Presentation on theme: "General Physics I: Day 14 Work & Kinetic Energy Exams back Thursday… sorry. Blame viruses (in general)."— Presentation transcript:
General Physics I: Day 14 Work & Kinetic Energy Exams back Thursday… sorry. Blame viruses (in general).
Newton’s Laws - Complex & Powerful Newton’s work has occupied us for a while Using these we can predict the motion of nearly any object… as long as we know all the forces on it! But reality quickly becomes very hard to handle with Newton’s laws: Collision of bowling ball & pins Interactions of atoms in a gas Etc… 2
Conservation Laws Another approach: Look for what is not changing A quantity that doesn’t change even while interesting things are happening is called a conserved quantity. At the top of the list is… The Law of the Conservation of Energy Energy is never created or destroyed! 3
Warm-Up: Weight Lifting Work A weight lifter picks up a barbell and 1. lifts it chest high 2. holds it for 30 seconds 3. puts it down. How does the work in each part rank? ~12% → W1 > W3 > W2 ~44% →W1 = W3 > W2 ~31% →W1 > W2 > W3 ~8% →W2 > W1 > W3 ~6% →W3 > W2 > W1 4 Yes… if ignoring sign Correct (including sign) Blending phys/intuition
Work – 1 st Try Not the same as the daily definition of “work”. Need a technical definition to define energy later. Work done: Product of the distance moved times the part of the force parallel to the motion: Also called the dot product, or scalar product Remember that this translates to But… real forces don’t have steady mag. & dir. 5
Work – 2 nd Try Consider a real force, like a “bench press”. How do we calculate work here? We must account for each little bit of work in which the force is basically constant (could be small). Thanks to calculus, we can: … Better! But this assumes the force is always in the direction of motion. 6
Work – 3 rd Try Need an integral in which both relative direction and magnitude can vary! Any real examples? The math here is a bit tougher, but we can write it easily This won’t be necessary very often, usually direction is fixed and only the strength varies. 7 from HyperPhysics. Photo of shuttle launch (taken by James Prichard for the Palm Beach Post)
Energy Energy is not easy to define... Energy is the ability to do work. Specifically, the amount of energy an object has is equal to how much work that object could possibly do. Forms of energy: 9 Kinetic energy (translational & rotational) Gravitational energy Elastic energy Thermal energy (a.k.a. internal energy) Electromagnetic energy Chemical energy Nuclear energy, Rest energy Some energies: “potential energy”, symbol U. Later!
Kinetic Energy Kinetic energy (and all energy) is a scalar! We can calculate kinetic energy using Kinetic energy is never negative (v 2 ). See the units? [kg·m 2 /s 2 ] is called a Joule. 15
Work-Energy Theorem Work is the transfer of energy from one system to another by mechanical forces. Any change in the kinetic energy must be the result of a net force doing work: This connects between the concepts of force and the concepts of energy! 16
WarmUp: Estimating Work “a. Bugatti Veyron Super Sport 10.9m/s^2 b.1500 N*M” “a) Bugatti Veyron 16.4 Super Sport goes from 0-60 in 2.4 s. The cars acceleration is (26.8 m/s)/2.4s = 11.67 m/s/s. b)W = (1888kg)(11.67m/s/s)(64.32m) W = 1356037.12 J” “a) 1990 ford mustang gt 166.4m/s^2 b)553839J” “07 Chevy Silverado. 3.67[m.s^2].5(2540.12kg)(26.8^2) =91.22kJ” 18
WarmUp: Estimating Work “1990 Dacia 1310 (0 to 100km/hr (27.8m/s) in 15 sec, m=940 kg) a =.54 m/s^2 work = 1/2*940*.54^2 = 137 J” “A.I looked up an Enzo Ferrari which had a 0- 60[mph] time of 3.14[s] B. To find the total work I used the work energy theorem equation. knowing the initial velocity is 0 then k1=0 as well. I was left with W=K2=.5*mv^2. Looking up the mass I found the mass of the car is 1255[kg]. Plugging these values in I found the total work to be 450696[J]” 19
WarmUp: Estimating Work “a) The Tesla Roadster Sport electric car goes 0-60 mph in 3.7 seconds, which gives an acceleration of 7.28 m/s^2 b) If the car's mass is 1225 [kg] and acceleration is 7.28 [m/s^2], then ignoring air resistance and static friction, Force= 7.28 * 1225= 8920 [N]. Displacement= (1/2)(7.28[m/s^2])(3.7 [s])^2= 50 [m]. Work= 8920[N] * 50[m]= 446 [kj]” 20
Sample Problem The only force on a 2.0 kg canister that is moving on a frictionless table has a magnitude of 5.0 N. The canister initially has a velocity of 4.0 m/s in the positive x direction, and some time later has a velocity of 6.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time? 21
A certain amount of work is required to get a car moving at 10 mph. To then get the car from 10 mph to 20 mph (not from 0 to 20 mph) takes A)the same amount of work. B)twice as much work. C)three times as much work. D)four times as much work. 22
Coming up… Thursday (10/9) → 6.4 – 7.1 Warm-Up due Wednesday by 10:00 PM Homework 6 due Sunday by 11:59 PM Exams back on Thursday Come pick up Journal Part II if you don’t have one 23
Your consent to our cookies if you continue to use this website.