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Chapter 10 - PowerPoint PPT Presentation


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Chapter 10. Energy, Work, & Simple Machines. Energy. The ability to produce change. Energy. The ability to do work. Types of Energy. Kinetic Potential. Kinetic Energy (K). The energy of motion. Potential Energy (U). Stored energy. Kinetic Energy. v f 2 = v i 2 + 2ad

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Presentation Transcript
chapter 10

Chapter 10

Energy, Work, & Simple Machines

energy
Energy
  • The ability to produce change
slide3

Energy

  • The ability to do work
types of energy
Types of Energy
  • Kinetic
  • Potential
kinetic energy k
Kinetic Energy (K)
  • The energy of motion
kinetic energy
Kinetic Energy
  • vf2 = vi2 + 2ad
  • vf2 - vi2 = 2ad
slide8

Kinetic Energy

  • a = F/m
  • vf2- vi2 = 2Fd/m
slide9

Kinetic Energy

½ mvf2- ½ mvi2

= Fd

slide10

Kinetic Energy

K = ½ mv2

work w
Work (W)
  • The process of changing the energy of a system
slide13

Work

  • The product of force times displacement
slide14

Work

  • W = Fd
slide15

Work-Energy

Theorem

  • W = DK
slide18
Calculate the work required to push a 500.0 kg box 250 m at a constant velocity.m = 0.20 between the box & the floor.
constant force at an angle
Constant force at an Angle

Direction of applied force

a

Direction of movement

slide21
Calculate the work done when mowing the lawn when a boy applied a 50.0 N force at a 37o from horizontal for 2.0 km.
slide22

Calculate the work done when a girl pulls a 4.0 kg box with a rope at a 37o from horizontal for 2.0 m. m = 2.5

power
Power
  • The rate of doing work
slide24

Power

  • P = W/t
slide26

A 10.0 Gg crate is accelerated by a cable up a 37o incline for 50.0 m in 2.5 hrs. m = 0.20

Calculate: FT, W, & P

slide27

A 50.0 g box is accelerated up a 53o incline for 50.0 m at 250 cm/s2. m = 0.20

Calculate: FA, vf,W, P, K, & U at the top of the ramp

machines
Machines
  • Devices used to ease force one has to apply to move an object by changing the magnitude and direction of the force.
slide29

Machines

  • Machines do not reduce the work required, but do reduce the force required.
slide30

Machines

  • The force applied is called the effort force (Fe).
slide31

Machines

  • The force exerted by the machine is called the resistant force (Fr).
mechanical advantage
Mechanical Advantage
  • The ratio of resistant force to effort force
in an ideal situation
In an Ideal Situation
  • 100 % of the work input into a system would be transferred to output work, thus:
slide35
Wo = Wi or

Frdr = Fede or

Fr/Fe= de/dr

efficiency
Efficiency
  • The ratio of output work to input work times 100 %
efficiency1
Efficiency =

Wo

Wi

X 100 %

slide39

Efficiency =

MA

IMA

X 100 %

simple machines
Simple Machines

Lever Inclined plane

Wedge Wheel & Axle

Screw Pulley

lever
Lever

Fr

Fe

de

dr

slide42

Fr

Fe

de

dr

IMA = de/dr = length de/length dr

slide44

de

Fr

Fe

dr

a

IMA = de/dr =

length hyp/hyp sin a

slide45

Wedge

½ Fr

Fe

½ Fr

slide46

½ Fr

a

Fe

½ Fr

IMA = de/dr = cot ½ a

slide47

Screw

Fe

Fr

slide48

Pulley

Fe

Fr

slide49

IMA = the number

of lines

pulling up

Fe

Fr

slide51

IMA = ratio

of effort wheel

radius/resistant

wheel radius

Fr

Fe

slide52

A 100.0 Mg trolley is pulled at 750 cm/s up a 53o inclined railway for 5.0 km. m = 0.20

Calculate: FA,W, P, K, & U at the top of the ramp

slide53
An alien exerts 250 N on one end of a 18 m lever with the fulcrum 3 m from a 1200 N load. Calculate: IMA, MA, & efficiency
slide54

A 350 N force is applied to push a 50.0 kg box up a 20.0 m ramp at 37o from horizontal. Calculate: IMA, MA, & efficiency

slide55

A pulley with an efficiency of 80.0 % with 5 interconnecting ropes lifts a 100.0 kg load. Calculate:

IMA, MA, & FA

slide56

A 1.0 m handle is connected to 5.0 cm wheel. The efficiency of this system is 90.0 %. Calculate IMA, MA, & the force required to pull a 500 kg object connected to the wheel.

slide57

A 100.0 cm handle is connected to 5.0 cm wheel with teeth connecting it to another 50.0 cm wheel connected to a 2.5 cm axle. A cable is connected to the axle. The efficiency of this system is 90.0 %. Calculate IMA & MA

slide58

A sledge hammer is used to apply 25 kN drive a 2.0 cm x 10.0 cm wedge into a board. Calculate the force on the board if the efficiency is 75 %.

slide59
Design a system of simple machines that can lift at least 100,000 times the force applied by a human. Assume 90 % efficiency.
slide60

The front sprockets on a 21 speed bike are 24 cm, 18 cm, & 15 cm in diameter. The back sprockets range from 12 cm to 4.0 cm. Determine the ratio of highest to lowest gears.

slide61

On the same bike, the wheels are 80.0 cm in diameter. Calculate the speed in the lowest & highest gears if a person can pedal at 1.0 revolution per second.

slide62

A 100.0 kg block (m = .20) slides from rest down a 50.0 m ramp at 37o from horizontal. At the bottom of the ramp, it collides with a 25 kg box (m = .25) & stops. Calculate:

slide64

The 1.0 m crank is turned lifting the box to a height of 50.0 cm in 5.0 minutes with an efficiency of 90 %. Calculate: IMA, MA, di, FA, Wo, Wi, & P.

10.0 m

1.0 Mg

2.0 m

r = 5.0 cm

slide65

A 50.0 Mg elevator

is raised 200.0 m in 3.0 minutes at a constant speed. Calculate: FAupward, W, & P

slide66

A 200.0 kg sled (m = 0.10) slides from rest down a 500.0 m incline at 37o from horizontal. Calculate: F,F//, Ff, Fnet, a, t, vf, Wo, P, & Kmax