Work, Energy, and Power H.W. Ch. 5 Read and Write for the whole chapter. ½ page per section All problems are due. Odds may be done separately and copied from me in class.

Work Underlined words are WOD When I do a pushup, I move. When Chuck Norris does a push up, what happens?

Work Underlined words are WOD When Chuck Norris does a push up, what happens? The earth moves. What happens when my son Luke does a push up? Real answer: Use Newtons 3 rd law.

Work Underlined words are WOD Luke-a-meter: The distance the earth moves when my 100 lb son does a push up.

Work Demo Have a student walk across the front of the room, carrying the 10 kg brick, at a constant speed. Did they do work?

Work Work = Force ∙ displacement or W = F d cos Θ (I have memorized “Work is Force times distance.”) Guess, what are the units? Note: Dot Product => mult by cos of angle between Force and displacement. So W = F d cos Θ Which means forces only do work in the direction of motion.

Work Work = Force ∙ displacement Units = N m = Joules

Work Demo Remember the moles we shot into space? We applied a 50N force to them over a distance of 2 cm. Work on Mole = F x d = 50N x.02m x cos 0 = 1 Joule Have a student walk across the front of the room, carrying the 10 kg brick, at a constant speed. How much work did they do? Guesses in Joules???

Work Work = Force ∙ displacement Units = N m = Joules Force and displacement are both vectors. How do you multiply vectors?

The Dot Product One way of multiplying vectors is “the dot product”. If A & B are vectors, A ∙ B = |A| * |B|* cos [Θ] Θ = the angle between A and B |A| is the length of A, |B| is the length of B The result of the dot product is a SCALAR, not a vector. Order of the vectors is important. Θ Goes from 2 nd to 1 st vector.

The Dot Product A ∙ B = |A| * |B|* cos [Θ] If B is at 30 degrees and A is at 70 degrees, Then Θ is angle from 30 to 70 = 40 deg Θ can be negative. Order of the vectors is important. Θ Goes from 2 nd to 1 st vector: from Displacement to Force. 30 Deg 70 Deg

Dot Product examples A = 0, B = 0 A ∙ B = __________ C = 30, D = 60 C ∙ D = __________ E = 90, F = 135 E ∙ F = __________ G = 120, H = 60 G ∙ H = __________ I = 180, J = 90 I ∙ J = __________ K = 100, L = 235 K ∙ L = __________ M = 45, N = 225 M ∙ N = __________

Dot Product examples A = 0, B = 0 A ∙ B = ___12_____ C = 30, D = 60 C ∙ D =___10.4____ E = 90, F = 135E ∙ F = ___8.5____ G = 120, H = 60 G ∙ H = ____6_____ I = 180, J = 90 I ∙ J = ____0_____ K = 100, L = 235 K ∙ L = __-8.5 ____ M = 45, N = 225 M ∙ N = _-12______

Work Work = Force ∙ displacement Units = N m = Joules So, if Force & Displacement are – In the same direction, work is + – In the opposite direction, work is - – Perpendicular, work is 0 Memorize this table: Force: SameOpposite90deg angle Disp.: Work:Positive Negative Zero, Zed, Zilch Nada, Nothing, Zippo, Donut Hole

Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m.

Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m. Remember: Force is a vector F Fx = F cos θ Motion

Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m.

Work (example 1) a 1.02 kg rock is in free fall. It falls for 5 meters. How much work does gravity do on the rock?

Work (example 1) a 1.02 kg rock is in free fall. It falls for 5 meters. How much work does gravity do on the rock? Why use 1.02kg instead of 1.00 kg??? (seems like a weird number?)

Work (example 1) a 1.02 kg rock is in free fall. It falls for 5 meters. How much work does gravity do on the rock? Why use 1.02kg? (Answer: Makes for a round number when multiplied by gravity.)

Work Example: a 1.02 kg rock is in free fall. It falls for 5 meters. How much work does gravity do on the rock? W = F*∆y = m*g*∆y = 1.02kg*9.80m/s 2 *5m = 50 Nm = 50 Joules Gravity is downward, so is ∆y, so work is positive. Also, make sure the units are correct.

Work Example: a 1.02 kg rock is in free fall. It falls for 5 meters. So how much work do you have to do on the rock to lift it up the two flights of stairs to 5m height? W = F*∆y = m*g*∆y = 1.02kg*9.80m/s 2 *5m = 50 Nm = 50 Joules Same. What comes out is what goes in!!!

Work What if there are multiple forces acting on an object? We can find the work done by each force, and we can find the total (net) work. Step 1: Find the work each force does. Step 2: Add those works together.

Work A 1.02 kg rock is accelerated upwards by a string for a displacement of 30 meters. The tension in the string is 20 N. What is the work done by each force? What is the was the net force during that 30 m? How many forces? 2 Tension in string Gravity

Work A 1.02 kg rock is accelerated upwards by a string. The tension in the string is 20 N. It travels 30 m up. 1.) What is the work done by tension? 2.) What is the work done by gravity during that same 30 m? 3.) What is the net work?

Work A 1.02 kg rock is accelerated upwards by a string. The tension in the string is 20 N. What is the work done by the string's Tension to move the rock up 30 meters? Work T = F d =T*∆y cos [angle between them] = 20 N * 30 m * cos(0) = 600 J

Work A 1.02 kg rock is accelerated upwards by a string. The tension in the string is 20 N. What is the work done by gravity during that same 30 m? Work g = mg* ∆y * cos (180) = 10N*30m * (-1) = -300 Joules

Work A 1.02 kg rock is accelerated upwards by a string. The tension in the string is 20 N. What is the total amount of work done during that 30 m?

Work A 1.02 kg rock is accelerated upwards by a string. The tension in the string is 20 N. What is the total amount of work done during that 30 m? W net = W T + W g W net = 600 J – 300 J = 300 J

Work A 1.02 kg rock is accelerated upwards by a string. The tension in the string is 20 N. Bonus question: What is the was the net force during that 30 m?

Work A 1.02 kg rock is accelerated upwards by a string. The tension in the string is 20 N. Bonus question: What is the was the net force during that 30 m? W net = F net * ∆y F net = W net /∆y = 300J / 30m = 10 N

Example 3 Accelerating a Crate The truck is accelerating at a rate of m/s 2. The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it?

The angle between the displacement and the normal force is 90 degrees. The angle between the displacement and the weight is 270 degrees. Another way to think of this is:

Calculate work in the direction of force. In other words, because there is no movement in the y direction, there is no work done by “y forces”.

The angle between the displacement and the applied force is 0 degrees. ΣF = ma So net force is just from friction.

Work Demo Revisited. Have a student walk across the front of the room, carrying the 10 kg brick, at a constant speed. How much work did they do? Any new Guesses???

What is work, really? Conceptually, what is work?

What is work, really? Conceptually, what is work? Work is the change of energy. From one form to another or one object to another. Add the following to your work def: WORK: The amount of energy added to an object.

Types of Energy There are two types of energy that we're going to be talking about in Mechanics. Energy can easily be changed between these two forms. MEMORIZE these equations: Kinetic Energy: Energy of motion. KE = ½ mv 2 Potential Energy(Gravitational): Energy of position. PE = mgh (Sometimes called GPE instead of PE) Together, they are called Mechanical Energy: ME = KE + PE Talk about 1. other types of PE. 2. mechanical engineers and what they do.

Lets look at units. Add Units to WOD. KE = ½ mv 2 kg * (m/s) 2 (kg *m/s 2 )*m N * m Add: “Units are Joules.” to WOD. PE = mgh kg *m/s 2 * m N *m Both equal Joules

Work Demo Revisited. Have a student walk across the front of the room, carrying the 10 kg brick, at a constant speed. How much work did they do? Any new Guesses???

Work Demo Revisited. Have a student walk across the front of the room, carrying the 10 kg brick, at a constant speed. How much work did they do? What was the ME of the brick before moving? After moving? What is the equation for the change in ME of the brick? This is the amount of work done to it.

Work Demo Revisited. Why do we talk about ME? Add to WOD: Work can also be defined as the change in ME of an object. Work is in Joules. Energy is in Joules. Work can be thought of as Work is also “adding energy” to an object. Most often, the equation used is: Work = Change in KE and we ignore PE. Add to WOD: W = ΔKE

Work Demo Revisited. Have a student walk across the front of the room, carrying the 10 kg brick, at a constant speed. How much work did they do? What is the equation for the change in KE of the brick? Change in KE = KE 2 – KE 1 = (½ mv 2 2 – ½ mv 1 2 ) but v 2 = v 1 so the change in KE = 0. So no work is done because no kinetic energy is added to the package. (But, don’t tell the UPS man he is doing no work if you want to get your packages.)

Work Demo Revisited. Have a student walk across the front of the room, carrying the 10 kg brick, at a constant speed. How much work did they do? What is the equation for the change in PE of the brick? Change in PE = PE 2 – PE 1 = (mgh 2 – mgh 1 ) But h 2 = h 1, so no PE is added or change in PE = 0. So no work is done because no potential energy is added to the package. (Again, don’t tell the UPS man he is doing no work if you want to get your packages.)

Work Demo Revisited. Have a student walk across the front of the room, carrying the 10 kg brick, at a constant speed. How much work did they do? What about Work = Force. Displacement. Again, the angle between them is 90 degrees. The cos 90 is 0 and Work = F x D Cos 90 = 0 So no work is done because the force and displacement are 90 degrees to each other. (STILL, don’t tell the UPS man he is doing no work if you want to get your packages.)

Work Demo Revisited. Question: I lift the 10 kg steel brick 1 foot. I walk forward at a constant speed and quickly hand it to a student. The students carries the brick around the track outside for 5 miles at the same constant speed. Who does the most work? Me or the student? Answer: Who added energy to the brick?

Work Demo Revisited. Revisit the red box motion. Think this time in terms of Work as being a change in Kinetic energy. Me Pushing the red box (on ice) adds Kinetic Energy, when speeding it up. Removes Kinetic Energy when slowing it down. Now add friction and discuss.

2 Important Ideas Law of conservation of energy - 1.) Energy cannot be created or destroyed, but it can change form. 2.) The total amount of energy in the universe is constant. (Amount does not change.) Mechanical Energy (KE + PE) is not always constant. Things like friction can dissipate it, usually into heat. Discuss E = mC^2

Work Energy Theorem Work(net) = ∆K.E. = K.E. f – K.E. i = ½ m v f 2 - ½ mv i 2 Does not MEAN ∆ KE = ½ m (v f - v i ) 2 Because (v f - v i ) 2 Does not equal v f 2 - v i 2 Right is: W = ½ m (v f 2 - v i 2 )

Work Energy Theorem Work “done against a non-friction force” = Potential Energy.”

Work Energy Theorem Work “done against a non-friction force” = Potential Energy.” Recall “Work = Force ∙ displacement” Well, m*g = force height = displacement (vertical) So, work “done against gravity” = mgh