by a height h above a reference point, we say that the object's potential energy has been increased by an amount Mgh. There are other forces that do recoverable work for which we associate a potential energy. One such example is the work done on an object by a spring. We'll talk more about that on Wednesday.
This takes us back to our starting point, when talking about the motion of a ball being thrown upward, a force (gravity) does work on the ball causing its energy to change from kinetic to potential energy. At its lowest point, the energy of the ball is purely kinetic; at its highest point, the energy of the ball is purely potential. In between the energy is split between kinetic and potential energy.
Let's deal with the question of utility of these concepts by doing some examples. As you'll see, one way physicists deal with a given situation is to analyze what energy changes are taking place.
Example 1 A 1.0-kg ball is thrown straight upwards with an initial velocity of 20 m/s. What is the maximum height that the ball reaches?
Initially, if we choose the point the ball is thrown as our reference point, then we can say that the ball has only kinetic energy (KE). At the maximum height, the ball is neither going up nor coming down, it is instantaneously at rest. This means that its kinetic energy is zero. The energy has been transformed into gravitational potential energy (PE)!
Initial KE of ball = 1/2 Mv2 = 1/2 (1.0 kg)(20 m/s)2 = 200 J
KE of ball at maximum height = 0
Initial PE of ball = 0 (this is our point of reference)
PE of ball at maximum height = Mgh
Conservation of energy implies that the ball's initial KE was transformed into PE, or
Mgh - 0 = delta_KE = 200 J, or
h = 200 J / Mg = 200 J / 9.8 N = 20 m
There's a couple of things to note here
when evaluating the expressions, all quantities must be expressed in SI units! Otherwise, we'll get the wrong answer. You can check your answer by replacing 1 J = 1 N.m = 1 kg.m2/s2.
nowhere did we talk about forces and motion in this analysis. We simply looked at what happened to the ball's energy!
Example 2 Let's use our concepts of energy to consider a 1,000-kg car, starting from resting, speeding up at a constant rate to its final speed of 20 m/s.
We know the following
Initial KE of the car = 0 , because the car is initially at rest.
Final KE of the car = 1/2 Mv2 = 1/2 (1000 kg)(20 m/s)2 = 4 x 105 J
- work done on car = 0.4 MJ (note the use of the power-of-ten prefix, mega-)
Power = work done on car / time interval = 0.4 MJ / 2 second = 0.2 MW
recall that 1 horsepower = 746 W, so this car must have at least a 270 horsepower engine!
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Last updated: 8 Sep 1999