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Physics Chapter 10

Work, Energy, and Simple Machine

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Next Generation Science Standards

HS-PS3-1.Create a computational model to calculate the change in the energy of one component in a system when the change in energy of the other component(s) and energy flows in and out of the system are known.

HS-PS3-2.Develop and use models to illustrate that energy at the macroscopic scale can be accounted for as either motions of particles or energy stored in fields.

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Physics

Turn in Chapter 9 homework, worksheet, and Lab Report

Lecture Q&A

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Work, W

Work: the product of the force exerted on an object and the distance (displacement) the object moves in the direction of the force.

Valid only when F and d are in the same direction. F: force

d: displacement, more exactly, d

Unit:

[W] =

W Fd

[F] · [d] = N · m = Joule = J

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825 , 35F N d m

) ?a W

W

) 2 825 1650 , 35 , ?b F N N d m W

41650 35 5.8 10W Fd N m J

Example: Pg261pp2Together, two students exert a force of 825 N in pushing a car a distance of 35 m.a) How much work is done?b) If the force was doubled, how much work would they do pushing the car the same distance?

4825 35 28875 2.9 10Fd N m J J

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W

7.5 , 30.0min, 8.2

) ?b

b

m kg t d m

a W

F

) 645 , ?g g bb F N W

F

W

Practice: Pg261pp3A 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?b) If the climber weighs 645 N, how much work does she do lifting herself and the backpack?

27.5 9.8 73.5b

mm g kg N

s

73.5 8.2 602.7 603Fd N m J J

73.5 645 718.5N N N

718.5 8.2 5891.7 5890Fd N m J J

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PracticeWhat work is done by a forklift raising a 583-kg box 1.2 m?

583 , 1.2 , ?m kg d m W

F

W

2583 9.8 5713

mmg kg N

s

5713 1.2 6856 6900 6.9Fd N m J J kJ

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Practice:

If you push twice as hard against a stationary brick wall, the amount of work you do

a) doubles.

b) is cut in half.

c) remains constant but non-zero.

d) remains constant at zero.

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What if

F and d are not in the same direction?

F: force d: displacement (more accurately, d) : angle between F and d

cosW Fd

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Positive and Negative Work

< 90o W ____ 0

W = 0

F = 0

d = 0

= 90o

Work is a scalar—it has no direction.

Negative sign does not indicate direction.

A positive work is always larger than a negative work.

cosW Fd

= 90o W ____ 0 > 90o W ____ 0

>

=

<

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Example: Pg262pp5Two people lift a heavy box a distance of 15 m. They use ropes, each of which makes an angle of 15o with the vertical. Each person exerts a force of 225 N. How much work do they do?

Fd

W

15 , 15 , 2 225 450 , ?od m F N N W

cos 450 15 cos15 6520oFd N m J

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Practice: Pg262pp7A rope is used to pull a metal box 15.0 m across the floor. The rope is held at an angle of 46.0o with the floor and a force of 628 N is applied to the rope. How much work does the force on the rope do?

46.0o

F

15.0 , 46.0 , 628 , ?od m F N W

W cos 628 15.0 cos 46.0 6540oFd N m J

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Practice:

If you walk 5.0 m horizontally forward at a constant velocity carrying a 10-N book, the amount of work you do is

a) more than 50 J.

b) equal to 50 J.

c) less than 50 J, but more than 0 J.

d) zero.

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Work Done by Weight (Gravity)

When an object goes up or down, the work done by weight (gravity) is given by

where m: mass of object g = 9.8 m/s2

Δh: change in height of object Δh > 0 when going up W ____ 0

W mg h

Δh < 0 when going down W ____ 0 Δh = 0 when going horizontally W ____ 0

>

<

=

cos cos

still works.

W Fd mgd

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13 , 275 , 25 , 25om kg d m F N

Practice: Pg262-8A bicycle rider pushes a bicycle that has a mass of 13 kg up a steep hill.

The incline is 25o and the road is 275 m long, as shown in Figure 10-4. The rider pushes the bike parallel to the road with a force of 25 N.

a. How much work does the rider do on the bike?b. How much work is done by the force of gravity on the bike?

F = 25 N h

) Fda

W

) ?gb W h

gW

0 , ?o W

cos 25 275 cos0 6875oFdFd N m J

sin 275 sin 25 116od m m 4

213 9.8 116 1.48 10

mmg h kg m J

s

d = 275 m

gF mg

gd 90 90 25 115o o o o

gW cosg gdF d cos gdmgd 42

13 9.8 275 cos115 1.48 10omkg m J

s

Fg

gd

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Practice:

You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity of 0.50 m/s. What is the work done by the weight of the book?

a) 10Jb) -10 Jc) 5.0 Jd) -5.0 J

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Power, P

Power: (time) rate of doing work, or (time) rate of energy transfer

WP

tAverage power:

W: Work done t: time, duration, for the work done (more exactly, Δt)

Unit:

1 horse power = 1 hp = 746 W

or E

Pt

P W

tJ

sWatt W

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Instantaneous Power

• F: force• v: velocity : angle between F and v

cosP Fv

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Solution: Pg264pp9

F = 575N, d = 20.0m, t = 10.0s, P = ?

W

P

31.15 10 W

Example: Pg264pp9A box that weighs 575 N is lifted a distance of 20.0 m straight up by a rope. The job is done in 10.0 s. What power is developed in watts and kilowatts?

4575 20.0 1.15 10Fd N m J 4

31.15 101.15 10

10.0

W JW

t s

1000

kW

W

1.15kW

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Practice:

Does the centripetal force acting on an object do work on the object?

a) Yes, since a force acts and the object moves, and work is force times distance.

b) Yes, since it takes energy to turn an object.c) No, because the object has constant speed.d) No, because the force and the velocity of the

object are perpendicular.

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P = 65kW = 65000 W, d = 17.5 m, t = 35.0 sF = ?

v

P 5650001.3 10

0.500

P WN

mvs

P 665000 35.0 2.275 10Pt W s J

W 6

52.275 101.3 10

17.5

W JN

d m

Practice: Pg264pp12An electric motor develops 65 kW of power as it lifts a loaded elevator 17.5 m in 35.0 s. How much force does the motor exert?

17.50.500

35.0

d m m

t s s

Fv F

W

tW

Fd F

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Kinetic Energy, KE

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2KE mv

m: mass

v: velocity

Unit:

Energy or ability to do work because of motion.

KE 2m v

2m

kgs

2

2

mkg

s

2

mkg m

s

N m J

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Practice:

Car J moves twice as fast as car K, and car J has half the mass of car K. The kinetic energy of car J, compared to car K is

a) The same.

b) 2 to 1.

c) 4 to 1.

d) 1 to 2.

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Work-Kinetic Energy Theorem

Wnet: total work done on an object KE: kinetic energy of that object

Work and energy are related by

netW KE f iKE KE

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10 , 100 , 8.5 , 0, ?i fm kg F N d m v v

W

Fd

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2 fFd mv

fv

2fv

ExampleYou push a 10-kg desk, initially at rest, with a force of 100 N a distance of 8.5 m across the classroom. What is the final speed of the desk?

f iKE KEKE

2 21 1

2 2f imv mv

2Fd

m

2 2 100 8.513

10

Fd N m m

m kg s

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Practice: A rock of mass 2.0 kg is thrown with a speed of 10 m/s at an angle of 40o above the horizontal at the edge of a 30-m high cliff. With what speed does it hit the ground?

2.0 , 40 , 30 , 10 , ?oi i f

mm kg h m v v

s

f iW KE KE KE

mg h

2 22 f ig h v v

2 2 2f iv v g h

2 2f iv v g h

2 21 1

2 2f imv mv

2

210 2 9.8 30 26m m m

ms s s

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Practice:

If the net work done on an object is positive, then the object’s kinetic energy

a) decreases.

b) remains the same.

c) increases.

d) is zero.

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Practice:

An arrow of mass 0.020 kg is shot horizontally into a bale of hay, striking the hay with a velocity of 60 m/s. It penetrates a depth of 0.20 m before stopping. What is the average stopping force acting on the arrow?

a) 45 Nb) 90 Nc) 180 Nd) 360 N

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Simple Machine

Simple Machine: anything that can change the force (magnitude or direction) to accomplish a task

Does not change the amount of work done. (Then why use a machine?)

Example:– Lever, pulley, wheel-and-axle,

inclined plane, wedge, and screw

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Mechanical Advantage, MA

Fi or Fe: input force or effort force

o

i

r

eF

FMA

F

F

– Force you actually exert on a machine– Force you actually exert when using a machine

Fo or Fr: output force or resistance force– Force exerted by the machine on the load. – Force you need to exert if not using a machine.

MA has no unit.

MA > 1: machine increases your force

– MA has a unit of 1, or MA is dimensionless.

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Input and Output Work

Wi: input work, work done by you when using machine

Fi: input (effort) force

di: input (effort) displacement, displacement of your hand

Wo: output work, work given out by machine. (Work you need to do if not using the machine.)

Fo: output (resistance) force

do: output (resistance) displacement, displacement caused by machine, displacement when not using machine

i i iW Fd

o o oW F d

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Ideal Machine

Ideal Machine:

iI

o

dIMA MA

d Ideal Mechanical Advantage:

i oW W

is valid for ideal and non-ideal.o

i

FMA

F

is valid only for ideal machine.i

o

dIMA

d

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Efficiency, eff

Efficiency: ratio of the work done by the machine to the work put into the machine; how much of the input work converted into output work by the machine

100%o

i

WEff

W

100%MA

EffIMA

Efficiency has no unit.

Also

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di = 0.20m, do = 5.0cm = 0.05m, Fo = 1.7 104 N, Fi = 1.1 104 N

) ?a IMA

) ?b MA ) ?c eff

IMA

MAeff

Example: Pg272pp25A sledge hammer is used to drive a wedge into a log to split it. When the wedge is driven 0.20 m into the log, the log is separated a distance of 5.0 cm. A force of 1.7 104 N is needed to split the log, and the sledge exerts a force of 1.1 104 N .a) What is the IMA of the wedge?b) Find the MA of the wedge.c) Calculate the efficiency of the wedge as a machine.

0.204.0

0.05i

o

d m

d m

4

4

1.7 101.5

1.1 10o

i

F N

F N

100%MA

IMA

1.5100% 38%

4.0

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224 9.8 / 235 , 16.5 , 129 , 33.0o o i iF mg kg m s N d m F N d m

) ?a MA

) ?b eff IMA

MA

oW

iW

eff

Eff

Practice: Pg272pp26A worker uses a pulley system to raise a 24.0-kg carton 16.5 m. A force of 129 N is exerted and the rope is pulled 33.0 m.a) What is the MA of the pulley system?b) What is the efficiency of the system?

2351.82

129o

i

F N

F N

235 16.5 3878o oF d N m J

129 33.0 4257i iFd N m J

38780.911 91.1%

4257o

i

W J

W J

33.02.00,

16.5i

o

d m

d m

100%MA

IMA

1.82100% 91%

2.00

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Inclined Plane

IMA

hd

IMA i

o

d

d

d

h

1

sin