To begin, a little information about units: Milliliters, liters, gallons and ounces measure (liquid) volume.

Transcription

1 6.4: Work To begin, a little information about units: You know about feet and tablespoons, meters and gallons, hours and pounds... These are all units of measurement. Some measure distance, some measure weight, some volume and some time. Some units also measure money, price or cost, but these are only "standard" in one place at a time. Units like feet, meters, miles or millimeters all measure distance. A meter is the same the world over, as is a foot. Pounds, kilograms and ounces measure weight. Milliliters, liters, gallons and ounces measure (liquid) volume. Seconds, minutes, hours and days (among others) measure time. In science, consistency and predictability are highly valued, as they make work across continents and across disciplines more manageable. For this reason, the scientific community has agreed upon a set of units that it uses consistently; that makes this set of units a "standard". The base units in this system are Measurement Unit Name Unit Symbol length meter m L mass kilogram kg M time second s T current ampere A I temperature kelvin K \theta amount (of mole mol N Dimensional Symbol

2 amount (of substance) mole mol N amount (of light) candela cd J This international system of units (or International System of Units, since it is a recognized entity unto itself) is called the "SI" system. SI is short for (the French) Système international, and we refer to its units as "SI units". To measure something very long, we can either say that it is 1,000 meters long, or we can describe it as being 1 10 U meters long. Each power of 10 that is a multiple of 3 (10 U, 10 X 10 YZ[, and so on) has a special name. "Kilo" in Greek means 1,000; "milli" is Latin for 1,000. One of these was chose to represent "multiply by 1,000", the other represents "divide by 1,000". Thus we have a kilometer which is 1,000 meters, and a millimeter Z which is of a meter. Z Any unit other than a multiple of 1 meter, 1 second or 1 mole are called derived amounts, or derived units, and you get them simply by multiplying by a power of 10. For example, for length, some of the derived units are: yoctometer 10 Y[b 1 ym micrometer 10 YX 1 μm millimeter 10 YU 1 mm kilometer 10 U 1 km zettameter 10 [Z 1 Zm We can take measurements larger or smaller than any standard unit by applying the appropriate prefix to the unit name. This table was unabashedly stolen from a Wikipedia article on the subject:

3 Source: You likely recognize some of these from what has become know as "the metric system"; it's true that the metric system has these various units in it, but the standard unit is always only ever the meter (for length), and only powers of 10 that are multiples of 3 are considered "standard" for use in the system, "hecto", "deca", "deci" and "centi" being notable Z exceptions added to make using the practicable (e.g. or 10 Y[ meters; Z the power of 10 is not a multiple of 3.) Some derived units have special names, and they are often described in terms of one another; for example one pascal is one newton per square meter (N/m [ ). A few of these are given here: Measurement Unit Name Unit Symbol angle radian rad frequency hertz Hz

4 frequency hertz Hz pressure pascal Pa work (energy) joule J power watt W temperature degree (Celsius) C magnetic flux density tesla T force (weight) newton N Now we are in a position to talk about Work: In a physics class, you will typically start a unit on force by looking at the amount of effort it takes to move an object a certain distance. We call this effort "work", and we describe it mathematically by the equation W = Fd, where W stands for work, F for force and d for distance. Force is generally measured in a unit called Newtons (for Isaac Newton), and distance can be measured either in "SI" units or "Imperial" units (e.g. pounds, inches or ounces). For simplicity, to help you learn about them, and because they are much more common in the sciences, I will use SI units in this section. For example, if we are asked, "How much work is required to move a 2.3 kg box from the floor to the surface of a 0.8 m table?" we would need to determine the amount of force required to move the box, and the distance over which that force would need to be applied. Gravity is a kind of force, and so we would need to overcome this force; weight is a kind of force as well. In fact, when you combine weight and gravity, you get mass. So mass is a kind of force. But mass isn't the only force; the speed - specifically the acceleration - with which it is being moved is also a factor - literally. Since acceleration is the second derivative of the position of an object (i.e. the position

5 is the second derivative of the position of an object (i.e. the position function s(t), we have a formula: F = m } ~, where "F" stands for force, "m" for mass (weight - but not } exactly - for the non-physicists among us), "s" represents "s(t)", the position function, and "t" is the variable for "time". Put simply, force is the relationship between the mass of an object and its acceleration (the second derivative of its position function). Force is measured (in SI units) in newtons. In physics, the word "work" means the amount of effort required to do something: to lift, push, pull, stretch, throw, etc. To determine this, we need not just the force, but the displacement (the distance over which the force needs to be applied). So we have another formula: W = Fd Work is the product of force and distance. Now we can address our original question: "How much work is required to move a 2.3 kg box from the floor to the surface of a 0.8 m table?" Acceleration due to gravity is 9.8 m/s [. This means that the velocity of a falling object will change by 9.8 meters per second (as long as it's not impeded by any kind of resistance). F = m d[ s dt [ so for this question, since "all" we have to do to lift our box is to overcome gravity: F = (2.3 kg) 9.8 m s [

6 F = kg m s [ The units here are awkward and not very friendly, which is why we use the standard SI unit "the newton" for force: 1 Š ~ } = 1N Now our force equation looks like this: F = N Much better! But that's just force; we were asked to determine the amount of work required. W = Fd So for our current situation, W = N 0.9m W = Nm - newton-meters. So work problems are very straighforward... as long as the force is constant. But that's no fun; force is seldom constant. We can still use the basic premise above over very small intervals, assuming the force is constant over each small interval, and then add up the results. In other words, as long as the force doesn't change much over 1 mm, we can just find the work done over that one millimeter, and all the other millimeters in the distance in question, and add them up. Using calculus, of course! W = lim Ž Ž f(x )Δx = f(x) dx

7 Ž Z š Example: A tank has the shape of an inverted circular cone with height 10 m and base radius 4 m. It is filled with water to a height of 8 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 Š } ). Answer: The top of the tank is our destination (like the top of the table was in our earlier example), so the first drop of water out of the tank has to travel 2 m, but the last drop has to travel 10 m. If we think of each particle of water as lying in a 1-particle thick layer, then the first layer has to move 2 meters, and the last layer 10 m. Thus, a representative layer will have to move somewhere between 2 and 10 meters, and this gives us our overall interval: [2,10]. A representative layer has an area of A = πr [, but the radius is different for each layer. Since we're talking about a representative layer, we can say that it must travel x m to the top of the tank, and that it has radius r. In other words, the ith layer has radius r and has to travel x m. A particle isn't very thick, but there is some thickness to it. We'll think of this thickness as the thickness of the layer, which causes us to be lifting, not an area, but a volume, and so we can talk about the change in the height of the water from the top of one layer to the bottom of that layer, and we'll call that our Δx. Here is a diagram of what we have discussed so far:

8 The volume of one layer is V = πr [ Δx, but what is r? Using similar triangles, as we did in related rates problems, we can find the relationship between known values for a height and radius and the height and radius of a representative layer: By similar triangles, 4 = 10 r 10 x Solving this proportion yields: 4(10 x ) = 10(r ) 4(10 x ) = r 10 r = 2 (10 x ) 5 r = 2 5 (10 x )

9 r = 5 (10 x ) Now we can say that our representative layer has volume V = π 2 [ 5 (10 x ) Δx The density of water is 1000 Š }, so the mass of this layer is m = 1000 π 2 [ 5 (10 x ) Δx or m = 160π(10 x ) [ Δx The force required to move one layer must also overcome the force of gravity; therefore F = m g [160π(10 x ) [ Δx](9.8) = 1568π(10 x ) [ Δx Since W = Fd We have W = F x = 1568π(10 x ) [ Δx x or W = 1568πx (10 x ) [ Δx To find the total work required to empty the tank, we sum all the layers, which gives us Ž

10 lim Ž Ž 1568πx (10 x ) [ Δx Z which we interpret as an integral Z_ 1568πx(10 x) [ dx [ Z_ 1568π 100x 20x [ + x U dx [ One could evaluate the integral by hand and then perform the arithmetic on a calculator, but equally, if one is going to use a calculator, one might as well do so at this point: On the TI-84 Plus, MATH 9: fnint( 100x -20x^2+x^3, X, 2, 10 ) Result: π Result: , or 3.36x10^6 J Homework: #21

11 #21 Be prepared to turn in this question - not as a HW assignment, but as part of your quiz for this unit.