Chapter 7: Work
7.1 Systems and Environments: system, system boundary, environment 7.2 Work Done by a Constant Force: work, Newton meter, joule, Work is an energy transfer 7.3 The Scalar Product of Two Vectors: scalar product, dot product, distributive law of multiplication 7.4 Work Done by a Varying Force : spring force, force constant, spring constant, Hooke's Law 7.5 Kinetic Energy and the Work-Kinetic Energy Theorem: kinetic energy, work-kinetic energy theorem						7.6 The Non-Isolated System-Conservation of Energy: nonisolated system, isolated system, internal energy, work, mechanical waves, heat, matter transfer, electrical transmission, electromagnetic radiation, conservation of energy 7.7 Situations Involving Kinetic Friction 7.8 Power 7.9 Energy and the Automobile
7.1 Systems and Environments Motion
Our book uses two systems to solve problems… · The individual object is defined as a system and how it’s influenced by the environment · All object involved in the movement, etc, is defined as a system. All examples I present are of the latter type…all objects involved are defined as the system.
7.2 Work Done by a Constant Force
WORK The work done on a system by an external agent exerting a constant force is the product of the magnitude of the force and the magnitude of the component of the displacement vector in the direction of the force or Work = F•Δr Work = FΔr cosθ
During lecture I raised a 0.1 kg eraser through a vertical displacement of meter. (A)   How much work did "I" do on the eraser? (B)   How much work was done on the eraser? (C)   How much work was done on the eraser When lifting at a 45° angle through 1 meter?				(A-answer) "I" did 1 N-m of Work on the eraser. (B-ans) NO work was done on the eraser. Don't forget gravity!!!, which was equal and opposite (+1 (me) + -1(gravity) = 0 Nm) (C-ans) “I” do work in the vertical direction, but gravity does work in the opposite direction for a total of 0 Nm in the veritical. No force opposed the horizontal component. So NO work is done.
This is a good time to include my use of units. I have said repeatedly that I used the unit Nm instead of Joule most of the time in 131 lecture and lab. It isn't until late in the 131 lecture course do I start exclusively using Joules instead of Newton meters....why?
Loading a Truck It's SUMMER!!!, also time for a summer job You have an intern at an engineering firm, but your boss is in an bind. Your boss needs to get a truck loaded ASAP and no one is around, so you volunteer. You are loading the truck. Just as the truck is loaded another engineer comes out and says, “You loading it in the WRONG truck.” The other engineer instructs you to move the load to the truck in the next bay. You start unloading the original truck. One half hour later your boss come out just as you finish unloading the truck and says, "What's have you been doing? You haven't done any work!!!" You lower your head and say "I give up..." and sigh.				Is your boss correct? Of course he is...isn't he always? Actually Work is the force applied through the total displacement What was your displacement? ZERO...so he was correct...you did NO WORK!!! Work is the force applied (or component of) to an object through the distance the object is displaced Units: Newton-meters. The unit describes the physical action
Example: Work with Constant Force A throw the ball with displacement of r = (5i + 6j) meters in a storm. While in flight the wind exerts a constant Force of F = (10i + 8j) N on the object during the duration of the balls path. How much work was done upon the ball? Work = F•Δr = (10i•5i + 10i•6j + 8j•5i + 8j•6j) Nm = Work = F•Δr = (50 + 0 + 0 + 48) Nm = 98 Nm Please NOTE that Force caused by a wind is WORK. Air friction (or any type of friction) is a non-conservative force. All non-conservative forces reduce the Kinetic Energy of the system. Wind actually can cause a sailboat to move. Friction causes things to reduce their speed or NOT to move.
7.3 The Scalar Product of Two Vectors
The Scalar Product of Two Vectors The dot product as pertains to work Work = F•r = F r cos θ Also the unit vectors i, j, and k are perpendicular to each other thus i•i = j•j = k•k = 1 and i•j = j•k = i•k = 0 This second statement dealing with orthogonal components follows the same logic for the x-dir and y-dir components for ballistic objects. (The vertical component allows you to determine the time in flight, and the horizontal component (with time) yields the range of the object. This phrase is the same in kinematics.)
The physical meaning of the Dot Product: F•r The projection of vector r on vector F. Commonly used in 3D programming for collision detection and path finding. If the projection of vector r is less than the magnitude of F, then the object at displacement r is within vector F ’s influence and must be considered.
7.4 Work Done by a Varying Force
Example: Work with Varying Force The Mar's rovers, Spirit and Opportunity, traveled to Mars from Earth. Earth is about 1.5 x 10^11 m from the sun while Mars distance from the sun is about 2.3 x 10^11 m from the sun. Kepler's 2nd law is simplified to F = 1.3 x 10^11/ r² by supplying "G", the mass of the Sun and Rover, and applying the sign for direction. How much work was done on the rovers by the sun?			Kepler analyzed Tycho Brahe's data and discovered that the force between two objects is F = G Mm/r², thus inversely proportional to square of the distance. Work = F ∫dr Work = 1.3 x 10^11∫dr/ r² from xinit = 1.5 x 10^11 m to xfinal = 2.3 x 10^11 m Work = -1.3 x 10^11 (1/r) from xinit = 1.5 x 10^11 m to xfinal = 2.3 x 10^11 m Work = -1.3 x 10^11 (1 / 2.3x10^11 – 1 / 1.5x10^11) Work = -3 x 10^10 Nm
From previous lecture and demonstrations we know that force applied to return a spring to its equilibrium position is directly proportional and opposite to the displacement, thus calling it a Restoring Force. F = -kx As above if the force varies during the path we must integrate or Work = ∫F•dx Work = -k ∫x dx Work = ½kxi² - ½kxf²						Example Work with Varying Force due to a spring If 1 Nm of work is required to stretch the spring from equilibrium to 10 cm how much more work is required to bring the spring back to 20 cm? If 1 Nm of work is applied to the spring, the spring exerts equal but opposite force through a distance of -1 Nm. Work = 1 Nm Work = ½kxi² - ½kxf² - 1 = 0 - ½k*0.1² k = 200 N/m
Simple Harmonic Motion Lab has been performed by all which allows the spring constant, k to be calculated. In lab you found it to be about 10 N/m.							“k” measures the stiffness of the spring. The stiffer the spring, the greater value of “k”.
7.5 Kinetic Energy and the Work-Kinetic Energy Theorem
Work - Kinetic Energy Relationship Work is force applied (or component of) on an object displaced through a distance. Work = F•x and if varies Work = ∫Fdx Newton's 2nd law ΣF = m a So Work = ∫m a dx                 = m ∫ (dv/dt) dx				At this point we need to apply the chain rule...                 = m ∫ (dv/dx) (dx/dt) dx The definition of instantaneous velocity is dx/dt thus we re-write the equation                 = m ∫ (dx/dt) (dv/dx) dx                 = m ∫ v dv Work = ½mvf² - ½mvi²
Work = Δ Kinetic Energy (the energy of motion) In the case in which work is done on a system and the only change in the system is in its speed, the work done by the force equals the change in kinetic energy of the system.
7.6 The Non-Isolated System-Conservation of Energy
Energy transferred out of the system by means other than the object moving. Raise of internal energy, sound waves, air transferring internal energy frictional heat from the system					Power plants (moving turbine) transferring mechanical energy into electrical energy. Electromagnetic energy (light transfer) etc
7.7 Situations Involving Kinetic Friction
What role does friction play in the Work - KE Relationship? Friction ALWAYS opposes motion, thus friction will reduce KE. ΔKE is equal to the opposing Frictionkinetic through xtotal distance plus ΣWorkother forces ΔKE = -Ffkd + ΣWorkother forces Note Friction does NOT involve displacement, but the total distance the object moves.
Don't confuse the action (foot attempting to go backwards) and the reaction (whole body moving forwards). Newton's Third law (Action & Reaction)
If you walk what's the reaction?	Reaction Your foot takes a step forward, exerts a force down and backwards. Friction is equal and opposite to the horizontal (backwards) component and your foot is not allowed to go backwards (the motion). What's the reaction? Your move forwards. Friction was in the forward horizontal direction, the reaction is your body moves forwards.
If you push on a wall, what's the reaction?	Reaction The wall pushes back with an equal and opposite force
If you stand on the floor, what's the reaction?	Reaction The floor pushes back with an equal and opposite force.
Airplane movement Action The propeller forces air through the turbine housing.	Reaction The air on the front side of the housing was at rest. Now the air is being propelled at a great velocity behind the housing in the backward direction. Thus the reaction is a force on the propeller in the forward direction which mounted to the airplane, and the entire plane moves forward.
7.8 Power
Look at a clock (preferably with a second hand). Do you think it takes energy to move the second hand? · So it’s using energy all the time · And it’s moving every second		The clock at the front of the room is plugged into the wall (auto updates every day). Where does the plug eventually lead? Ans: The power plant. So the clock uses power!!! So that means that Power = ΔKE / time (energy all the time) Or P = Work / time
Example: If a 200 kg piano is carried up to the second floor in 1 minute…approximately what is the total energy required and the combined power output by the piano movers during this time?		We’ll estimate that the piano is raised about 3 meters. We know the piano weighs 2000 N (FW = mg = 200kg 10 m/s^2).
		So the work is Fapplied d Work = 2000 N * 3 m Total energy is 6000 Nm							Power = Work / time P = 6000 Joules / 60 seconds Power = 100 Watts
7.9 Energy and the Automobile
The internal chemical energy one gallon of gasoline is 1.34 x 10^8 Joules. Let’s pretend the largest production SUV is 100% efficient (the average automobile is less than 10% efficient). You are going to the top of Mt Ranier, which his 14,410 feet. How much gas will be used by this 8000 lb vehicle? Just for fun Gallon of diesel is 1.55 x 10^8 Joules (16% more) Gallon of propane is 0.91 x 10^8 Joules (32% less)								8000lb (4.5 N/lb) = 36,000N 14410 ft (1 m / 3.3 ft) = 4400 m Work = F d Work = 36000 * 4400 Work = 1.6 x 10^8 Joules 1.6x 0^8 J / 1.34x10^8 J/gal Gas used = 1.2 gallons
Now let’s pretend this same SUV is a gas electric hybrid and the same trek will be done solely using the electric engine. If the solar insolation rate for SoCal is 7 kW-hr/m^2/day and the 3 m^2 roof of this SUV is lined with solar panels of 15% efficiency. How many days are required to charge the batteries of this SUV to make this same trip? 1 kW-hr = 1000 J/s*hr (1000 J/s / kW) (3600 s / 1day) = 3.6 x 10^6 Joules 7 kW-hr/m^2/day (3.6 x 10^6 J/ 1 kW-hr) * 3 m^2 = 0.72x10^8 J /day 0.72x10^8 J /day * x = 1.6 x 10^8 Joules x = 2.2 days So we now know 1.2 Gallons of gas is equivalent to 2.2 days charging in the sun for 3 m^2 of solar panels
Intro to Chapter 8: Potential Energy
What role does Potential Energy play in the Work - KE Relationship? First off, Potential Energy usually implies Gravitational Potential Energy in a system. This is the amount of energy that could be delivered (or transferred) at the bottom of its path due to the work by gravity. (Once again...gravity acts towards the center of the earth thus only the vertical component of the path is considered.) Referring back to the Eraser Example, the 0.1 kg eraser was held 1 meter above the table in the classroom then subsequently displaced horizontally about 1 meter so to be directly 2 meters above the floor. (A)   While above the table, what is the PE of the eraser? (B)   While above the floor, what is the PE of the eraser? (C)   How much PE will the eraser have immediately before striking the ground (D)   Where will the energy have gone? (E)   How fast will the eraser be traveling before striking the ground? (A-ans) The eraser weighs 1 N and while above the table it has the potential to travel through 1 meter before striking the table, thus the Eraser has 1 N-m of PE. (B-ans) Likewise while 2 meters above the floor the Eraser has 2 N-m of PE. (C-ans) Immediately before striking the floor, the height above the surface is 0 m...thus the Potential energy is 0 N-m (D-ans) All of the Potential energy has all transferred to motion or Kinetic energy due to the work exerted by gravity (neglecting non-conservative forces like air friction.) (E-ans) This can be solved from two different approaches TE = KE + PE At the top of its path: vinit = 0;   KEinit = 0;   PE = mghmax TE = 0 + 2 Nm     where TE remains constant (barring non-conservative forces) TE = 2Nm and at hmin PE = 0 thus KE = 2 Nm = ½mvf² - ½mvi² vf = (2*2/0.1)^½ = 6.3 m/s or Work = ΔKE ... where work is the work done by gravity Work by gravity = mg(hmax- hmin) = Δ ½mv² As you can see...these are essentially identical methods both dependant upon gravity. Our Book uses T = K + U, but usually I'll use TE, KE and PE respectively. In our book T represents Tension, Period, and Total energy, which sometimes becomes confusing to beginning Physics students. During lecture we already used kinematic equations to calculate velocity of an eraser dropped from 2 m. Did we arrive at the same value?
Example: Pendulum Bowling Ball Today we are attempting knock me over with a swinging bowling ball. First let's determine how fast the ball will be going at the bottom of its path. The ΔPE is the change of height from top position to the bottom position.				We also know that the TE = KE + PE (T = K + U) TE = KEtop + PEtop TE = 0 + mghmax = 45N * 1 m TE = 45 Nm = KEbottom + PEbottom = KEmax + 0 TE = 45 Nm = KEmax = ½mv² = ½4.5kg v² v = 4.5 m/s or about 10 mph. Above I'm saying the vertical displacement is 1 meter. So if I place my face on the other side so that my nose is at the same height, I should be safe. Let's try it!
PE isn't always Gravitational PE!!! How can a spring (in a pin ball machine) store energy and release it later to do work on an object? This energy is called Elastic Potential energy (PEs or Us) Us = ½kx² Below isn't a spring, but is this elastic PE?