Chapter 7Energy 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: cosθ Decreases when 0 < θ < 90o

Δr

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

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

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

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

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

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

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

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

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