Chapter 7 Energy and Energy Transfer

Introduction to Energy energy The concept of energy is one of the most important topics in science energy energy transfers transformations Every physical process that occurs in the Universe involves energy and energy transfers or transformations Energy is not easily defined!! Energy is not easily defined!!

Energy Approach to Problems The energy approach to describing motion is particularly useful when the force is not constant Conservation of Energy An approach will involve Conservation of Energy This could be extended to biological organisms, technological systems and engineering situations

7.1 Systems and Environments system A system is a small portion of the Universe We will ignore the details of the rest of the Universe to identify A critical skill is to identify the system valid system A valid system may be a single object or particle be a collection of objects or particles be a region of space vary in size and shape

Problem Solving Does the problem require the system approach? What is the particular system and what is its nature? Can the problem be solved by the particle approach? The particle approach is what we have been using to this time

Environment system boundary There is a system boundary around the system The boundary is an imaginary surface It does not necessarily correspond to a physical boundary environment The boundary divides the system from the environment The environment is the rest of the Universe

7.2 Work Done by a Constant Force work, W magnitude, F,the magnitude  r cos  The work, W, done on a system by an agent exerting a constant force on the system is the product of the magnitude, F, of the force, the magnitude  r of the displacement of the point of application of the force, and cos  where  is the angle between the force and the displacement vectors

Work, cont. W = F  r cos  (7.1) W = F  r cos  (7.1) The displacement is that of the point of application of the force does no work A force does no work on the object if the force does not move through a displacement The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application Work is a scalar quantity Work is a scalar quantity unitjoule (J) The unit of work is a joule (J) 1 joule = 1 NEWTON  1 meter 1 joule = 1 NEWTON  1 meter J = N · m J = N · m

Work Example normal force, n gravitational force, mg,do no work The normal force, n, and the gravitational force, mg, do no work on the object cos  = cos 90° = 0 Fdoes do The force F does do work on the object

Example 7.1 Conceptual Example F θ WF If the magnitude of F is held constant but the angle θ is increased, What happened with the work W done by F?DECREASES!!! Decreases 0 < θ < 90 o Since: cosθ Decreases when 0 < θ < 90 o ΔrΔr

Can exert a force & do no work! Can exert a force & do no work! Example: v Example: Walking at constant v with a grocery bag. W = F  r cosθ  r = 0, F ≠ 0 Could have  r = 0, F ≠ 0 W = 0  W = 0 F   r Could have F   r θ = 90º, cosθ = 0  θ = 90º, cosθ = 0 W = 0  W = 0 Example 7.2 Conceptual Example ΔrΔrΔrΔr

More About Work systemenvironment The system and the environment must be determined when dealing with work does work on The environment does work on the system by on Work by the environment on the system sign F  r The sign of the work depends on the direction of F relative to  r positiveprojectionFDr same direction Work is positive when projection of F onto Dr is in the same direction as the displacement negativ opposite direction Work is negative when the projection is in the opposite direction

More About Work, 2 Cause of the Displacement Cause of the Displacement not A force is not necessarily the cause of the object’s displacement. Example: Lifting an object. by F G F G upward! Example: Lifting an object. Work is done by Gravitational Force (F G ), although F G is not the cause of the object moving upward!

More About Work, final Displacement with Frictionless Displacement with Frictionless n mg An object is displaced on a frictionless horizontal surface, the normal force n and the gravitational force mg do no work on the object. Example: F P on Example: In the situation shown here, the force applied F P is the only force doing work on the object.

Work Is An Energy Transfer This is important for a system approach to solving a problem workpositive is transferred to If the work is done on a system and it is positive, energy is transferred to the system worknegative is transferred from If the work done on the system is negative, energy is transferred from the system interacts transfer of energy If a system interacts with its environment, this interaction can be described as a transfer of energy across the system boundary changeamount of energy stored This will result in a change in the amount of energy stored in the system

Example 7.3 Work Done ON a Box Given:m = 50 kg, F P = 100N, F fr = 50N, Δr = 40.0m Given: m = 50 kg, F P = 100N, F fr = 50N, Δr = 40.0m Find:BY Find: (a). Work done BY each force. ON (b). Net work Done ON the box.

Example 7.3 Work Done ON a Box, 2 ONW = F  r cosθ For each force ON the box: W = F  r cosθ(a) W G = mgΔrcos90 o = mg(40m)(0) = 0 W n = nΔrcos90 o = n(40m)(0) = 0 W P = F P Δrcos37 o = (100N)(40m)(cos37 o ) = 3200J W fr = F fr Δrcos180 o = (50N(40m)(–1)= – 2000J (b) 1. W net = W G + W n + W P + W fr = 1200J 2. W net = (F net ) x Δr = (F P cos37 o – F fr ) Δr  W net = (100Ncos37 o – 50N)40m = 1200J

Active Figure 7.4

Example 7.4 Work Done ON a Backpack net work ON Find The net work done ON the backpack. h = dcosθ From (a): h = dcosθ ΣFy = 0  From (b): ΣFy = 0  F H = mg F H = mg From (c): BYthe hiker: Work done BY the hiker: W H = F H dcosθ  W H = mgh

Example 7.4 Work Done ON a Backpack, 2 From (c): BYgravity: Work done BY gravity: W G = mg dcos(180 – θ) W G = mg d(–cosθ) = – mg dcosθ  W G =– mgh NET WORK on the backpack: NET WORK on the backpack: W net = W H + W G = mgh – mgh W net = 0

F G BYON centripetal acceleration F G exerted BY the Earth ON the Moon acts toward the Earth and provides its centripetal acceleration inward along the radius orbit F G   r F G   r (Tangent to the circle & parallel with v) The angle θ = 90 o  W E-M = F G  r cos 90 o = 0 This is why the Moon, as well as artificial satellites, can stay in orbit without expenditure of FUEL!!! This is why the Moon, as well as artificial satellites, can stay in orbit without expenditure of FUEL!!! Example 7.5 Conceptual Example: Work BY the Earth ON the Moon.

7.3 Scalar Product of Two Vectors A. B The scalar product of two vectors is written as A. B dot product It is also called the dot product A. B = A B cos  (7.2)  betweenA B  is the angle between A and B The scalar product is commutative A. B = B. A A. B = B. A The scalar product obeys the distributive law of multiplication A. (B + C) = A. B + A. C A. (B + C) = A. B + A. C

Dot Products of Unit Vectors (7.4) & (7.5) A B Using component form with A and B:(7.6) (Problem #6) (Problem #6)

Example 7.6 Work Done by a Constant Force Example 7.6 Work Done by a Constant Force (Example 7.3 Text Book) Given:  r = (2.0 î ĵ) m F = (5.0 î ĵ) N Calculate the following magnitudes:  r = (4 + 9) ½ = (13) ½ = 3.6 m  r = (4 + 9) ½ = (13) ½ = 3.6 m F = (25 + 4) ½ = (29) ½ = 5.4 N F = (25 + 4) ½ = (29) ½ = 5.4 N F: Calculate the Work done by F: W = F Δr =[(5.0 î ĵ) N][(2.0 î ĵ) m] W = F Δr = [(5.0 î ĵ) N][(2.0 î ĵ) m] (5.0 î 2.0 î î 3.0 ĵ ĵ 2.0 î ĵ 3.0 ĵ) N m = (5.0 î 2.0 î î 3.0 ĵ ĵ 2.0 î ĵ 3.0 ĵ) N m = [ ] J = 16 J = [ ] J = 16 J

Examples to Read!!! Examples to Read!!! Example 7.5 Example 7.5 (page 189) Example 7.6 Example 7.6 (page 193) Homework to be solved in Class!!! Homework to be solved in Class!!! Questions: 1 Questions: 1 Problems: 1, 6 (Done: See Eqn 7.6) Problems: 1, 6 (Done: See Eqn 7.6) Material for the Final