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 = ∫ f(x) dxx = a

x = b

Work = force • distance = 150 lb • 20 ft = 3000 ft-lbs

1 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)

∫ F(x) dx = Work

∫ (1/x²) dx = -(1/x) |

= (-1/6) – (-1) = 5/6

x = 1

x = 6

x = 1

x = 6

6-4 Example 1

∫ (250d) dd = 125d² |

= 125(36 – 9) = 3375 pounds

d =3

d = 6

d =3

d = 6

Step 2

3 6 9 12 15

0 to 3

3 to 6

750

natural length = 15

force • (distance increment)

Step 1 (find k) kd = 750 pounds k = 250 Work (Hooke’s Law)

6-4 Example 2

distance yth layer moves = 5 + (16 – y)= (21 – y)

weight = density • volume

∆F = weight = density • volume = 50 lb/ft³ • πx²∆y

∆yr = x

disk vol

Area • thicknessπx² • ∆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

y = 0

y = 8

r = 8

Tank

d = 5

½ fulloil

x

6-4 Example 3

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