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Mass training of trainers General Physics 1

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1 Mass training of trainers General Physics 1
May 20-22, 2017 The Law Conservation of Mechanical Energy & Power Prof. MARLON FLORES SACEDON Physics Instructor Visayas State University Baybay City, Leyte

2 Objectives To study and calculate WorkDone by the system.
To study and calculate Potential Energy of the system. To study and calculate Kinetic Energy by the system. To derive and apply the Work-Energy theorem. To derive and apply The Law of Conservation of Mechanical Energy. To study and calculate Average Power.

3 Conservation of Energy Energy is conserve.
The total energy for every point of travel are the same. p.1 𝐸 1 p.2 p.3 𝐸 2 𝐸 3 p.4 𝐸 4 𝐸 1 = 𝐸 2 = 𝐸 3 = 𝐸 4

4 What is Work? In physics and engineering, work is accomplished only when a force acts on a body and this force is able to move the body.

5 What is Work? In physics and engineering, work is accomplished only when a force acts on a body and this force is able to move the body. Work Done is defined is either of two ways: 1. It is the product of the force and the component of the displacement in the direction of the force. 2. The product of the displacement and the component of the force in the direction of the displacement. 𝑦 π‘₯ 𝑦 π‘₯ 𝐹 m πœƒ 𝐹 𝑠 m πœƒ 𝑠 Def’n 1: π‘Š=(𝐹)(𝑠 π‘π‘œπ‘ πœƒ) Def’n 2: π‘Š=(s)(πΉπ‘π‘œπ‘ πœƒ) π‘Š=πΉπ‘ π‘π‘œπ‘ πœƒ π‘Š= 𝐹 βˆ™ 𝑠 Note: Work (π‘Š) is scalar physical quantity.

6 Units of Work What is Work?
In physics and engineering, work is accomplished only when a force acts on a body and this force is able to move the body. Work Done is defined is either of two ways: 1. It is the product of the force and the component of the displacement in the direction of the force. 2. The product of the displacement and the component of the force in the direction of the displacement. Units of Work James Prescott Joule FRS (24 December 1818 – 11 October 1889) was an English physicist and brewer, born in Salford, Lancashire. Joule studied the nature of heat, and discovered its relationship to mechanical work

7 What is Work? In physics and engineering, work is accomplished only when a force acts on a body and this force is able to move the body. Work Done is defined is either of two ways: 1. It is the product of the force and the component of the displacement in the direction of the force. 2. The product of the displacement and the component of the force in the direction of the displacement. p.1 p.2 Work is done on the object. m 𝐹 𝑾= 𝑭 βˆ™ 𝒔 𝑾=π‘­π’”π’„π’π’”πœ½ 𝑠 𝑾=π‘­π’”π’„π’π’”πŸŽ 𝑾=𝑭𝒔 𝐹 m πœƒ p.1 p.2 Work is done on the object. 𝑾=π‘­π’”π’„π’π’”πœ½ If: πœƒ= 30 π‘œ 𝑠=20π‘š Calculate workdone by the force π‘Š 𝐹 = 100𝑁 20π‘š π‘π‘œπ‘  30 π‘œ 𝑠 π‘Š 𝐹 =1, π‘π‘š or π‘Š 𝐹 =1, 𝐽

8 𝑾=π‘­π’”π’„π’π’”πœ½ 𝑠 Calculate Work done on the object If: P=100𝑁 @ πœƒ= 30 π‘œ
𝑦 π‘₯ 𝑦 π‘₯ Calculate Work done on the object If: πœƒ= 30 π‘œ 𝑠=20π‘š, πœ‡ π‘˜ =0.15 𝑾=π‘­π’”π’„π’π’”πœ½ 𝑁 𝑃 m=40kg πœƒ 𝑓 𝑀 𝑠 πœ‡ π‘˜ =0.15 Step 5: Calculate Work done on the object by summing up all work done by forces acting on the object. Ξ£ 𝐹 𝑦 =𝑁+ 𝑃 𝑦 βˆ’π‘šπ‘”=0 Step 2: Calculate Work done by gravity 𝑁=π‘šπ‘”βˆ’ 𝑃 𝑦 =π‘šπ‘”βˆ’π‘ƒπ‘ π‘–π‘›πœƒ 𝑁= 40π‘˜π‘” π‘š 𝑠 2 βˆ’100𝑠𝑖𝑛30 =342.4𝑁 π‘Š 𝑔 =π‘šπ‘”π‘ π‘π‘œπ‘ πœƒ π‘Š 𝑔 = 40π‘˜π‘” π‘š 𝑠 π‘š π‘π‘œπ‘  270 π‘œ =0 π‘Š π‘œπ‘π‘— = π‘Š 𝑃 + π‘Š 𝑔 + π‘Š 𝑁 + π‘Š 𝑓 𝑓= 𝑁 =51.36𝑁 Step 3: Calculate Work done by Normal force N π‘Š π‘œπ‘π‘— =1, βˆ’1,027.2 π‘Š π‘œπ‘π‘— = 𝐽 π‘Š π‘œπ‘π‘— = π‘Š 𝑃 + π‘Š 𝑔 + π‘Š 𝑁 + π‘Š 𝑓 π‘Š 𝑁 =π‘π‘ π‘π‘œπ‘ πœƒ Another way to calculate Work done on the object by π‘Š π‘œπ‘π‘— =Ξ£ 𝐹 𝑛𝑒𝑑 βˆ™π‘  π‘Š 𝑁 =342.4𝑁 20π‘š π‘π‘œπ‘  90 π‘œ =0 Step 1: Calculate Work done by pulling force 𝑃 Step 4: Calculate Work done by friction force 𝑓 Ξ£ 𝐹 𝑛𝑒𝑑 =π‘ƒπ‘π‘œπ‘ πœƒβˆ’π‘“ π‘Š 𝑃 =π‘ƒπ‘ π‘π‘œπ‘ πœƒ Ξ£ 𝐹 𝑛𝑒𝑑 =100π‘π‘œπ‘ 30βˆ’51.36=35.25N π‘Š 𝑃 = 100𝑁 20π‘š π‘π‘œπ‘  30 π‘œ π‘Š 𝑓 =π‘“π‘ π‘π‘œπ‘ πœƒ π‘Š π‘œπ‘π‘— =(35.25N)(20π‘š) π‘Š 𝑃 =1, 𝐽 π‘Š 𝑓 =51.36𝑁 20π‘š π‘π‘œπ‘  180 π‘œ =βˆ’1,027.2𝐽 π‘Š π‘œπ‘π‘— = 𝐽

9 What is Energy? Energy (𝐸) is often associated with work.
When work is done on the body, a change is produced in the body. 1. Change in its motion or inertia 2. Change in position 3. Change in temperature (work to overcome friction) The change produced in the body by the application of work is often times called energy so that energy may be thought of as the ability or capacity to do work. An agent is said to possess energy if it is also to do work. Energy being the maximum work, then the units of energy is the same as the units of work. There are many forms of energy: Mechanical, Chemical, Electrical, Heat or Thermal, Electromagnetic, Nuclear, Solar, etc. Our daily observations make us realize that transformations occur from one form of energy to another.

10 Forms of mechanical energy:
1. Potential Energy (π‘ˆ) – energy at rest 2. Kinetic Energy (𝐾) - energy in motion Kinds of P-E: 1. Gravitational ( π‘ˆ 𝑔 ) - energy possessed due to its position or elevation. 2. Elastic ( π‘ˆ 𝑒𝑙 ) - due to its state of strain or configuration/ 3. Chemical ( π‘ˆ π‘β„Žπ‘’π‘š ) - due to its composition

11 m Gravitational Potential Energy Calculating work done by gravity
π‘Š=πΉπ‘ π‘π‘œπ‘ πœƒ Def’n. of Work done 𝑀=π‘šπ‘” p.1 𝑦 1 π‘Š 𝑔 =π‘šπ‘”β„Žπ‘π‘œπ‘ πœƒ β„Ž=βˆ’( 𝑦 2 βˆ’ 𝑦 1 ) π‘Š 𝑔 =π‘šπ‘” βˆ’ 𝑦 2 βˆ’ 𝑦 1 cos0 π‘Š 𝑔 =βˆ’π‘šπ‘” 𝑦 2 βˆ’ 𝑦 1 (1) π‘Š 𝑔 =βˆ’ π‘šπ‘” 𝑦 2 βˆ’π‘šπ‘” 𝑦 1 p.2 𝑦 2 Def’n of Grav’l potential energy π‘ˆ 𝑔 =π‘šπ‘”π‘¦ π‘Š 𝑔 =βˆ’ π‘ˆ 2 βˆ’ π‘ˆ 1 𝑦 π‘₯ π‘œ π‘Š 𝑔 =βˆ’βˆ†π‘ˆ Therefore, workdone by gravity is equal to the negative change in grav’l potential energy.

12 Elastic potential Energy
𝐹=π‘˜π‘₯ Where: F = applied force [N, dyne, poundal] x = stretched/ compressed length [m, cm, ft] k = spring constant [N/m, dyne/cm, poundal/ft Definition of elastic potential energy. π‘ˆ 𝑒𝑙 = 1 2 π‘˜ π‘₯ 2 Work done by elastic π‘Š 𝑒𝑙 =βˆ’ βˆ†π‘ˆ 𝑒𝑙 =βˆ’ π‘ˆ 𝑒𝑙 2 βˆ’ π‘ˆ 𝑒𝑙 1 =βˆ’ π‘˜ π‘₯ 2 2 βˆ’ π‘˜ π‘₯ 1 2

13 Kinetic Energy (𝐾) 𝑦 π‘₯ π‘œ 𝑦 π‘₯ π‘œ Suppose the net external force 𝐹=500𝑁, displacement 𝑠=20π‘š, & moving mass π‘š=15π‘˜π‘”. What is the final velocity of moving object if it starts from rest.? 𝑣 1 𝑣 2 m 𝐹 𝑠 π‘Š=βˆ†πΎ π‘Š=πΉπ‘ π‘π‘œπ‘ πœƒ Def’n. of Work done Substitute Eq.2 in Eq.1 π‘Š=π‘š 𝑣 2 2 βˆ’ 𝑣 π‘Š= 𝐾 2 βˆ’ 𝐾 1 π‘Š=πΉπ‘ π‘π‘œπ‘ 0 π‘Š=𝐹𝑠 π‘Š= 1 2 π‘š 𝑣 2 2 βˆ’ 1 2 π‘š 𝑣 1 2 π‘Š= 1 2 π‘š 𝑣 2 2 βˆ’ 1 2 π‘š 𝑣 1 2 But: 𝐹=π‘šπ‘Ž 500𝑁 20π‘š = π‘˜π‘” 𝑣 2 2 βˆ’0 π‘Š=π‘šπ‘Žπ‘  Eq.1 Def’n of Kinetic energy: 𝐾= 1 2 π‘š 𝑣 2 From Kinematics: 𝑣 𝑓 2 = 𝑣 𝑖 2 +2π‘Žπ‘  𝑣 2 = 2(500𝑁) 20π‘š 15π‘˜π‘” =36.51 π‘š 𝑠 𝑣 2 2 = 𝑣 π‘Žπ‘  π‘Š= 𝐾 2 βˆ’ 𝐾 1 This is Work-Energy Theorem π‘Žπ‘ = 𝑣 2 2 βˆ’ 𝑣 π‘Š=βˆ†πΎ Eq.2

14 For Gravitational Forces only For the effect of other Forces
π‘Š π‘‘π‘œπ‘‘ =π‘Š 𝑔 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ π‘Š 𝑔 =βˆ’βˆ†π‘ˆ Eq.1 Work Done by gravity π‘Š=βˆ†πΎ Work-Energy Theorem π‘Š 𝑔 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = π‘Š π‘‘π‘œπ‘‘ For gravitational forces only βˆ’βˆ†π‘ˆ+ π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ =βˆ†πΎ π‘Š 𝑔 =βˆ†πΎ Eq.2 βˆ’ π‘ˆ 2 βˆ’ π‘ˆ 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = 𝐾 2 βˆ’ 𝐾 1 m Substitute Eq.2 in Eq.1 βˆ’ π‘ˆ 2 + π‘ˆ 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = 𝐾 2 βˆ’ 𝐾 1 βˆ†πΎ=βˆ’βˆ†π‘ˆ 𝐾 1 + π‘ˆ 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = 𝐾 2 + π‘ˆ 2 𝐾 2 βˆ’ 𝐾 1 =βˆ’ π‘ˆ 2 βˆ’ π‘ˆ 1 1 2 π‘š 𝑣 π‘šπ‘” 𝑦 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = 1 2 π‘š 𝑣 π‘šπ‘” 𝑦 2 𝐾 2 βˆ’ 𝐾 1 =βˆ’ π‘ˆ 2 + π‘ˆ 1 𝐾 1 + π‘ˆ 1 = 𝐾 2 + π‘ˆ 2 The law of conservation of mechanical energy But: 𝐸=𝐾+π‘ˆ 𝐸 1 = 1 2 π‘š 𝑣 π‘šπ‘” 𝑦 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ 𝑬 𝟏 = 𝑬 𝟐 𝐸 2 = 1 2 π‘š 𝑣 π‘šπ‘” 𝑦 2 This equation means that total energy 𝐸 is constant through out the travel. Therefore the total initial energy is equal to the total final energy, hence total energy is constant. 𝑬 𝟏 = 𝑬 𝟐

15 Calculate the final velocity of the box.
1 2 π‘š 𝑣 π‘šπ‘” 𝑦 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = 1 2 π‘š 𝑣 π‘šπ‘” 𝑦 2 Application problem 𝑣 1 =0 30kg Assume the bottom of the plane be our ref. point, hence 𝑦=0 at this point. πœƒ= 40 π‘œ 𝐿=20π‘š πœ‡ π‘˜ =0.20 0+π‘šπ‘” 𝑦 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = 1 2 π‘š 𝑣 Work Done by the other π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ is only friction 0+π‘šπ‘” 𝑦 1 +(π‘“πΏπ‘π‘œπ‘  180 π‘œ )= 1 2 π‘š 𝑣 𝑣 2 =? 0+π‘šπ‘” 𝑦 1 βˆ’π‘“πΏ= 1 2 π‘š 𝑣 0+π‘šπ‘” 𝑦 1 βˆ’ πœ‡ π‘˜ π‘šπ‘”π‘π‘œπ‘ πœƒβˆ™πΏ= 1 2 π‘š 𝑣 Calculate the final velocity of the box. 0+π‘šπ‘” πΏπ‘ π‘–π‘›πœƒ βˆ’ πœ‡ π‘˜ π‘šπ‘”π‘π‘œπ‘ πœƒβˆ™πΏ= 1 2 π‘š 𝑣 𝑬 𝟏 = 𝑬 𝟐 𝑔 πΏπ‘ π‘–π‘›πœƒ βˆ’ πœ‡ π‘˜ π‘”π‘π‘œπ‘ πœƒβˆ™πΏ= 𝑣 2 2 𝐾 1 + π‘ˆ 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = 𝐾 2 + π‘ˆ 2 𝑣 2 = 2𝑔𝐿 π‘ π‘–π‘›πœƒβˆ’ πœ‡ π‘˜ π‘π‘œπ‘ πœƒ 𝑣 2 =13.86 π‘š 𝑠

16 Summary Work (π‘Š) can be done if their is force applied to the object and able to move the object. The product of the displacement and the component of the force in the direction of the displacement. π‘Š=πΉπ‘ π‘π‘œπ‘ πœƒ The unit of Work: Joule (𝐽), erg, foot-poundal Work is scalar physical quantity. π‘Š= 𝐹 βˆ™ 𝑠 Energy (𝐸) is the ability to do work, hence Energy have the same unit with Work. Two kinds of mechanical energy: The Potential (π‘ˆ) and Kinetic (𝐾) Energy. Definition of gravitational potential energy. π‘ˆ=π‘šπ‘”y Definition of elastic potential energy. π‘ˆ 𝑒𝑙 = 1 2 π‘˜ π‘₯ 2 Definition of kinetic energy. 𝐾= 1 2 π‘š 𝑣 2 Work done by gravity π‘Š 𝑔 =βˆ’βˆ†π‘ˆ=βˆ’ π‘ˆ 2 βˆ’ π‘ˆ 1 =βˆ’ π‘šπ‘” 𝑦 2 βˆ’π‘šπ‘” 𝑦 1 Work done by elastic π‘Š 𝑒𝑙 =βˆ’ βˆ†π‘ˆ 𝑒𝑙 =βˆ’ π‘ˆ 𝑒𝑙 2 βˆ’ π‘ˆ 𝑒𝑙 1 =βˆ’ π‘˜ π‘₯ 2 2 βˆ’ π‘˜ π‘₯ 1 2 Work-Energy theorem π‘Š=βˆ†πΎ The total energy at any point of travel is constant 𝐸 1 = 𝐸 2 = 𝐸 3 = 𝐸 𝑛 … The law of conservation of mechanical energy π‘š 𝑣 π‘šπ‘” 𝑦 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = 1 2 π‘š 𝑣 π‘šπ‘” 𝑦 2

17 PROBLEM

18 ASSIGNMENT #1

19 ASSIGNMENT #2

20 ASSIGNMENT #3 A 2.00kg package is released on a 53.1 incline, 4.00 m from a long spring with force constant 1.20x102 N/m that is attached at the bottom of the incline (Fig). The coefficients of friction between the package and incline are Β΅s = and Β΅k = The mass of the spring is negligible. (a) What is the maximum compression of the spring? (b) The package rebounds up the incline. How close does it get to its original position? (c) What is the change in the internal energy of the package and incline from the point at which the package is released until it rebounds to its maximum height?

21 Power What is Power? The rate of doing work.
It is the amount of energy consumed per unit time. Having no direction, it is a scalar quantity. In SI system, the unit of power is the joule per second (J/s), known as the watt in honor of James Watt, the eighteenth-century developer of the steam engine. 𝑃 π‘Žπ‘£ = π‘Š βˆ†π‘‘ Where: 𝑃 π‘Žπ‘£ = average power, in (watt) or in 𝐽 𝑠 π‘Š = quantity of work or work done, in (𝐽) βˆ†π‘‘ = time interval, in (𝑠) 𝑃 π‘Žπ‘£ = πΉβˆ†π‘  βˆ†π‘‘ 𝑃 π‘Žπ‘£ =πΉβˆ™ βˆ†π‘  βˆ†π‘‘ 1 𝐻𝑝=746 π‘€π‘Žπ‘‘π‘‘π‘  1 𝐻𝑝=550 π‘“π‘‘βˆ’π‘™π‘ 𝑠 = 2545 Btu/h 𝑃 π‘Žπ‘£ =πΉβˆ™π‘£ Note: Horse power (𝐻𝑝) is another unit of power

22 Power Problem: A 50kg marathon runner runs up the stairs to the top of of Chicago’s 443-m-tall Willis Tower, the tallest building in the United States (Figure, right). To lift herself to the top in 15 minutes, what must be her average power output? Express your answer in watts, in kilowatts, and in horse power. π‘Š=π‘šπ‘”β„Ž=50π‘˜π‘” π‘š 𝑠 π‘š = 217, 𝐽 Solutions: Another method: Solve the velocity, then multiply by the required force. Time to reach the top is 15 min β„Ž=443π‘š βˆ†π‘‘=15π‘šπ‘–π‘› 60 𝑠 1 π‘šπ‘–π‘› =900 𝑠 𝑣= βˆ†π‘  βˆ†π‘‘ = 443 π‘š 900 𝑠 = π‘š/𝑠 𝑃 π‘Žπ‘£ = 217, 𝐽 900 𝑠 = π‘Š 𝑃 π‘Žπ‘£ = π‘Š 1π‘˜π‘Š 1000 π‘€π‘Žπ‘‘π‘‘π‘  =0.241 π‘˜π‘Š 𝑃 π‘Žπ‘£ =πΉβˆ™π‘£=π‘šπ‘”βˆ™π‘£ 𝑃 π‘Žπ‘£ =50π‘˜π‘”(9.81 π‘š 𝑠 2 )( π‘š 𝑠 ) = π‘Š 𝑀=π‘šπ‘” 𝑃 π‘Žπ‘£ = π‘Š 1 𝐻𝑝 746π‘Š =0.324 𝐻𝑝

23 Power Assignment 1: A 20 Hp engine is used to lift gravel from the ground to the top of a building 60 ft high. Neglecting loss of energy due to friction how many tons of gravel can be lifted in 50 minutes. Assignment 2: What weights can a 6 Hp engine pulls along a level road at 15 mi/hr if the coefficient of friction between the weight and the road is 0.2? Assignment 3: A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is What average power is produced by friction as the rock stops? Assignment 4: When its 75-kW (100-hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

24 Problems 1. A force of 8 𝑙𝑏𝑠. pulls a body along a horizontal surface to a distance of 10 𝑓𝑑. a) How much work is done, b) If the force acts at an angle of 30o above the horizontal, how much work is done? 2. A 100𝑔 object is dragged with a uniform velocity along a plane inclined 30o with the horizontal by a force parallel to the inclined. If the coefficient friction between the object and the plane is 0.20, how much work is done when the object is moved a distance of 40 π‘π‘š along the plane? 3. A man weighing 120 𝑙𝑏𝑠, climb up a stairway inclined 45π‘œ consisting of 20 𝑠𝑑𝑒𝑝𝑠 each step 6" high. What is his potential energy at the top. 4. A body a mass 10 𝑠𝑙𝑒𝑔𝑠 is thrown with a velocity of 6 𝑓𝑑/𝑠. along a horizontal floor. The coefficient of friction between the body and the floor 𝑖𝑠 Find a) Kinetics Energy and velocity of the body after travelling a distance of 2 𝑓𝑑. b) How far will the body travel before it comes to rest.

25 Problems 5. A body weighing 64 𝑙𝑏 slides down from rest at the top of a plane 18 𝑓𝑑 long and inclined 30 π‘œ above the horizontal. The coefficient of friction is Find the velocity of the body as it reaches the bottom of the plane. 6. What weights can a 6 𝐻𝑝 engine pulls along a level road at 15 π‘šπ‘–/β„Žπ‘Ÿ if the coefficient of friction between the weight and the road is 0.20?

26 Summary Work (π‘Š) can be done if their is force applied to the object and able to move the object. The product of the displacement and the component of the force in the direction of the displacement. π‘Š=πΉπ‘ π‘π‘œπ‘ πœƒ The unit of Work: Joule (𝐽), erg, foot-poundal Work is scalar physical quantity. π‘Š= 𝐹 βˆ™ 𝑠 Energy (𝐸) is the ability to do work, hence Energy have the same unit with Work. Two kinds of mechanical energy: The Potential (π‘ˆ) and Kinetic (𝐾) Energy. Definition of gravitational potential energy. π‘ˆ 𝑔 =π‘šπ‘”y Definition of kinetic energy. 𝐾= 1 2 π‘š 𝑣 2 Work done by gravity π‘Š 𝑔 =βˆ’βˆ†π‘ˆ Work-Energy theorem π‘Š=βˆ†πΎ The total energy at any point of travel is constant 𝐸 1 = 𝐸 2 = 𝐸 3 = 𝐸 𝑛 … The law of conservation of mechanical energy π‘š 𝑣 π‘šπ‘” 𝑦 1 + π‘Š π‘œπ‘‘β„Žπ‘’π‘Ÿ = 1 2 π‘š 𝑣 π‘šπ‘” 𝑦 2 Average power: The rate of doing work. It is the amount of energy consumed per unit time. 𝑃 π‘Žπ‘£ = βˆ†π‘Š βˆ†π‘‘

27 eNd


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