1 / 9

Lesson 6-4

Lesson 6-4. Work. Quiz. Homework Problem: Reading questions:. Objectives. Determine the amount of work done in constant force, variable force and spring problems. Vocabulary. Indefinite Integral – is a function or a family of functions

julian-mcintyre
Télécharger la présentation

Lesson 6-4

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lesson 6-4 Work

  2. Quiz • Homework Problem: • Reading questions:

  3. Objectives • Determine the amount of work done in constant force, variable force and spring problems

  4. Vocabulary • Indefinite Integral – is a function or a family of functions • Distance – the total distance traveled by an object between two points in time • Displacement – the net change in position between two points in time

  5. Work In Physics, Work is a force times a distance. If the force is constant, then the problem is just algebra; however, if the force is variable based on the distance, then the problem falls into calculus. where f(x) is the variable force and x is the distance. Work = force • distance = 150 lb • 20 ft = 3000 ft-lbs x = b Work = ∫ f(x) dx x = a

  6. 1 6 x = 6 ∫ F(x) dx = Work ∫ (1/x²) dx = -(1/x) | = (-1/6) – (-1) = 5/6 x = 1 x = 6 x = 1 6-4 Example 1 Note: Gravity is really a 1/x² type of force (not -32 ft/sec²) (we treat it like a constant because of the distances involved)

  7. 6-4 Example 2 3 to 6 0 to 3 750 natural length = 15 3 6 9 12 15 ∫ (250d) dd = 125d²| = 125(36 – 9) = 3375 pounds Step 1 (find k) kd = 750 pounds  k = 250 Work (Hooke’s Law) d = 6 d = 6 Step 2 d =3 d =3 force • (distance increment)

  8. 6-4 Example 3 Tank weight = density • volume d = 5 distance yth layer moves = 5 + (16 – y)= (21 – y) r = 8 disk vol ½ full oil r = x ∆y x Area • thickness πx² • ∆y ∆F = weight = density • volume = 50 lb/ft³ • πx²∆y (x – h)² + (y – k)² = r² x² + (y – 8)² = 64 x² = 64 - (y – 8)² x² = 16y - y² Need to eliminate variable with secondary equation y = 8 y = 8 50π∫ (y³ – 37y² + 336y) dy = 50π(¼y4– (37/3)y³ + 168y²) | = 50π (832 – 0) = 273066.65π ≈ 857,864 ft-lbs y = 0 y = 0 ∆W = ∆F • d = (50π(16y – y²)∆y) • (21 – y)

  9. Summary & Homework • Summary: • For constant force: Work is force times distance • For variable force: Work is the integral of force times distance • Homework: • pg 463-464, Day 1: 1, 2, 3, 7 Day 2: 4, 10, 19, 24

More Related