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Chapter 7 Energy of a System EXAMPLES

Chapter 7 Energy of a System EXAMPLES. 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: cos θ Decreases when 0 < θ < 90 o. Δ r. Example 7.2 Conceptual Example.

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Chapter 7 Energy of a System EXAMPLES

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  1. Chapter 7Energy of a System EXAMPLES

  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: cosθDecreases when 0 < θ < 90o Δr

  3. Example 7.2 Conceptual Example • 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 Δr

  4. Example 7.3 Work Done ON a Box If: m = 50 kg, FP = 100N, Ffr = 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 Δr cosθ (a) WG = mgΔrcos90o = mg(40m)(0) = 0 Wn = nΔrcos90o = n(40m)(0) = 0 WP = FPΔrcos37o = (100N)(40m)(cos37o) = 3200J Wfr= FfrΔrcos180o = (50N(40m)(–1)= = – 2000J (b) • Wnet= WG+Wn + WP+ Wfr= 1200J • Wnet=(Fnet)xΔr = (FPcos37o – Ffr) Δr  Wnet= (100Ncos37o – 50N)40m= 1200J

  5. Example 7.4 Work Done ON a Backpack • Find The Net Work done ONthe backpack. • From (a): h = dcosθ • From (b): ΣFy = 0  FH = mg • From (c): • Work doneBY the hiker: WH=FHdcosθWH=mgh • From (c): • Work doneBY gravity: WG = mg dcos(180 – θ) WG = mg d(–cosθ) = – mg dcosθ WG =–mgh • NET WORK on the backpack: Wnet = WH+ WG = mgh – mgh Wnet= 0

  6. Example 7.5 Work BY the Earth ON the Moon • FG exerted BY the Earth ON the Moon acts toward the Earth and provides its centripetal acceleration inward along the radius orbit • FG  Δr (Tangent to the circle & parallel with v) The angle θ = 90o WE-M = FGΔr cos 90o = 0 • This is why the Moon, as well as artificial satellites, can stay in orbit without expenditure of FUEL!!!

  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: • Δr= (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 = AR + AT 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 Δx =(12)(3)J = 36J W = ΔK = ½ mvf2 – 0 36J = ½(6.0kg)vf2 = (3kg)vf2 Vf =(36J/3kg)½ = 3.5m/s

  10. Example 7.9 Work to Stop a Car • Wnet = Fdcos180°= –Fd= –Fd Wnet = K= ½mv22 – ½mv12 = –Fd  – Fd = 0 – ½m v12  dv12 • If the car’s initial speed doubled, the stopping distance is 4 times greater. Then: d = 80 m

  11. Material for the Final Exam • 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

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