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Chapter 7Energy of a System EXAMPLES
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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 < 90o
Δr
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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 = 0Could have F Δr θ = 90º, cosθ = 0
W = 0
Example 7.2 Conceptual Example
Δr
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If: m = 50 kg, FP = 100N, Ffr = 50N, Δr = 40.0mFind: (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)
1. Wnet = WG+ Wn + WP+ Wfr = 1200J
2. Wnet = (Fnet)x Δr = (FPcos37o – Ffr) Δr Wnet = (100Ncos37o – 50N)40m = 1200J
Example 7.3 Work Done ON a Box
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Find The Net Work done ON the backpack.
From (a): h = dcosθ From (b): ΣFy = 0 FH = mg From (c):
Work done BY the hiker:WH = FH dcosθ WH = mgh
From (c): Work done BY 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
Example 7.4 Work Done ON a Backpack
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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 = FG Δr cos 90o = 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
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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
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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
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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 = 36JW = ΔK = ½ mvf
2 – 0 36J = ½(6.0kg)vf
2 = (3kg)vf2
Vf =(36J/3kg)½ = 3.5m/s
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Example 7.9 Work to Stop a Car
Wnet = Fdcos180°= –Fd = –Fd
Wnet = K = ½mv22 – ½mv1
2 = –Fd
– Fd = 0 – ½m v12 d v1
2
If the car’s initial speed doubled, the stopping distance is 4
times greater. Then: d = 80 m
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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