ch10.1 – energy and work energy – the ability to produce change

Ch10.1 – Energy and Work

Energy – the ability to produce change.



Kinetic Energy – energy of motion

KE = ½ m∙v2 Units:

Ex1) What is the kinetic energy associated with a freight train car travelling 2 m/s with a mass of 1,000 kg?

Ex2) What is the kinetic energy associated with a 10 kg bowling ball rolling at 20 m/s ?




KE = ½ m ∙ v2 Units: kg ∙ m2 = kg ∙ m ∙ m = N∙m = J s2 s2 (Joule)








KE = ½( 1000 kg )( 2 m/s )2 = 2000 N∙m = 2,000 N.m







KE = ½( 1000 kg )( 2 m/s )2 = 2000 N∙m = 2,000 N.m


KE = ½ ( 10 kg )( 20 m/s )2 = 2000 J

Work – force applied over a distance

W = F ∙ d Force and distance must be in the same direction

Ex3) How much work is done when a person pushes a shopping cart with a force of 50 N for 500 meters?

Work – force applied over a distance W = F ∙ d Force and distance must be in the same direction


W = F ∙ d = ( 50N )( 500m ) = 25,000 N∙m

Work – force applied over a distance

W = F ∙ d Force and distance must be in the same direction


W = F ∙ d = ( 50N )( 500m ) = 25,000 N∙m

Work – Energy Theorem W = ∆KE

Ex4) A hockey player hits a puck, applying a force of 4,000 N over a distanceof 0.5 m, accelerating the puck up to 40 m/s. What is the mass of the puck?

Work – force applied over a distance W = F ∙ d Force and distance must be in the same direction


W = F ∙ d = ( 50N )( 500m ) = 25,000 N∙m

Work – Energy Theorem W = ∆KE

Ex4) A hockey player hits a puck, applying a force of 4,000 N over a distance of 0.5 m, accelerating the puck up to 40 m/s. What is the mass of the puck?

F = 4000N W = ∆KE d = 0.5m vF

2 = vi2 + 2ad F ∙ d = KE f – KEi

vF = 40 m/s F ∙ d = ½ mvf2 – ½ mvi

2 vi = 0 F = m ∙ a ( 4000 )( .5 ) = ½ m( 40 )2

m = ? m = 2.5 kg

1. an object is lifted, work is done overcoming gravity W = (+)

4

3

2

1

1. an object is lifted, work is done F dovercoming gravity W = (+)

2. an object is dropped, gravity does the work to bring it back down. W = (–)

4

3

2

1



3. an object is pushed across a surface, F d work is done overcoming friction.

4

3

2

1



3. an object is pushed across a surface, F d F d work is done overcoming friction.

4. an object is carried horizontally,no work is done on the object. 4

3

2

1



3. an object is pushed across a surface, F d F d work is done overcoming friction.

4. an object is carried horizontally, Fno work is done on the object. d

What about work done in circular motion lab? Top View

vel

Fc

4

3

2

1



3. an object is pushed across a surface, F d F d work is done overcoming gravity.

4. an object is carried horizontally, Fno work is done on the object. d

So what about in between?Force at an angleA tall person pushing a cart:

d

θ

F W = F.d.cosθ

4

3

2

1

Ex5) A sailor pulls a boat 30.0 m along a dock using a rope that makes a 25° angle with the horizontal. How much work does he do if he exerts a force of 255 N on the rope? F = 255 N

25° d = 30 m

Ex5) A sailor pulls a boat 30.0 m along a dock using a rope that makes a 25° angle with the horizontal. How much work does he do if he exerts a force of 255 N on the rope? F = 255 N

25° d = 30 m

W = F ∙ d ∙ cosθ = ( 255 N )( 30 m )( cos25° ) = 6933 J

HW #7) An airplane passenger carries a 215 N suitcase up the stairs, a displacement of 4.20 m vertically and 4.60 m horizontally.

a) How much work does the passenger do?

b) The same passenger carries it back downstairs. How much work now?

4.20m

4.60m

Ch10 HW#1 1 – 8

HW #7) An airplane passenger carries a 215 N suitcase up the stairs, a displacement of 4.20 m vertically and 4.60 m horizontally.

a) How much work does the passenger do?

b) The same passenger carries it back downstairs. How much work now?

a) W = F..d

4.20m = 215N.4.20m

= 903 J

4.60m b) W = - 903 J

Ch10 HW#1 1 – 8

Chapter 10 HW #1 1 – 8

1. A student lifts a box that weighs 185N. The box is lifted 0.800m. How much work?

d = .800 m

Fg = 185 N

2. Two students exert a force of 825N in pushing a car 35m. How much work do they do on the car?

d = 35m F = 825N


1. A student lifts a box that weighs 185N. The box is lifted 0.800m. How much work? F F = Fg

d = .800 m Wg = Fg ∙ d = ( 185 N )( .800 m ) = 148 J

Fg = 185 N


d = 35m F = 825N


1. A student lifts a box that weighs 185N. The box is lifted 0.800m. How much work? F F = Fg

d = .800 m Wg = Fg ∙ d = ( 185 N )( .800 m ) = 148 J

Fg = 185 N


d = 35m W = F ∙ d = ( 825 N )( 35 m ) = 28,875

JF = 825N

3) A 0.180kg ball falls 2.5m. How much work does the force of gravity do on the ball?

d = 2.5 m Fg = 1.8 N

4) A forklift raises a box 1.2m doing 7000J of work on it. What is the mass of the box? F d

Fg


d = 2.5 m Fg = 1.8 N Wg = Fg ∙ d = - 4.4 J


Fg


d = 2.5 m Fg = 1.8 N Wg = Fg ∙ d = - 4.4 J


W = F ∙ d ( W = mgd ) 7000 J = F ∙ 1.2 m F = 5833 N

m = 583 kg Fg

5. You and a friend each carry identical boxes to a room one floor above you and down the hall. You carry your box up a set of stairs then down the hallway. Your friend carries a box down the hall then up another stairwell. Who does more work?

6. How much work does the force of gravity do when a 24N object falls a distance of 3.5m?

W = F ∙ d d = 3.5 m

Fg = 24 N


same vertical

distance = same work


W = F ∙ d d = 3.5 m

Fg = 24 N


same vertical

distance = same work


W = F ∙ d d = 3.5 m = ( 24 N )( 3.5 m )

Fg = 24 N = - 84.0 J

8. A rope is used to pull a metal box 15.0 m across the floor. The rope is held at an angle of 46.0° with the floor, and a force of 628N is used. How much work is done on the box? F = 628 N

46° W = F · d · cosθ Fx

d = 15 m

8. A rope is used to pull a metal box 15.0 m across the floor. The rope is held at an angle of 46.0° with the floor, and a force of 628N is used. How much work is done on the box? F = 628 N

46° W = F · d · cosθ Fx = 628 N · 15 m · cos46°

d = 15 m = 6544 J

Ch10.2 – Work Under a Varying Force

Work = Area under the Force/Distance graph

Ex1) A bow string is pulled back 0.5 meters. The force to pull it increases with distance from 0 N to 20 N as shown. How much work is done?

10

5

20

15

F (N)

0.1 0.2 0.3 0.4 0.5d (m)

Ch10.2 – Work Under a Varying Force

Work = Area under the Force – Distance graph

Ex1) A bow string is pulled back 0.5 meters. The force to pull it increases with distance from 0 N to 20 N as shown. How much work is done?

Work = Area

= ½ b · h

= ½ ( .5 m )( 20 N )

= 5 J

10

5

20

15

F (N)

0.1 0.2 0.3 0.4 0.5d (m)

Ex2) How much work is done by this erratic Force?

Work = Area → + +

10

5

20

15

F (N)

1 2 3 4 5

d (m)

Ex2) How much work is done by this erratic Force? Work = Area → + +

= (½ b·h) + (l·w) + (½ b·h) = ½(2)(10) + (3)(10) + ½(2)(10)

= 50 J 10

10

10

5

20

15

F (N)

1 2 3 4 5d (m)

Ex3) To compress a large coil spring 10 cm requires a force that increases linearly from 10 N to 50 N. How much work is done on the spring?

20

10

50

40

30F (N)

.02 .04 .06 .08 .10d (m)

Ex3) To compress a large coil spring 10 cm requires a force that increases linearly from 10 N to 50 N. How much work is done on the spring?

Work = Area = 1 + 2

= ½ b·h + l·w40 = ½(.1)(40) + (.1)(10)

= 3 J

10

20

10

50

40

30F (N)

.02 .04 .06 .08 .10

d (m)

2

1

Power – the rate at which work is done

Power = work time

Units:

Ex1) An electric motor lifts an elevator 9.0 m in 15.0 seconds by exerting a force of 12,000 N. What power does it produce? F d = 9 m

Fg = 12,000 N

Power – the rate at which work is done

Power = work time

P = W or P = F·d or P = F · v t t t

Units: J/s → Watt

Ex1) An electric motor lifts an elevator 9.0 m in 15.0 seconds by exerting a force of 12,000 N. What power does it produce? F P = F·d = ( 12,000 N )( 9.0 m ) = 7,200 W

t ( 15.0 sec ) d = 9 m

Fg = 12,000 N

Ex2) Through a set of pulleys, a 10 kg mass is lifted .25 m in 0.5 seconds. How much power was required?

F

FG

Ch10 HW#2 9 – 12

Ex2) Through a set of pulleys, a 10 kg mass is lifted .25 m in 0.5 seconds. How much power was required?

F P = F·d = ( mg )·d = ( 100 N )( .25 m ) = 50 W t t ( .5 s )

FG

Ch10 HW#2 9 – 12

Lab10.1 Power

- due tomorrow

- go over Ch10 HW#2 before lab

Ch10 HW#2 9 – 12 9. A 575N box is lifted 20.0m in 10 sec. What is the power required?

10.A 645N rock climber climbs 8.2m in 30 min. a. How much work?

b. Power?

Ch10 HW#2 9 – 12 9. A 575N box is lifted 20.0m in 10 sec. What is the power required?

P = F·d = (575N)(20.0m) = t (10s)

10.A 645N rock climber climbs 8.2m in 30 min. a. How much work? W = F·d = (645N)(8.2m) =

b. Power? P = W = 5280 J =

t 1800s

11. An electric motor develops 65 kw of power as it lifts a loaded elevator 17.5 m in 35 seconds. How much force does the motor exert?P = 65,000 W P = W

d = 17. 5 m t t = 35 sec P = F·d

F = ? t

12. Pushing a stalled car, it takes 210N to get it moving, and the force decreases at a constant rate until it reaches 40N by 15m. How much work is done during this interval?

11. An electric motor develops 65 kw of power as it lifts a loaded elevator 17.5 m in 35 seconds. How much force does the motor exert?P = 65,000 W P = W

d = 17. 5 m t t = 35 sec P = F·d

F = ? tF =

12. Pushing a stalled car, it takes 210N to get it moving, and the force decreases at a constant rate until it reaches 40N by 15m. How much work is done during this interval?

Work = Area = 1 + 2

= ½ b·h + l·w= ½(15m)(170N) + (15m)(40N)=

100

50

200

150

F (N)

5 10 15d (m)

1

2

13. In the tractor pull competition, the trailer is set up so that as the tractorpulls the trailer, the trailer shifts its mass forward, increasing the drag, and thus increasing the force required to pull. If ur not sure what I’m referring to, Youtube it! A graph of the Force required vs distance is shown:How much work is done by the tractor?

20000

15000

10000

5000

F (N)

10 20 30 40 50d (m)

13. In the tractor pull competition, the trailer is set up so that as the tractorpulls the trailer, the trailer shifts its mass forward, increasing the drag, and thus increasing the force required to pull. If ur not sure what I’m referring to, Youtube it! A graph of the Force required vs distance is shown:How much work is done by the tractor?

Work = Area = 1 + 2 = ½ b·h + l·w = ½(50m)(5000N)+(50m)(15000N) =

20000

15000

10000

5000

F (N)

10 20 30 40 50d (m)

2

1

Ch10.3 – Machines

- make work “feel” easier by changing forces, either magnitude or direction.

Ch10.3 – Machines


- you apply an input force, Fin, the machine multiplies the force lifting

the object, called the output force, Fout.

Ch10.3 – Machines




- this multiplying of forces is at the expense of distance moved.

The input distance, din, is greater than the output distance, dout.

Ch10.3 – Machines






- Machines conserve energy:

Win = Wout Fin

.din = Fout.dout

Ch10.3 – Machines- make work “feel” easier by changing forces,

either magnitude or direction.





- Machines conserve energy:

Win = Wout Fin

.din = Fout.dout

- Machines are rated by their mechanical advantage:

Real Mechanical Advantage:

Ideal Mechanical Advantage:

in

out

F

FRMA

out

in

d

dIMA

- as machines become more complex (more moving parts), they lose efficiency:

Efficiency =

%100

or

%100

IMA

RMA

W

W

in

out

- as machines become more complex (more moving parts), they lose efficiency:

Efficiency =

- Six types of simple machines (On test)1. Levers (3 classes, Lab10.2)2. Pulleys (Lab10.3)3. Incline Planes4. Wedge5. Screw6. Wheel and axle

- Compound Machines – combination of 2 or more simple machinesExs: axe, bike, block and tackle system

%100

or

%100

IMA

RMA

W

W

in

out

Ex1) A 1st class lever is set up to lift a 10N object as shown. What does the scale read?

Ex2) A 2nd class lever is set up as shown.a. What does the scale read?b. What is the ideal MA?

10N

10N

Ex1) A 1st class lever is set up to lift a 10N object. 6m .2m as shown. What does the scale read?

Fin.din = Fout

.dout

Fin.(.6m) = (10N).(.2m)

Fin = 3.3N Fin Fout

Ex2) A 2nd class lever is set up as shown.a. What does the scale read?b. What is the ideal MA?

10N

10N

Ex1) A 1st class lever is set up to lift a 10N object. 6m .2m as shown. What does the scale read?

Fin.din = Fout

.dout

Fin.(.6m) = (10N).(.2m)

Fin = 3.3N Fin Fin Fout

Ex2) A 2nd class lever is set up as shown. a. What does the scale read? d in = .8m dout = .3m b. What is the ideal MA?

a) Fin.din = Fout

.dout

Fin.(.8m) = (10N).(.3m)

Fin = 3.75N

b) Fout

10N

10N

7.23.

8.

m

m

d

dIMA

out

in

Ex3) A 3rd class lever is set up as shown.What does the scale read?

Ex4) A pulley system is set up as shown.The scale reads ___N when is lifts a 10Nobject. The scale moves ___mwhen the object moves 0.05m.a. What is the IMA?

b. What is the RMA?

c. What is the efficiency?

10N

10N

Ex3) A 3rd class lever is set up as shown.What does the scale read? din =.4m

Fin.din = Fout

.dout

Fin.(.4m) = (10N).(.8m)

Fin = 20N dout = .8m

Ex4) A pulley system is set up as shown.The scale reads 3.5N when is lifts a 10Nobject. The scale moves .20mwhen the object moves 0.05m.a. What is the IMA?

b. What is the RMA?


Fout Fin

10N

10N

Ex3) A 3rd class lever is set up as shown.What does the scale read? din =.4m

Fin.din = Fout

.dout

Fin.(.4m) = (10N).(.8m)

Fin = 20N dout = .8m

Ex4) A pulley system is set up as shown.The scale reads 3.5N when is lifts a 10Nobject. The scale moves .20mwhen the object moves 0.05m.a. What is the IMA?

b. What is the RMA?


Fout Fin

Ch10 HW#3 13 – 16

10N

10N

405.

2.

m

m

d

dIMA

out

in

85.25.3

10

N

N

F

FRMA

in

out

%71%1004

85.2%100 Eff

IMA

RMA

Lab10.2 – Levers

- due tomorrow

- Ch10 HW#3 due at beginning of period

Lab10.3 – Pulleys

- due tomorrow

Ch10 HW#3 13 – 16 13. A sledgehammer drives a wedge into a piece of wood. The wedge is driven .20m into a log, and the log separates by 0.05m. A force of 1.9x104 N splits the log. What is the input force?

14.A pulley system raises a 240N carton 16.5m. A force of 129N is exerted on the ropeand it is pulled 33.0m.a. What is IMA?b. What is RMA?c. Efficiency?

Ch10 HW#3 13 – 16 13. A sledgehammer drives a wedge into a piece of wood. The wedge is driven .20m into a log, and the log separates by 0.05m. A force of 1.9x104 N splits the log. What is the input force? Fin

.din = Fout.dout

Fin.(.2m) = (1.9x104N).(.05m)

Fin =

14.A pulley system raises a 240N carton 16.5m. A force of 129N is exerted on the rope and it is pulled 33.0m.a. What is IMA?b. What is RMA?c. Efficiency?

m

m

d

dIMA

out

in

5.16

33

N

N

F

FRMA

in

out

129

240

%100 EffIMA

RMA

15. A boy exerts a force of 225N on a lever to raise a 1.25x103 N rock a distance of 13cm. How far did the boy move the lever?

Fin.din = Fout

.dout

16. A lab group makes a lever out of a meterstick. The fulcrum is at 30cm. The object is at the 10cm mark. The scale reads 2.5N and is located at 90cm.

What is the weight of the object? din = .6m dout = .2m

Fin.din = Fout

.dout

Fin = 2.5N Fout = ?

?

15. A boy exerts a force of 225N on a lever to raise a 1.25x103 N rock a distance of 13cm. How far did the boy move the lever?

Fin.din = Fout

.dout

(225N).(din) = (1.25x103N).(13cm)

din =16. A lab group makes a lever out of a meterstick. The fulcrum is at 30cm.The object is at the 10cm mark. The scale reads 2.5N and is located at 90cm.

What is the weight of the object? din = .6m dout = .2m

Fin.din = Fout

.dout

(2.5N).(60cm) = Fout.(20cm)

Fout =

Fin = 2.5N Fout = ?

?

Ch11 – Energy

Kinetic Energy KE = ½ mv2 (Work = ∆KE )

Ex1) An 875 kg car speeds up from 22.0 – 44.0 m/s. What were the initial and final energies of the car? How much work was done on the car to increase the speed?

Ch11 – Energy

Kinetic Energy KE = ½ mv2 (Work = ∆KE )

Ex1) An 875 kg car speeds up from 22.0 – 44.0 m/s. What were the initial and final energies of the car? How much work was done on the car to increase the speed?

KEi = ½ ( 875 kg )( 22 m/s )2 = 211,750 J

KEf = ½ ( 875 kg )( 44 m/s )2 = 847,000 J

W = ∆KE = KEF – KEi = 847,000 – 211,750 = 635,250 J

HW #2) A rifle can shoot a 4.20 g bullet at a speed of 965 m/s.

vi = 0 vf = 965 m/s

b. KEf of bullet:

c. What work was done on the bullet? d. If the work is done over a distance of 0.75 m what was the average force?

e. The bullet comes to rest, penetrating 1.5 cm into metal. What is the magnitude and direction of the force the metal exerts?

d = .015 m KEi =____J KEf = 0 J vi = 965 m/s vf = 0 F = ?

HW #2) A rifle can shoot a 4.20 g bullet at a speed of 965 m/s.

vi = 0 vf = 965 m/s KEi = 0 KEf

b. KEf of bullet KEf = ½ mvf2 = ½ ( .0042 kg )( 965 m/s )2 = 1956 J

c. What work was done on the bullet? W = ? W = ∆KE = KEf – KEi = 1956 J d. If the work is done over a distance of 0.75 m what was the average force?

W = F·d F = W = 1956J = 2608 N d .75 m

e. The bullet comes to rest, penetrating 1.5 cm into metal. What is the magnitude and direction of the force the metal exerts?

W = ∆KE = 0 – 1956 J d = .015 m W = F·d

KEi = 1956 J KEf = 0 J F = W = -1956 J = -130,400 N vi = 965 m/s vF = 0 d .015 m F = ?

Potential Energy – stored energy

Gravitational Potential Energy – energy due to the position above the ground

PEG = mgh (sometimes called Ug)

Ex2) Lift a 2kg book from the floor to a shelf 2.10m up.a. What is the PEG relative to the floor?b. What is the PEG relative to your head 1.65 m

above the floor?

Elastic PE – energy stored in a spring. PES = ½kx2

Ch11 HW#1 1 – 6

Potential Energy – stored energy

Gravitational Potential Energy – energy due to the position above the ground

PEG = mgh (sometimes called Ug)

Ex2) Lift a 2kg book from the floor to a shelf 2.10m up. a. What is the PEG relative to the floor? PEG = (2kg)(9.8m/s2)(2.10m) b. What is the PEG relative to your head 1.65 m = 41.2 J

above the floor? PEG = (2kg)(9.8m/s2)(0.45m) = 8.8 J

Elastic PE – energy stored in a spring. PES = ½kx2

Ch11 HW#1 1 – 6

Chapter 11 HW #1 1 – 6 1. A compact car with a mass of 875 kg is traveling at 22 m/s. a. What is the kinetic energy of the car?

KE = ½ mv2 = b. If the same car were traveling at 44 m/s, what would be its kinetic energy?

KE = ½ mv2 = c. If the car doubled its mass to 1750 kg and traveled at 22 m/s,

what would be its kinetic energy?

KE = d. Which has a bigger effect on KE, doubling the mass or doubling the speed?

2. In class

3. A comet with a mass of 7.85x1011 kg strikes Earth at a speed of 25.0 m/s. a. Find the kinetic energy of the comet.

KE = b. Compare that energy to the 4.2x1015 J of energy

that was released from largest nuke every built.


KE = ½ mv2 = ½ ( 875 kg )( 22 m/s )2 = 211,750 J b. If the same car were traveling at 44 m/s, what would be its kinetic energy?

KE = ½ mv2 = ½ ( 875 kg )( 44 m/s )2 = 847,000 J c. If the car doubled its mass to 1750 kg and traveled at 22 m/s,


KE = d. Which has a bigger effect on KE, doubling the mass or doubling the speed?

2. In class

3. A comet with a mass of 7.85x1011 kg strikes Earth at a speed of 25.0 m/s. a. Find the kinetic energy of the comet.







KE = ½ mv2 = ½ ( 1750 kg )( 22 m/s )2 = 423,500 J d. Which has a bigger effect on KE, doubling the mass or doubling the speed?

847,000 J > 423,500 J => doubling the speed2. In class

3. A comet with a mass of 7.85x1011 kg strikes Earth at a speed of 25,000m/s. a. Find the kinetic energy of the comet.







KE = ½ mv2 = ½ ( 1750 kg )( 22 m/s )2 = 423,500 J d. Which has a bigger effect on KE, doubling the mass or doubling the speed?

847,000 J > 423,500 J => doubling the speed2. In class

3. A comet with a mass of 7.85x1011 kg strikes Earth at a speed of 25000m/s. a. Find the kinetic energy of the comet.

KE = ½ ( 7.85x1011 kg )( 25,000 m/s )2 = 2.45x1020 J b. Compare that energy to the 4.2x1015 J of energy


(2.45x1020 J )/(4.2x1015 J) = 58,407 nukes!

4. A 500 kg boulder sits precariously at the edge of a cliff 50 m tall. What is its potential energy?

PEg = mgh =

5. You lift a 10 kg weight to a height of 1.5 m. a. How much potential energy does it now have?

PEg = mgh = b. Using the formula: W = f·d, how much work did you do on the weight?

W = Fg·d =( W = ∆PE )

6. A 90 kg climber climbs 45 m up a vertical wall. a. How much potential energy does the climber now have?

b. If the climber continues climbing to a height of 85 m,

how much potential energy does he now have?


PEg = mgh = ( 500 kg )( 9.8 m/s2 )( 50 m ) = 245,000 J


PEg = mgh = b. Using the formula: W = f·d, how much work did you do on the weight?

W = Fg·d =( W = ∆PE )





PEg = mgh = ( 500 kg )( 9.8 m/s2 )( 50 m ) = 245,000 J


PEg = mgh = ( 10 kg )( 9.8 m/s2 )( 1.5 m ) = 147 J b. Using the formula: W = f·d, how much work did you do on the weight?

W = Fg·d = ( m·g )·d = ( 10 kg )( 9.8 m/s2 )( 1.5 m) = 147 J( W = ∆PE )





PEg = mgh = ( 500 kg )( 9.8 m/s2 )( 50 m ) = 245,000 J


PEg = mgh = ( 10 kg )( 9.8 m/s2 )( 1.5 m ) = 147 J b. Using the formula: W = f·d, how much work did you do on the weight?

W = Fg·d = ( m·g )·d = ( 10 kg )( 9.8 m/s2 )( 1.5 m) = 147 J( W = ∆PE )


PEg = mgh = ( 90 )( 9.8 )( 45 ) = 39,690 J b. If the climber continues climbing to a height of 85 m,


PEg = mgh = ( 90 )( 9.8 )( 85 ) = 74,970 J

Ch11.2 – Conservation of EnergyMechanical Energy – combination of PE and KE

If no friction, and no energy lost to heat,

MEi = MEf PEi + KEi = PEf + KEf

Ex1) A large chunk of ice with a mass of 15kg falls from a roof 8m above the ground.How fast is it moving when it’s just about to hit the ground?

Ch11.2 – Conservation of EnergyMechanical Energy – combination of PE and KE

If no friction, and no energy lost to heat,

MEi = MEf PEi + KEi = PEf + KEf

Ex1) A large chunk of ice with a mass of 15kg falls from a roof 8m above the ground.How fast is it moving when it’s just about to hit the ground?

PEi + KEi = PEf + KEf PEi PEi = KEf

mgh = ½mvf2

(9.8)(8) = ½vf2

KEf vf = 12.5 m/s

HW#7) A bike rider approaches a hill at a speed of 8.5 m/s. The total mass is 85kg.a. Drawb. Find KEc. How high up hill? h=?d. Does mass matter?

HW#7) A bike rider approaches a hill at a speed of 8.5 m/s. The total mass is 85kg.a. Draw PEf b. Find KEc. How high up hill? h=?d. Does mass matter?

KEi

b. KEi = ½mv2 = ½(85kg)(8.5 m/s)2 = 3070 J

c.

HW#7) A bike rider approaches a hill at a speed of 8.5 m/s. The total mass is 85kg.a. Draw PEf b. Find KEc. How high up hill? h=?d. Does mass matter?

KEi

b. KEi = ½mv2 = ½(85kg)(8.5 m/s)2 = 3070 J

c. PEi + KEi = PEf + KEf KEi = PEf ½mvi

2 = mghh = 3.6m

d. Mass cancels

Ex2) What type of energy does the object have at the indicated positions?

1. Pendulum 2. Comet

____ ____

_____ _____

HW#12) A moon rock is dropped from a cliff on the moon 50m tall. Since there is no atmosphere, there’s no air resistance. How fast when it reaches the bottom of the cliff?

(gm = 1.63 m/s2 )

PEi

KEf

HW#12) A moon rock is dropped from a cliff on the moon 50m tall. Since there is no atmosphere, there’s no air resistance. How fast when it reaches the bottom of the cliff?

(gm = 1.63 m/s2 )

Pei PEi + KEi = PEf + KEf PEi = KEf mgh = ½mvf

2 (1.63)(50) = ½vf

2 vf = 12.8 m/s

KEf

Ch11 HW#2 7 – 12

Lab11.1 – Lab11.1 Conservation of Energy

- due tomorrow

- Ch11 HW#2 7 – 12

Chapter 11 HW #2 7 – 127) ( in class )8) Tarzan, mass of 85 kg, swings down from a vine 4 m above the ground.

a) How fast does he go before he hits the ground? PEi PEi + KEi = PEf + KEf 4m KEF

b) Does the answer depend on his mass? No!9) A skier starts from rest at top of a 45 m hill, skies down to the bottom,

then up a 40m hill. PEi a) How fast is he going at the bottom?

KEi + PEi = KEf + PEf PE + KE

45m 40m

b) How fast at the top of the second hill? KEf

(use bottom as initial) KEf = KE + PE


a) How fast does he go before he hits the ground? PEi PEi + KEi = PEf + KEf 4m mgh = ½ mvf

2 KEF vf

2 = ( 2gh ) vf = b) Does the answer depend on his mass? No!9) A skier starts from rest at top of a 45 m hill, skies down to the bottom,



45m 40m

b) How fast at the top of the second hill? KEf

(use bottom as initial) KEf = KE + PE


a) How fast does he go before he hits the ground? PEi PEi + KEi = PEf + KEf 4m mgh = ½ mvf

2 KEF vf

2 = ( 2gh ) vf = b) Does the answer depend on his mass? No!9) A skier starts from rest at top of a 45 m hill, skies down to the bottom,



mgh = ½ mvf2 45m 40m

vf

2= (2gh) = b) How fast at the top of the second hill? KEf

(use bottom as initial) KEf = KE + PE ½ mvi

2 = ½ mvf2 + mgh

½(30)2 = ½vf2 + (10)(40)

vf =

10. A ball is dropped from a roof 6 m high. How fast before it hits the ground?

PEi + KEi = PEF + KEF

PEi

KEf vF = ?

11. A ball is thrown upwards at 20 m/s, how high up will it go?

PEf PEi + KEi = PEF + KEF

KEi

Ch11.3 – Conservation in Collisions and Explosions- Momentum is conserved in collisions, KE is not.

Ex1) Lab cart1 has a mass of 2kg and is traveling at 1m/s collides with cart 2, with a mass of 1kg initially at rest. If the 2 carts stick together, what is their speed? Compare the KEi to the KEf.

+ m1v1i + m2v2i = (m1 + m2)vf

Some KE gets converted into Thermal Energy (heat).

2kg 2kg1kg 1kg

Ch11.3 – Conservation in Collisions and Explosions- Momentum is conserved in collisions, KE is not.

Ex1) Lab cart1 has a mass of 2kg and is traveling at 1m/s collides with cart 2, with a mass of 1kg initially at rest. If the 2 carts stick together, what is their speed?Compare the KEi to the KEf.

+

m1v1i + m2v2i = (m1 + m2)vf (2)(1) + (1)(0) = (3)vf

vf = 0.67 m/s

KEi = ½m1v1i2 + ½m2v2i

2 = ½(2)(1)2 + ½(1)(0)2 = 1 J

1 - 0.67 =0.33J (lost)KEf = ½(mTotal)vf

2 = ½(3)(.67)2 = 0.67 J


2kg 2kg1kg 1kg

Ex2) In an accident on a slippery road, a car with a mass of 575kg moving at 15m/s smashes into a car with a mass of 1575kg, moving at 5m/s in the same direction. The 2 cars stick together, what is their speed?

+

m1v1i + m2v2i = (m1 + m2)vf

Compare the KEi to the KEf.

1575 1575575 575

Ex2) In an accident on a slippery road, a car with a mass of 575kg moving at 15m/s smashes into a car with a mass of 1575kg, moving at 5m/s in the same direction.The 2 cars stick together, what is their speed?

+

m1v1i + m2v2i = (m1 + m2)vf (575)(15) + (1575)(5) = (2150)vf vf = 7.7 m/s

Compare the KEi to the KEf.

KEi = ½m1v1i2 + ½m2v2i

2 = ½(575)(15)2 + ½(1575)(5)2 = 84,000 J

KEf = ½(mTotal)vf2 = ½(2150)(7.7)2 = 63,000 J

84,000 – 63,000 = 21,000J (lost)


1575 1575575 575

HW#13) A 2g bullet, moving at 538m/s , strikes a 0.250kg piece of wood at rest.The bullet embeds itself in the wood, and the two move along the table.If the table is frictionless, what is their final speed?

m1v1i + m2v2i = (m1 + m2)vf

Find KEi and KEf

What percentage of KE is lost to heat?

HW#13) A 2g bullet, moving at 538m/s , strikes a 0.250kg piece of wood at rest.The bullet embeds itself in the wood, and the two move along the table.If the table is frictionless, what is their final speed?

m1v1i + m2v2i = (m1 + m2)vf (.002)(538) + (.250)(0) = (.252)vf vf = 4.3 m/s

Find KEi and KEf KEi = ½m1v1i

2 + ½m2v2i2

= ½(.002)(538)2 + ½(.25)(0)2 = 289 J

KEf = ½(mTotal)vf2 = ½(.252)(4.3)2 = 0.67 J

What percentage of KE is lost to heat?

Ch11 HW#3 13 – 16

%99%100289

3.2289

x

J

JJ

Lab11.2 – Momentum and KE in Collisions

- due tomorrow

- Ch11 HW#3 due @ beginning of period

Ch11 HW#3 13 – 16 13. (In class)14. An 8g bullet is fired horizontally into a 9kg block of wood at rest. After the collision, they move off together at 10cm/s . What was the initial speed of the bullet?

m1v1i + m2v2i = (m1 + m2)vf

15. Suppose that Superman with a mass of 104kg is at rest, and gets struck by a 4.2g bullet moving at 835m/s. The bullet drops straight down. How fast does Superman move?

m1v1i + m2v2i = m1v1f + m2v2f

16. A 10kg bowling ball is rolling at 5 m/s when it collides with a 5kg stationary pin.The pin flies off at 4m/s, the ball at 2m/s.

a. KEi =b. KEf =c. Where did the energy go? Heat and sound


m1v1i + m2v2i = (m1 + m2)vf (.008)(v1i ) + (9)(0) = (9.008)(.10) vf = 112.6m/s


m1v1i + m2v2i = m1v1f + m2v2f






m1v1i + m2v2i = m1v1f + m2v2f (.0042)(835) + (104)(0) = (.0042)(0) + (104)(v2f) v2f = 0.034m/s






m1v1i + m2v2i = m1v1f + m2v2f (.0042)(835) + (104)(0) = (.0042)(0) + (104)(v2f) v2f = 0.034m/s


a. KEi = ½(10)(5)2 = 125J b. KEf = ½(10)(2)2 + ½(5)(4)2 = 60J c. Where did the energy go? Heat and sound

Ch10 and 11 Test Review1. Brutus raises 240kg of weights a distance of 2.35m. a. How much work is done?

b. How much work done while holding weights above his head?

c. How much work done lowering back down?

d. Does Brutus do any work if he drops the weights?

e. How much power if he lifts them in 2.5s?

Bonus ex) A car with a mass of 500kg speeds up from 20m/s to 40m/s. How much work is done?


W= Fd = (2400N)(2.35m) = 5527J b. How much work done while holding weights above his head?

0J c. How much work done lowering back down?

W = -5527J d. Does Brutus do any work if he drops the weights?

0J, gravity does the work e. How much power if he lifts them in 2.5s?

P =W/t = 5527J/2.5s = 2167W



W= Fd = (2400N)(2.35m) = 5527J b. How much work done while holding weights above his head?

0J c. How much work done lowering back down?

W = -5527J d. Does Brutus do any work if he drops the weights?

0J, gravity does the work e. How much power if he lifts them in 2.5s?

P =W/t = 5527J/2.5s = 2167W


W = ∆KE = KEf - KEi = ½mvf

2 – ½mvi2

= ½(500)(40)2 – ½(500)(20)2 = 300,000 J

2. Graph of displacement vs time. Calc work done. Calc power if work is done in 2s?

20

10

40

30

F (N)

1 2 3 4 5

d (m)

2. Graph of displacement vs time. Calc work done. Calc power if work is done in 2s?

Work = (area) = ∆1 + ∆2 + 3 + 4 = ½bh + ½bh + lw + lw

= ½(4)(30) + ½(2)(20) + (4)(50) + (1)(20)

= 255J

b. P = W/t = 255J/2s = 127.5Watts

20

10

40

30

F (N)

1 2 3 4 5

d (m)

1

2

3

4

3. A pulley system lifts a 1345N weight a distance of 0.975m. The person pulls the rope a distance of 3.90m.

a. How much force does the person exert?

b. What is the IMA?

4. A 30kg gun fires a 50g bullet with a muzzle velocity of 310 m/s. a. What is the recoil velocity?

b. What are KE’s of bullet and gun after shot?

5. 420N Kelli is sitting atop a 4m tall slide. How fast at bottom?


a. How much force does the person exert?Fin.din = Fout

.dout

Fin.(3.90m) = (1345N).(.975m)

b. What is the IMA? Fin = 336N

4. A 30kg gun fires a 50g bullet with a muzzle velocity of 310 m/s. a. What is the recoil velocity?



4975.

90.3

m

m

d

dIMA

out

in



.dout

Fin.(3.90m) = (1345N).(.975m)


4. A 30kg gun fires a 50g bullet with a muzzle velocity of 310 m/s.

a. What is the recoil velocity? m1v1 = m2v2 (30)v1 = (.05)(310)

v1 = .52 m/s



4975.

90.3

m

m

d

dIMA

out

in



.dout

Fin.(3.90m) = (1345N).(.975m)




v1 = .52 m/s


KEb = ½(.05)(310)2 = 2400J KEg = ½(30)(.52)2 = 4J


4975.

90.3

m

m

d

dIMA

out

in



.dout

Fin.(3.90m) = (1345N).(.975m)




v1 = .52 m/s


KEb = ½(.05)(310)2 = 2400J KEg = ½(30)(.52)2 = 4J


PEi PEi + KEi = PEf + KEf

mgh = ½ mvf2

KEf vf = 8.8 m/s

4975.

90.3

m

m

d

dIMA

out

in

ch10.1 – energy and work energy – the ability to produce change

Documents

work force

nm work energy theoremw

f d force

nm work energy theorem

ke f ke

distance w

ad f d

nm slide