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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∙v 2 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 ?

Ch10.1 – Energy and Work Energy

– the ability to produce change.

Kinetic Energy

– energy of motion KE = ½ m ∙ v 2 Units: kg ∙ m 2 s 2 = kg ∙ m ∙ m = N∙m = s 2

Ex1)

What is the kinetic energy associated with a freight train car travelling 2 m /s

(Joule) 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 ?

Ch10.1 – Energy and Work Energy

– the ability to produce change.

Kinetic Energy

– energy of motion KE = ½ m ∙ v 2 Units: kg ∙ m 2 s 2 = kg ∙ m ∙ m = N∙m = s 2

Ex1)

What is the kinetic energy associated with a freight train car travelling 2 m /s

(Joule) with a mass of 1,000 kg? KE = ½( 1000 kg )( 2 m /s ) 2 = 2000 N∙m = 2,000 N .

Ex2)

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

Ch10.1 – Energy and Work Energy

– the ability to produce change.

Kinetic Energy

– energy of motion KE = ½ m ∙ v 2 Units: kg ∙ m 2 s 2 = kg ∙ m ∙ m = N∙m = s 2

Ex1)

What is the kinetic energy associated with a freight train car travelling 2 m /s

(Joule) with a mass of 1,000 kg? KE = ½( 1000 kg )( 2 m /s ) 2 = 2000 N∙m = 2,000 N .

Ex2)

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

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

Ex3)

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

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

Ex3)

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

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?

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 – Energy Theorem

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

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 v F v i = 40 = 0 m / s v F 2 = v i 2 + 2ad F ∙ d = KE f F = m ∙ a – KE i F ∙ d = ½ mv f 2 – ½ mv ( 4000 )( .5 ) = ½ m( 40 ) 2 i 2 m = ? m = 2.5 kg

1. an object is lifted, work is done overcoming gravity W = (+) 1 3 4 2

1. an object is lifted, work is done overcoming gravity W = (+) 2. an object is dropped, gravity does the work to bring it back down. W = ( –) F 1 d 3 4 2

1. an object is lifted, work is done overcoming gravity W = (+) 2. an object is dropped, gravity does the work to bring it back down. W = ( –) F 1 d 3. an object is pushed across a surface, work is done overcoming friction.

3 4 2 F d

4. an object is carried horizontally, no work is done on the object. 3 F 4 d 2 F d

4. an object is carried horizontally, no work is done on the object. 3 F F 4 d

What about work done in circular motion lab?

F 2 Top View d d F c vel

4. an object is carried horizontally, 3 F F d no work is done on the object.

So what about in between?

Force at an angle

A tall person pushing a cart: 4 2 F d d θ d F

W = F

d

cos θ

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.

How much work does the passenger do?

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.

How much work does the passenger do?

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

4.60m

4.20m

a) W = F ..

d = 215N .

4.20m = 903 J 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 F g = 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

Chapter 10 HW #1 1 – 8

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

F = F g

d = .800 m

W g = F g ∙ d = ( 185 N )( .800 m ) = 148 J

F g = 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

Chapter 10 HW #1 1 – 8

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

F = F g

d = .800 m

W g = F g ∙ d = ( 185 N )( .800 m ) = 148 J

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

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

d = 2.5 m F g = 1.8 N

W g = F g ∙ d = - 4.4 J 4) A forklift raises a box 1.2m doing 7000J of work on it. What is the mass of the box?

F d F g

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

d = 2.5 m F g = 1.8 N

W g = F g ∙ d = - 4.4 J 4) A forklift raises a box 1.2m doing 7000J of work on it. What is the mass of the box?

F d

W = F ∙ d ( W = mgd ) 7000 J = F ∙ 1.2 m F = 5833 N m = 583 kg

F g

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 F g = 24 N

= same 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 F g = 24 N

= same work

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

F d = 3.5 m g = 24 N

W = F ∙ d = ( 24 N )( 3.5 m ) = - 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 ° F x d = 15 m

W = F · d · cosθ

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 F ° x d = 15 m

W = F · d · cosθ = 628 N · 15 m · cos46° = 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. 20 How much work is done?

15 F (N) 10 5 0.1 0.2 0.3 0.4 0.5

d (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. 20 How much work is done?

Work = Area 15 = ½ b · h F (N) 10 = ½ ( .5 m )( 20 N ) 5 = 5 J 0.1 0.2 0.3 0.4 0.5

d (m)

Ex2)

How much work is done by this erratic Force?

20 Work = Area → + + F (N) 15 10 5 1 2 3 4 5 d (m)

Ex2)

How much work is done by this erratic Force?

20 Work = Area → + + = (½ b·h) + (l·w) + (½ b·h) = ½(2)(10) + (3)(10) + ½(2)(10) 15 = 50 J 10 F (N) 10 5 10 1 2 3 4 5 d (m)

Ex3)

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

40 F (N) 30 20 10 .02 .04 .06 .08 .10

d (m)

Ex3)

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

F (N) 40 30 20 1 40 Work = Area = 1 + 2 = ½ b·h + l·w = ½(.1)(40) + (.1)(10) = 3 J 10 2 10 .02 .04 .06 .08 .10

d (m)

Power – the rate at which work is done

Power = work time Units:

Ex1) F

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?

d = 9 m F g = 12,000 N

Power – the rate at which work is done

Power = work time P = W t or P = Units: J/s → Watt F·d t or P = F · v t

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?

P = F·d = ( 12,000 N )( 9.0 m ) = 7,200 W t ( 15.0 sec )

d = 9 m F g = 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 F G 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?

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

F G 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

F = ?

P = W d = 17. 5 m t t = 35 sec P = F·d 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 = ?

t F = 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?

200 F (N) 150 100 1 Work = Area = 1 + 2 = ½ b·h + l·w = ½(15m)(170N) + (15m)(40N) = 50 2 5 10 15 d (m)

13. In the tractor pull competition, the trailer is set up so that as the tractor pulls 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 F (N) 10000 5000 10 20 30 40 50 d (m)

20000 Work = Area = 1 + 2 = ½ b·h + l·w F (N) 15000 10000 = ½(50m)(5000N)+(50m)(15000N) = 5000 1 2 10 20 30 40 50 d (m)

Ch10.3 – Machines

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

Ch10.3 – Machines

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

- you apply an input force, F in , the machine multiplies the force lifting the object, called the output force, F out .

Ch10.3 – Machines

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

- you apply an input force, F in , the machine multiplies the force lifting the object, called the output force, F out .

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

The input distance, d in , is greater than the output distance, d out .

Ch10.3 – Machines

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

- you apply an input force, F in , the machine multiplies the force lifting the object, called the output force, F out .

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

The input distance, d in , is greater than the output distance, d out .

- Machines conserve energy: W in F in .

d in = W out = F out .

d out

Ch10.3 – Machines

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

- you apply an input force, F in , the machine multiplies the force lifting the object, called the output force, F out .

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

The input distance, d in , is greater than the output distance, d out .

- Machines conserve energy: W in = W out F in .

d in = F out .

d out - Machines are rated by their mechanical advantage: Real Mechanical Advantage: Ideal Mechanical Advantage:

IMA RMA



F out F in



d in d out

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

W out

 100 %

W in

RMA

 100 %

IMA

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

W out W in

 100 % or

RMA

 100 %

IMA

- Six types of simple machines (On test) 1. Levers (3 classes, Lab10.2) 2. Pulleys (Lab10.3) 3. Incline Planes 4. Wedge 5. Screw 6. Wheel and axle - Compound Machines – combination of 2 or more simple machines Exs: axe, bike, block and tackle system

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

Ex2) A 2 nd class lever is set up as shown.

a. What does the scale read?

b. What is the ideal MA?

10N 10N

Ex1) A 1 st class lever is set up to lift a 10N object. 6m .2m

as shown. What does the scale read?

10N F in .

d in = F out .

d out F in .

(.6m) = (10N) .

(.2m) F in = 3.3N

F in F out Ex2) A 2 nd class lever is set up as shown.

a. What does the scale read?

b. What is the ideal MA?

10N

Ex1) A 1 st class lever is set up to lift a 10N object. 6m .2m

as shown. What does the scale read?

F in .

d in = F out .

d out F in .

(.6m) = (10N) .

(.2m) F in = 3.3N

10N F in F in Ex2) A 2 nd class lever is set up as shown.

a. What does the scale read? d in b. What is the ideal MA?

= .8m d out = .3m

F out 10N a) F in .

d in = F out .

d out F in .

(.8m) = (10N) .

(.3m) F in = 3.75N

IMA



d in d out

 .

 2 .

7 F out

Ex3) A 3 rd class lever is set up as shown.

What does the scale read?

10N Ex4) A pulley system is set up as shown.

The scale reads ___N when is lifts a 10N object. The scale moves ___m when the object moves 0.05m.

a. What is the IMA?

b. What is the RMA?

c. What is the efficiency?

10N

Ex3) A 3 rd class lever is set up as shown.

What does the scale read?

F in .

d in = F out .

d out F in .

(.4m) = (10N) .

(.8m) F in = 20N 10N Ex4) A pulley system is set up as shown.

The scale reads 3.5

N when is lifts a 10N object. The scale moves .20

m when the object moves 0.05m.

a. What is the IMA?

b. What is the RMA?

c. What is the efficiency?

d in =.4m d out = .8m

10N F out F in

Ex3) A 3 rd class lever is set up as shown.

What does the scale read?

d in =.4m F in .

d in = F out .

d out F in .

(.4m) = (10N) .

(.8m) F in = 20N 10N Ex4) A pulley system is set up as shown.

The scale reads 3.5

N when is lifts a 10N object. The scale moves .20

m when the object moves 0.05m.

a. What is the IMA?

IMA



d in d out

b. What is the RMA?

RMA



F out

c. What is the efficiency?

F in

  .

3 .

  4 2 .

85 Eff 

RMA

 100 % 

IMA

2 .

85  100 % 4  71 % d out = .8m

10N F out F in

Ch10 HW#3 13 – 16

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 1.9x10

4 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 rope and 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 1.9x10

4 N splits the log. What is the input force?

F in .

d in = F out .

d out F in .

(.2m) = (1.9x10

4 N) .

(.05m) F in = 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?

IMA



d d in out

 33

16 .



RMA



F out

 

F in

240

129

Eff 

RMA

 100 %

IMA



15. A boy exerts a force of 225N on a lever to raise a 1.25x10

3 N rock a distance of 13cm. How far did the boy move the lever?

F in .

d in = F out .

d out 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?

d in = .6m d out = .2m

F in .

d in = F out .

d out ?

F in = 2.5N

F out = ?

15. A boy exerts a force of 225N on a lever to raise a 1.25x10

3 N rock a distance of 13cm. How far did the boy move the lever?

F in .

d in = F out .

d out (225N) .

(d in ) = (1.25x10

3 N) .

(13cm) d in = 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?

d in = .6m d out = .2m

F in .

d in = F out .

d out (2.5N) .

(60cm) = F out .

(20cm) F out = ?

F in = 2.5N

F out = ?

Ch11 – Energy

Kinetic Energy KE = ½ mv 2 (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 mv 2 (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?

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

= ½ ( 875 kg )( 44 m /s ) 2 = 847,000 J W = ∆KE = KE F – KE i = 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 .

v i = 0 v f = 965 m /s b. KE f 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 v i KE i =____J KE f = 965 m /s v f = 0 J = 0 F = ?

HW #2)

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

v i = 0 KE i = 0 v f = 965 m /s KE f b. KE f of bullet KE f = ½ mv f 2 = ½ ( .0042 kg )( 965 m / s ) 2 = 1956 J c. What work was done on the bullet? W = ? W = ∆KE = KE f – KE i = 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 v i KE i = 1956 J = 965 m / s KE f v F = 0 J F = W = -1956 J = -130,400 N = 0 d .015 m F = ?

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

PE

= mgh

(sometimes called U g ) Ex2) Lift a 2kg book from the floor to a shelf 2.10m up.

a. What is the PE G b. What is the PE G relative to the floor?

relative to your head 1.65 m above the floor?

Elastic PE – energy stored in a spring.

PE

= ½kx

2 Ch11 HW#1 1 – 6

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

PE G = mgh

(sometimes called U g ) Ex2) Lift a 2kg book from the floor to a shelf 2.10m up.

a. What is the PE G b. What is the PE G relative to the floor?

relative to your head 1.65 m above the floor?

PE G = (2kg)(9.8

m / s 2 )(2.10m) = 41.2 J PE G = (2kg)(9.8

m / s 2 )(0.45m) = 8.8 J Elastic PE – energy stored in a spring.

PE S = ½kx 2 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 .

b .

What is the kinetic energy of the car? KE = ½ mv 2 = If the same car were traveling at 44 m / s , what would be its kinetic energy?

KE = ½ mv 2 =

If the car doubled its mass to 1750 kg and traveled at 22 m / s , what would be its kinetic energy? KE =

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

2. In class 3.

A comet with a mass of 7.85x10

11 kg strikes Earth at a speed of 25.0 m / s .

Find the kinetic energy of the comet. KE =

Compare that energy to the 4.2x10

15 J of energy that was released from largest nuke every built.

Chapter 11 HW #1 1 – 6 1.

A compact car with a mass of 875 kg is traveling at 22 m / s .

a .

b .

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

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

If the car doubled its mass to 1750 kg and traveled at 22 m / s , what would be its kinetic energy? KE =

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

2. In class 3.

A comet with a mass of 7.85x10

11 kg strikes Earth at a speed of 25.0 m / s .

Find the kinetic energy of the comet. KE =

Compare that energy to the 4.2x10

15 J of energy that was released from largest nuke every built.

Chapter 11 HW #1 1 – 6 1.

A compact car with a mass of 875 kg is traveling at 22 m / s .

a .

b .

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

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

If the car doubled its mass to 1750 kg and traveled at 22 m / s , what would be its kinetic energy? KE = ½ mv 2 = ½ ( 1750 kg )( 22 m /s ) 2 = 423,500 J

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

847,000 J > 423,500 J

2. In class

=> doubling the speed

A comet with a mass of 7.85x10

11 kg strikes Earth at a speed of 25,000 m / s .

Find the kinetic energy of the comet. KE =

Compare that energy to the 4.2x10

15 J of energy that was released from largest nuke every built.

Chapter 11 HW #1 1 – 6 1.

A compact car with a mass of 875 kg is traveling at 22 m / s .

a .

b .

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

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

If the car doubled its mass to 1750 kg and traveled at 22 m / s , what would be its kinetic energy? KE = ½ mv 2 = ½ ( 1750 kg )( 22 m /s ) 2 = 423,500 J

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

847,000 J > 423,500 J

2. In class

=> doubling the speed

A comet with a mass of 7.85x10

11 kg strikes Earth at a speed of 25000 m / s .

Find the kinetic energy of the comet. KE = ½ ( 7.85x10

11 kg )( 25,000 m /s ) 2

Compare that energy to the 4.2x10

15 J of energy that was released from largest nuke every built.

(2.45x10

20 J )/(4.2x10

15 = 2.45x10

J) = 58,407 nukes!

20 J

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

PE g = mgh =

You lift a 10 kg weight to a height of 1.5 m.

How much potential energy does it now have?

PE g = mgh =

Using the formula: W = f·d, how much work did you do on the weight? W = F g ·d = ( W = ∆PE )

A 90 kg climber climbs 45 m up a vertical wall.

How much potential energy does the climber now have?

If the climber continues climbing to a height of 85 m, how much potential energy does he now have?

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

= mgh = ( 500 kg )( 9.8 m /s 2 )( 50 m ) = 245,000 J

You lift a 10 kg weight to a height of 1.5 m.

How much potential energy does it now have?

PE g = mgh =

Using the formula: W = f·d, how much work did you do on the weight? W = F g ·d = ( W = ∆PE )

A 90 kg climber climbs 45 m up a vertical wall.

How much potential energy does the climber now have?

If the climber continues climbing to a height of 85 m, how much potential energy does he now have?

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

= mgh = ( 500 kg )( 9.8 m /s 2 )( 50 m ) = 245,000 J

You lift a 10 kg weight to a height of 1.5 m.

How much potential energy does it now have?

PE g = mgh = ( 10 kg )( 9.8 m /s 2 )( 1.5 m ) = 147 J

Using the formula: W = f·d, how much work did you do on the weight? W = F g ·d = ( m·g )·d = ( 10 kg )( 9.8 m /s 2 ( W = ∆PE ) )( 1.5 m) = 147 J

A 90 kg climber climbs 45 m up a vertical wall.

How much potential energy does the climber now have?

If the climber continues climbing to a height of 85 m, how much potential energy does he now have?

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

= mgh = ( 500 kg )( 9.8 m /s 2 )( 50 m ) = 245,000 J

You lift a 10 kg weight to a height of 1.5 m.

How much potential energy does it now have?

PE g = mgh = ( 10 kg )( 9.8 m /s 2 )( 1.5 m ) = 147 J

Using the formula: W = f·d, how much work did you do on the weight? W = F g ·d = ( m·g )·d = ( 10 kg )( 9.8 m /s 2 ( W = ∆PE ) )( 1.5 m) = 147 J

A 90 kg climber climbs 45 m up a vertical wall.

How much potential energy does the climber now have?

PE g = mgh = ( 90 )( 9.8 )( 45 ) = 39,690 J

If the climber continues climbing to a height of 85 m, how much potential energy does he now have? PE g = mgh = ( 90 )( 9.8 )( 85 ) = 74,970 J

Ch11.2 – Conservation of Energy

Mechanical Energy – combination of PE and KE If no friction, and no energy lost to heat, PE i ME i + KE i = ME f = PE f + KE f 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 Energy

Mechanical Energy – combination of PE and KE If no friction, and no energy lost to heat,

PE i

ME i

+ KE i

= ME f

= PE f + KE f

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?

PE KE i f PE i + KE i = PE f + KE f PE mgh i = KE f = ½mv f 2 (9.8)(8) = ½v f 2 v f = 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. Draw b. Find KE c. How high up hill?

d. Does mass matter?

h=?

HW#7) A bike rider approaches a hill at a speed of 8.5

a. Draw m / s . The total mass is 85kg.

PE f b. Find KE c. How high up hill?

d. Does mass matter?

h=?

KE i b. KE i = ½mv 2 = ½(85kg)(8.5

m / s ) 2 = 3070 J c.

HW#7) A bike rider approaches a hill at a speed of 8.5

a. Draw m / s . The total mass is 85kg.

PE f b. Find KE c. How high up hill?

d. Does mass matter?

h=?

KE i b. KE i = ½mv 2 = ½(85kg)(8.5

m / s ) 2 = 3070 J c. PE i + KE i KE i ½mv i 2 = PE = PE f = mgh f h = 3.6m + KE f 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?

(g m = 1.63

m / s 2 ) PE i KE f

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?

(g m = 1.63

Pe i m / s 2 ) PE i + KE i = PE f + KE f PE mgh i = KE f = ½mv f 2 (1.63)(50) = ½v f 2 v f = 12.8 m / s KE f

Ch11 HW#2 7 – 12

Lab11.1 – Lab11.1 Conservation of Energy - due tomorrow - Ch11 HW#2 7 – 12

Chapter 11 HW #2 7 – 12 7)

( in class )

Tarzan, mass of 85 kg, swings down from a vine 4 m above the ground.

How fast does he go before he hits the ground?

PE

PE i + KE i = PE f + KE f

4m KE

F b)

Does the answer depend on his mass? No!

A skier starts from rest at top of a 45 m hill, skies down to the bottom, then up a 40m hill.

PE i

How fast is he going at the bottom? KE i + PE i = KE f + PE f PE + KE 45m 40m

How fast at the top of the second hill?

(use bottom as initial ) KE f = KE + PE KE f

Chapter 11 HW #2 7 – 12 7)

( in class )

Tarzan, mass of 85 kg, swings down from a vine 4 m above the ground.

How fast does he go before he hits the ground?

PE

PE i + KE i = PE f mgh + KE f = ½ mv f 2 v f 2 = ( 2gh )

4m

v f =

Does the answer depend on his mass? No!

KE

F 9)

A skier starts from rest at top of a 45 m hill, skies down to the bottom, then up a 40m hill.

PE i

How fast is he going at the bottom? KE i + PE i = KE f + PE f PE + KE 45m 40m

How fast at the top of the second hill?

(use bottom as initial ) KE f = KE + PE KE f

Chapter 11 HW #2 7 – 12 7)

( in class )

Tarzan, mass of 85 kg, swings down from a vine 4 m above the ground.

How fast does he go before he hits the ground?

PE

PE i + KE i = PE f mgh + KE f = ½ mv f 2 v f 2 = ( 2gh )

4m

v f =

Does the answer depend on his mass? No!

KE

F 9)

A skier starts from rest at top of a 45 m hill, skies down to the bottom, then up a 40m hill.

PE i

How fast is he going at the bottom?

KE i + PE mgh i = KE f + PE f = ½ mv f 2 45m v f 2 = (2gh) = How fast at the top of the second hill?

(use bottom as initial ) KE f ½ mv i 2 ½(30) 2 v f KE f = KE + PE = ½ mv f 2 = ½v f 2 + mgh + (10)(40) = PE + KE 40m

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

PE i + KE i = PE F + KE F

PE

KE

v

= ?

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

PE f PE i + KE i = PE F + KE F KE i

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 1 m / 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 KE i to the KE f .

m 2kg 1 v 1i 1kg + + m 2 v 2i = (m 2kg 1 1kg + m 2 )v f Some KE gets converted into Thermal Energy (heat).

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 1 m / 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 KE i to the KE f .

2kg + 1kg 2kg 1kg m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f (2)(1) + (1)(0) = (3)v f v f = 0.67 m / s KE i = ½m 1 v 1i 2 = ½(2)(1) 2 + ½m 2 v 2i 2 + ½(1)(0) 2 = 1 J KE f = ½(m Total )v f 2 = ½(3)(.67) 2 = 0.67 J 1 - 0.67 =0.33J (lost) Some KE gets converted into Thermal Energy (heat).

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

575 + 1575 575 1575 m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f Compare the KE i to the KE f .

Ex2) In an accident on a slippery road, a car with a mass of 575kg moving at 15 m / s smashes into a car with a mass of 1575kg, moving at 5 m / s in the same direction.

The 2 cars stick together, what is their speed?

575 + 1575 575 1575 m 1 v 1i + m 2 v 2i Compare the KE i to the KE f .

= (m 1 + m 2 )v f (575)(15) + (1575)(5) = (2150)v f v f = 7.7 m / s KE i = ½m 1 v 1i 2 + ½m 2 v 2i 2 = ½(575)(15) 2 + ½(1575)(5) 2 = 84,000 J KE f = ½(m Total )v f 2 = ½(2150)(7.7) 2 = 63,000 J 84,000 – 63,000 = 21,000J (lost) Some KE gets converted into Thermal Energy (heat).

HW#13) A 2g bullet, moving at 538 m / 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?

m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f Find KE i and KE f What percentage of KE is lost to heat?

HW#13) A 2g bullet, moving at 538 m / 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?

m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f (.002)(538) + (.250)(0) = (.252)v f v f = 4.3 m / s Find KE i and KE f KE i = ½m 1 v 1i 2 + ½m 2 v 2i 2 = ½(.002)(538) 2 + ½(.25)(0) 2 = 289 J KE f = ½(m Total )v f 2 = ½(.252)(4.3) 2 = 0.67 J What percentage of KE is lost to heat?

289

 2 .

J x

100 % 289

Ch11 HW#3 13 – 16

 99 %

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 10 cm / s What was the initial speed of the bullet?

. m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f 15. Suppose that Superman with a mass of 104kg is at rest, and gets struck by a 4.2g bullet moving at 835 m / s . The bullet drops straight down. How fast does Superman move?

m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 16. A 10kg bowling ball is rolling at 5 m / s The pin flies off at 4 m / s , the ball at 2 m / s . when it collides with a 5kg stationary pin.

a. KE i = b. KE f = c. Where did the energy go? Heat and sound

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 10 cm / s What was the initial speed of the bullet?

. m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f (.008)(v 1i ) + (9)(0) = (9.008)(.10) v f = 112.6

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

m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 16. A 10kg bowling ball is rolling at 5 m / s The pin flies off at 4 m / s , the ball at 2 m / s . when it collides with a 5kg stationary pin.

a. KE i = b. KE f = c. Where did the energy go? Heat and sound

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 10 cm / s What was the initial speed of the bullet?

. m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f (.008)(v 1i ) + (9)(0) = (9.008)(.10) v f = 112.6

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

m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (.0042)(835) + (104)(0) = (.0042)(0) + (104)(v 2f ) v 2f 16. A 10kg bowling ball is rolling at 5 = 0.034

m / s m / s when it collides with a 5kg stationary pin.

The pin flies off at 4 m / s , the ball at 2 m / s . a. KE i = b. KE f = c. Where did the energy go? Heat and sound

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 10 cm / s What was the initial speed of the bullet?

. m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f (.008)(v 1i ) + (9)(0) = (9.008)(.10) v f = 112.6

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

m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (.0042)(835) + (104)(0) = (.0042)(0) + (104)(v 2f ) v 2f 16. A 10kg bowling ball is rolling at 5 = 0.034

m / s m / s when it collides with a 5kg stationary pin.

The pin flies off at 4 m / s , the ball at 2 m / s . a. KE i = ½(10)(5) 2 = 125J b. KE f = ½(10)(2) 2 + ½(5)(4) 2 = 60J c. Where did the energy go? Heat and sound

Ch10 and 11 Test Review

1. 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 20 m / s How much work is done? to 40 m / s .

Ch10 and 11 Test Review

1. Brutus raises 240kg of weights a distance of 2.35m.

a. 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

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

Ch10 and 11 Test Review

1. Brutus raises 240kg of weights a distance of 2.35m.

a. 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

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

W = ∆KE = KE

= ½mv - KE

i f 2

– ½mv

i 2

= ½(500)(40)

– ½(500)(20)

= 300,000 J

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

40 F (N) 30 20 10 1 2 3 4 5 d (m)

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

40 F (N) 30 20 10 1 2 3 4 Work = (area) = ∆ 1 + ∆ 2 + 3 = ½bh + ½bh + lw + lw = ½(4)(30) + ½(2)(20) + + (4)(50) + (1)(20) 4 = 255J b. P = W/t = 255J/2s = 127.5Watts

1 2 3 4 5 d (m)

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?

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?

F in .

d in = F out .

d out F in .

(3.90m) = (1345N) .

(.975m) b. What is the IMA?

IMA



d in d out

 3 .

975

 4 F in = 336N 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?

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?

F in .

d in = F out .

d out F in .

(3.90m) = (1345N) .

(.975m) b. What is the IMA?

IMA



d in d out

 3 .

975

 4 F in = 336N 4. A 30kg gun fires a 50g bullet with a muzzle velocity of 310 m / s .

a. What is the recoil velocity?

m 1 v 1 = m 2 v 2 (30)v b. What are KE’s of bullet and gun after shot?

1 v 1 = (.05)(310) = .52

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

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?

F in .

d in = F out .

d out F in .

(3.90m) = (1345N) .

(.975m) b. What is the IMA?

IMA



d in d out

 3 .

975

 4 F in = 336N 4. A 30kg gun fires a 50g bullet with a muzzle velocity of 310 m / s .

a. What is the recoil velocity?

m 1 v 1 = m 2 v 2 (30)v 1 = (.05)(310) b. What are KE’s of bullet and gun after shot?

KE b = ½(.05)(310) 2 = 2400J KE g v 1 = .52

m / s = ½(30)(.52) 2 = 4J 5. 420N Kelli is sitting atop a 4m tall slide. How fast at bottom?

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?

F in .

d in = F out .

d out F in .

(3.90m) = (1345N) .

(.975m) b. What is the IMA?

IMA



d in d out

 3 .

975

 4 F in = 336N 4. A 30kg gun fires a 50g bullet with a muzzle velocity of 310 m / s .

a. What is the recoil velocity?

m 1 v 1 = m 2 v 2 (30)v 1 = (.05)(310) b. What are KE’s of bullet and gun after shot?

KE b = ½(.05)(310) 2 = 2400J KE g v 1 = .52

m / s = ½(30)(.52) 2 = 4J 5. 420N Kelli is sitting atop a 4m tall slide. How fast at bottom? KE f PE i PE i + KE mgh v f i = PE f + KE = ½ mv f 2 = 8.8 m /s f

Slide 1

Transcript Slide 1

W = F ∙ d

W = F ∙ d

W = ∆KE

W = F

d

cos θ

PE

= mgh

PE

= ½kx

PE

4m KE

PE

4m

KE

PE

4m

KE

PE

KE

v

= ?

W= Fd = (2400N)(2.35m) = 5527J

0J

W = -5527J

0J, gravity does the work

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

W= Fd = (2400N)(2.35m) = 5527J

0J

W = -5527J

0J, gravity does the work

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

W = ∆KE = KE

= ½mv - KE

– ½mv

= ½(500)(40)

– ½(500)(20)

= 300,000 J

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