1

Work, Power, Energy

GlencoeChapters 9,10,11

2

Ch 9 assignments

• In class samples:

• 1,2,4,13,15

• Assigned problems

• 7-9,17,20

3

Chapter 9 Momentum and Its Conservation

Momentum, = mass of an object times its velocity = mv units: kgm/s

Impulse is an object’s difference in final momentum and initial momentumF t = m v = f – i units: ns, kgm/s

4

Conservation of Momentum

This is used in collisions

where Newton’s third law

is related to conservation

of momentum.

See figure 9-6 on page 236

i = f

Ex. prob 2 on p. 237

#2, p237

5

6

Recoil occurs when objects are at rest initially.

Example problem 3 on page 240

General formula

i = f but i = 0

Ci +Di = Cf + Df

0 = Cf + Df

So: Cf = - Df

7

8

9

Chapter 10 Energy, Work, and Simple Machines

• Assigned problems:

• 29,52,57,60,66,81

10

Work is the transfer of energy by mechanical means.

• In this section you will calculate work and power used.

• Terms:

work energy kinetic energy

work-energy theorem

Joule power watt

11

Work = Force x distance cos

• W = Fd units: nm or joule, j

• Work is equal to a constant force exerted on an object in the direction of motion, times the object’s displacement

12

Kinetic Energy – the energy of motion

• KE = ½ m v 2

• units: kg (m/s) 2 = (kgm/s2) m

• = nm = joule

• Work Energy Theorem: W = KE

• Work = the change in kinetic energy of a system.

13

Work at an angle

• W = Fd cos • Work = the product of force and

displacement, times the cosine of the angle between the force and the direction of the displacement.

• How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m?

14

• W = Fdcos • W = 15 N (12 m) (cos 20)

• W = 169 170 J

20o

15 N

15

Power

Power = Work / time

j/s = watt or w

Power is “The rate of doing work.”

16

Sample: #2 on page 261

• Together, two students exert a force of 825 n in pushing a car a distance of 35 m.

• A. How much work do they do on the car?

• W = Fd = 825n (35m) =

• B. If the force was doubled, how much work would they do pushing the car the same distance?

• W = 2Fd = 2(825n)(35m) =

17

A rock climber wears a 7.5 kg backpack while scaling a cliff. After 30.0 min, the climber is 8.2 m above the starting point.

a.How much work does the climber do on the backpack?

W = Fd = mgd = 7.5kg(9.8m/s2)(8.2m)

b.How much power does the climber expend in this effort?

P = W = 603 J = 0.34 watt

t 1800 s

18

Machines

19

Mechanical Advantage

MA = resistance (or load) force effort force

20

• Calculated Mechanical Advantage:• Fr / Fe = 200nm / 300 nm = .67

21

Ideal Mechanical Advantage,

IMA = d effort / d resistance

IMA = 2.4 m / 1.2 m = 2

22

Efficiency

Efficiency, e = work out = MA x 100

work in IMA

23

Chapter 11 Energy & Its Conservation

• Kinetic Energy: KE = ½ m v2

• KE for a Spring: KE = ½ k (d)2

k = F / d• Potential Energy: PE = mgh• Rest Energy: E = mc2

• Mechanical Energy, E = KE + PE• Conservation of Energy (collisions, etc.)• KE i + PE i = KE f + PE f

24

Ch11 Assignment

• 11/6,55,59,73,74.79

25

The total energy of a closed, isolated system is constant. The energy can

change form.

20.0J

26

Examples: Collisions

• Elastic Collisions: ones in which initial KE = final KE

• Inelastic Collisions: ones in which energy is changed into another form (i.e., KE into heat or sound)

27

11/73

73a. W = Fd = 98.0 N (50.0 m)

W = 4.90 x 103 J

73b. PE = mgh = Fd

PE = W = 4.90 x 103 J

73c. KEbottom = PEtop

KEbottom = 4.90 x 103 J