1. Chapter 7 Energy of a SystemEXAMPLES

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

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

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

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

6. 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  Δr (Tangent to the circle & parallel with v) The angle θ = 90 o  W E-M = F G Δr cos 90 o = 0 This is why the Moon, as well as artificial satellites, can stay in orbit without expenditure of FUEL!!! Example 7.5 Work BY the Earth ON the Moon

7. Example 7.6 Work Done by a Constant Force (Example 7.3 Text Book) Given: Dr = (2.0 î + 3.0 ĵ) m F = (5.0 î + 2.0 ĵ) N Calculate the following magnitudes: Δ r = (4 + 9) ½ = (13) ½ = 3.6 m F = (25 + 4) ½ = (29) ½ = 5.4 N Calculate the Work done by F: W = F Δr = [(5.0 î + 2.0 ĵ) N][(2.0 î + 3.0 ĵ) m] = (5.0 î 2.0 î + 5.0 î 3.0 ĵ + 2.0 ĵ 2.0 î + 2.0 ĵ 3.0 ĵ) N m = [10 + 0 + 0 + 6] J = 16 J

8. Example 7.7 Net Work Done from a Graph (Example 7.4 Text Book) The Net Work done by this force is the area under the curve W = Area under the Curve W = A R + A T W = (B)(h) + (B)(h)/2 = (4m)(5N) + (2m)(5N)/2 W = 20J + 5J = 25 J

9. Example 7.8 Work-Kinetic Energy Theorem (Example 7.6 Text Book) m = 6.0kg first at rest is pulled to the right with a force F = 12N (frictionless). Find v after m moves 3.0m Solution: The normal and gravitational forces do no work since they are perpendicular to the direction of the displacement W = F Δ x = (12)(3)J = 36J W = Δ K = ½ mv f 2 – 0  36J = ½(6.0kg)v f 2 = (3kg)v f 2 V f =(36J/3kg) ½ = 3.5m/s

10. Example 7.9 Work to Stop a Car W net = Fdcos180°= –Fd = –Fd W net =  K = ½mv 2 2 – ½mv 1 2 = –Fd  – Fd = 0 – ½m v 1 2  d  v 1 2 If the car’s initial speed doubled, the stopping distance is 4 times greater. Then: d = 80 m

11. Examples to Read!!! Example 7.5 (page 175) Example 7.7 (page 179) Material from the book to Study!!! Objective Questions: 7-11-14 Conceptual Questions: 2-5-8 Problems: 1-6-9-14-15-21-30-33-42-44 Material for the Final Exam