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 Distance – the total distance traveled by an object between two points in time Displacement – the net change in position between two points in time

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

6-4 Example 1 ∫ F(x) dx = Work ∫ (1/x²) dx = -(1/x) | = (-1/6) – (-1) = 5/6 x = 1 x = 6 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) 6

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

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

Summary & Homework Summary: Homework: 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