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. 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

7. 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 =  Work (Hooke’s Law)
∫ (250d) dd = 125d² |
= 125(36 – 9) = 3375 pounds
d =3
d = 6
Step 2
force • (distance increment) 7

8. 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) = π
≈ 857,864 ft-lbs
y = 0
y = 8 8

9. 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 , Day 1: 1, 2, 3, Day 2: 4, 10, 19, 24