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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

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Lesson 6-4

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  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

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