Chapter 10

66
Chapter 10 Energy, Work, & Simple Machines

description

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 - PowerPoint PPT Presentation

Transcript of Chapter 10

Page 1: Chapter 10

Chapter 10

Energy, Work, & Simple Machines

Page 2: Chapter 10

Energy•The ability to produce change

Page 3: Chapter 10

Energy•The ability to do work

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Types of Energy

•Kinetic

•Potential

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Kinetic Energy (K)

•The energy of motion

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Potential Energy (U)

•Stored energy

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Kinetic Energy

•vf2 = vi

2 + 2ad

•vf2 - vi

2 = 2ad

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Kinetic Energy•a = F/m

•vf2- vi

2 = 2Fd/m

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Kinetic Energy

½ mvf2- ½ mvi

2

= Fd

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Kinetic Energy

K = ½ mv2

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Potential Energy

U = mgh

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Work (W)•The process of

changing the energy of a system

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Work•The product of

force times displacement

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Work

•W = Fd

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Work-EnergyTheorem

•W = K

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Calculate the work required to lift a 50.0 kg box to a height of 2.0 m:

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Calculate the work done when a 250 N force is applied to move a cart 40.0

km:

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Calculate the work required to push a

500.0 kg box 250 m at a constant velocity. = 0.20 between the

box & the floor.

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Constant force at an Angle

Direction of applied forceDirection of movement

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Constant force at an Angle

W = F(cos )d

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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.

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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. = 2.5

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Power•The rate of doing work

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Power•P = W/t

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A 25 Mg elevator rises 125 m in 5.0

minutes. Calculate: F, W, & P

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A 10.0 Gg crate is accelerated by a cable

up a 37o incline for 50.0 m in 2.5 hrs. = 0.20Calculate: FT, W, & P

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A 50.0 g box is accelerated up a 53o

incline for 50.0 m at 250 cm/s2. = 0.20

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

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Machines• Devices used to ease force

one has to apply to move an object by changing the magnitude and direction of the force.

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Machines• Machines do not reduce the

work required, but do reduce the force required.

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

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Machines•The force exerted by the machine is called the resistant force (Fr).

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Mechanical Advantage

•The ratio of resistant force to effort force

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Mechanical Advantage

Fr

Fe

MA =

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In an Ideal Situation•100 % of the work input into a system would be transferred to output work, thus:

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Wo = Wi or

Frdr = Fede or

Fr/Fe= de/dr

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Ideal Mechanical Advantage

de

dr

IMA =

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

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Efficiency =

Wo

Wi

X 100 %

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Efficiency =

MAIMA

X 100 %

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Simple MachinesLever Inclined plane

Wedge Wheel & Axle

Screw Pulley

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Lever

Fe

Fr

dedr

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Fe

Fr

dedr

IMA = de/dr = length de/length dr

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Inclined Plane

Fe

Fr

dr

de

Page 44: Chapter 10

Fe

Fr

dr

de

IMA = de/dr = length hyp/hyp sin

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Wedge

Fe

½ Fr

½ Fr

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Fe

½ Fr

½ Fr

IMA = de/dr = cot ½

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Screw

Fr

Fe

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Pulley

Fe

Fr

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Fe

Fr

IMA = the number of lines pulling up

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Wheel & Axle

FeFr

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FeFr

IMA = ratioof effort wheelradius/resistantwheel radius

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A 100.0 Mg trolley is pulled at 750 cm/s up a 53o

inclined railway for 5.0 km. = 0.20

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

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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

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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

Page 55: Chapter 10

A pulley with an efficiency of 80.0 %

with 5 interconnecting ropes lifts a 100.0 kg

load. Calculate:IMA, MA, & FA

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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.

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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

Page 58: Chapter 10

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 %.

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Design a system of simple machines that

can lift at least 100,000 times the force applied by a human. Assume 90

% efficiency.

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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.

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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.

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A 100.0 kg block ( = .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 ( = .25) &

stops. Calculate:

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Answer the questions on page 175 & work

Problem Section A on pages 175 & 176.

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r = 5.0 cm

10.0 m

2.0 m

1.0 Mg

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.

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A 50.0 Mg elevator is raised 200.0 m in

3.0 minutes at a constant speed.

Calculate: FAupward, W, & P

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A 200.0 kg sled ( = 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