AP Phys1 Ch4

Newton’s 1st Law (Galileo’s Law of Inertia) - A body set in motion and left undisturbed, will forever continue with that constant motion. (Becomes obvious in space.)At rest is just a special case of motion.All uniform motions are forms of constant velocity.A net force is required to make a change.

Chapter 4.1 – Newton’s Laws


Newton’s 2nd Law – A net force will cause an object to change its motion.A force causes a mass to accelerate. Fnet = maA force causes a mass to accelerate. Fnet = ma


Newton’s 2nd Law – A net force will cause an object to change its motion.A force causes a mass to accelerate . Fnet = ma

Newton’s 3rd Law – To every action (force) there is an equal andNewton’s 3rd Law – To every action (force) there is an equal andopposite reaction (force).

Forces always come in pairs, act in opposite directions.(Usually act on different objects.)

Exs:

Law of inertia - A body set in motion and left undisturbed, will forever continue at a constant speed. (Becomes obvious in space.)

Newton’s first law - every body continues in a state of rest or uniform motion in a straight line except when a force acts on it.(Rest is just a special case of motion.) Uniform motion and rest are forms of constant velocity. Requires a force to change motion.

Chapter 4

Requires a force to change motion.Force – push or pull

Agent of change, alters motion(4 fundamental forces)

Gravitaty, electromagnetic, strong and weak nuclear forces.)Weight – force of gravity.

Easy conversion: a mass of 1 kg = a force of 10 Newtons = 2.2 poundsForce is a vector. Have to add forces vectorally.Net force - resultant of all force vectors acting on an object,

decides actual motion.

1st Law:

Ex1) Determine the resultant, or net force, exerted on the stationary elephant by the two clowns. What is the tension on the rope attached to the elephant?

F1 = 300 N

F2 = 400 N


F1 = 300 N

F2 = 400 N

Fnet = F1 + F2

N500

40030022

=

+=


F1 = 300 N

F2 = 400 N

Fnet = F1 + F2

N500

40030022

=

+=

FE

Fnet = 500 N

FE = Fnet = 500 N

Ex2) Determine the NET force exerted on the ring by the 3 people:

F1 = 500 NF2 = 707 N

F = 966 N

o45

o30

F3 = 966 N


F1 = 500 NF2 = 707 N

F3 = 966 N

o45

o30

F1x = F1.cos

= 500N.cos45

= 353.6 N

F1y = F1.sin

= 500N.sin45

= 353.6 N

o45

θ

θ

F2x = F2.cos

= 707N.cos60θ

o60

= 707N cos60

= - 353.6 N

F2y = F2.sin

= 707N.sin60

= 612.3 N

F3x = 0 N

F3y = - 966 N

θ


F1 = 500 NF2 = 707 N

F3 = 966 N

o45

o30

F1x = F1.cos

= 500N.cos45

= 353.6 N

F1y = F1.sin

= 500N.sin45

= 353.6 N

o45

θ

θ

F2x = F2.cos

= 707N.cos60θ

o60

= 707N cos60

= - 353.6 N

F2y = F2.sin

= 707N.sin60

= 612.3 N

F3x = 0 N

F3y = - 966 N

Fnetx = F1x + F2x + F3x

= 353.6 + (-353.6) + 0

= 0 N

Fnety = F1y + F2y + F3y

= 353.6 + 612.3 + (-966)

= 0 N

Fnet = 0 N

θ

2nd Law:The force of gravity acting on an object is commonly referred

to as weight.(Fnet = ma)

Fg = mg

The force that balances out gravity, when on level ground, is referred to as the normal force, FN.

FN is always I to the surface.

Ex1) What is the weight of a 2 kg mass on earth?

Ex2) What is the mass of this 2 kg object on the moon?

Ex3) What is its weight on the moon, if gm = 1.67 m\s2 ?

3rd Law:To every action (force) there is an equal and opposite reaction (force).

Force always occurs in pairs, point opposite direction.

If forces always occur in pairs that cancel out, how do things move?

Label interaction pairs:Ex5) Person running Ex6) Person jumps Ex7) How does a nail work?

Ch4 HW#1 p128 9,10,11,14AND Weight/NTL WS

Lab4.1 Inertial Mass

- due tomorrow

- Ch4 HW#1 due @ beginning of period

Ch4 HW#1 p128, 9,10,11,14

9) The two ropes attached to the hook in Fig P9 are pulled on

with forces of 100 N and 200 N. What size force acting in

what direction would produce the same effect?

o30

200 N

100 N

o45

o45

9) The two ropes attached to the hook in Fig P9 are pulled on



F1x = F1.cosθ

= 100N.cos45°= 70.7 N

F1y = F1.sinθ

= 100N.sin45°= 70.7 N

o30

200 N

100 N

o45 = 70.7 N

F2x = F2.cosθ

= 200N.cos30°= 173 N

F2y = F2.sinθ

= 200N.sin30°= 100 N

o45

Hw # 9) The two ropes attached to the hook in Fig P9 are pulled on



F1x = F1.cosθ

= 100N.cos45°= 70.7 N

F1y = F1.sinθ

= 100N.sin45°= 70.7 N

Fnetx = F1x + F2x

= 70.7 + 173= 243.7 No

30

200 N

100 N

o45 = 70.7 N

F2x = F2.cosθ

= 200N.cos30°= 173 N

F2y = F2.sinθ

= 200N.sin30°= 100 N

Fnety = F1y + F2y

= 70.7 + 100= 170.7 N

o45

Hw # 9) The two ropes attached to the hook in Fig P9 are pulled on



F1x = F1.cosθ

= 100N.cos45°= 70.7 N

F1y = F1.sinθ

= 100N.sin45°= 70.7 N

Fnetx = F1x + F2x

= 70.7 + 173= 243.7 N

Fnety = F1y + F2y

= 70.7 + 100o

30

200 N

100 N

o45 = 70.7 N

F2x = F2.cosθ

= 200N.cos30°= 173 N

F2y = F2.sinθ

= 200N.sin30°= 100 N

= 70.7 + 100= 170.7 N

o45

227.1707.243 +=netF

Fnet = 298 N

o35

7.243

7.170tan

1 == −θ

Ch4 HW#1 p128, 9,10,11,14

10. The scale is being pulled on via three ropes.

What net force does the scale read?

o30

F = 200 N F3 = 200 N

o30

F1 = 200 NF2 = 200 N

F3 = 200 N

Ch4 HW#1 p128, 9,10,11,14



F1x = F1.sinθ

= 200N.sin 30°= -100 N

F1y = F1.cosθ

= 200N.cos30°= -173 N

o30

F = 200 N F3 = 200 N

o30

= -173 N

F3x = F2.sinθ

= 200N.sin30°= +100 N

F3y = F2.cosθ

= 200N.cos30°= -173 N

F2x = 0 NF2y = -200 N

F1 = 200 NF2 = 200 N

F3 = 200 N

Ch4 HW#1 p128, 9,10,11,14



o30

F = 200 N F3 = 200 N

o30

F1x = F1.sinθ

= 200N.sin 30°= -100 N

F1y = F1.cosθ

= 200N.cos30°= -173 N

Fnetx = -100 + 0 + (+100) = 0

Fnety = -173 + (-200) + (-173) = -546 N

Fnet = -546 N

F1 = 200 NF2 = 200 N

F3 = 200 N= -173 N

F3x = F2.sinθ

= 200N.sin30°= +100 N

F3y = F2.cosθ

= 200N.cos30°= -173 N

F2x = 0 NF2y = -200 N

11. What is the net force acting on the ring? (F1 @ 30° to x axis.)

F1 = 200N

F2 = 100NF3 = 200N

11. What is the net force acting on the ring? (F1 @ 30° to x axis.)

F1 = 200N

F2 = 100NF3 = 200N

F1x = F1.cosθ

= 200N.cos 30°=173 N

F1y = F1.sinθ

= 200N.sin30°= 100 N

F2x = 100 NF = 0 N

F3x = - 100 NF3y = 0 N

F2y = 0 NFnetx = 173 + (+100) + (-200) = +73N

Fnety = 100 + 0 + 0 = + 100 N

Fnet = 124 N (pythag)

14. Write expressions for the net horizontal and vertical forces due to the hands acting on the block. The hand on the left pushes downward diagonally.

14. Write expressions for the net horizontal and vertical forces due to the hands acting on the block. The hand on the left pushes downward diagonally.

F2x = F2.cosθ

= F.cos 45°=+ .707F

F2y = F2.sinθ

= F.sin45°= + .707F

F1x = F1.cosθ

= F.cos 45°=+ .707F

F1y = F1.sinθ

= F.sin45°= - .707F = + .707F= - .707F

Fnetx = (+.707F) + (+.707F) = 1.4F

Fnety = (-.707F) + (+.707F) = 0F

Fnet = 1.4F

Weight and Action/Reaction WS

1. I weigh 750 N. What is my mass? Fg = mg

2. What is my mass on the moon?

3. If the acceleration of gravity on the moon, gm = 1.64 m/s2,

what is my weight on the moon?

4. At 3g, 3 times the acceleration of gravity, 4. At 3g, 3 times the acceleration of gravity,

walking becomes impossible.

What is my affective weight at 3g?


750 N = (m)(10 m/s2)

2. What is my mass on the moon? m = 75 kg

m = 75 kg


what is my weight on the moon?





750 N = (m)(10 m/s2)


m = 75 kg


what is my weight on the moon? Fgm = (75kg)(1.64 m/s2)

= 123 N





750 N = (m)(10 m/s2)


m = 75 kg


what is my weight on the moon? Fgm = (75kg)(1.64 m/s2)

= 123 N



What is my affective weight at 3g? Fg3 = (75kg)(3.10 m/s2)

= 2250 N

5. At 12g loss of consciousness is eminent.


6. A 7.5 kg bowling ball is set on the ground.

With what force does the ground push up on the ball?

7. A 5 kg sword is suspended from a rope,

with what force does the rope hold up the sword?

8. Complete the action/reaction pairs:

5. At 12g loss of consciousness is eminent.


Fg12 = (75kg)(12.10m/s2)

= 9000 N






5. At 12g loss of consciousness is eminent. Fg12 = (75kg)(12.10m/s2)

What is my affective weight at 12g? = 9000 N



Fnet = Fg – FN

0 = 75N – FN FN = 75N0 = 75N – FN FN = 75N




5. At 12g loss of consciousness is eminent. Fg12 = (75kg)(12.10m/s2)

What is my affective weight at 12g? = 9000 N



Fnet = Fg – FN

0 = 75N – FN FN = 75N0 = 75N – FN FN = 75N



Fnet = Fg – FN

0 = 50N – FN FN = 50N


Head hits ball, ball hits head, etc.

Ch4.2 - Newton’s Second Law (Fnet = m.a)

Ex1) A 0.142 kg ball left a player’s hand at a speed of 20 m/s. If the straight throw lasted 0.020 sec, determine the magnitude of the force exerted on a ball, assuming it is constant. Fnet=?

HW #29) A bullet is fired from a gun with a 24cm barrel. Its muzzle speed is 350m/s and the mass of the bullet is 6.00g. Compute the average force exerted on the bullet by the expanding gas.

Ex2) The kid in fig 4.14a pulls a loaded wagon having a total mass of 100kg. He applies a constant force of 100N at 30°. Ignoring friction, compute the horizontal force on the wagon and the resulting acceleration.

Ch4 HW#2 p130 20,21,26,29,30,33

Ch4 HW#2 p130, 20,21,26,29,30,3320. Suppose that an air hockey puck, having a mass of 0.25 kg, is pushed along

by a 10.0 N force for 10.0 s. Determine its acceleration.

21. A 60 kg ice skater holds up a large sheet of cardboard that can catch the wind and drive her (frictionlessly) across the ice. While she’s moving at 0.5 wind and drive her (frictionlessly) across the ice. While she’s moving at 0.5 m/s, a wind that is constant and horizontal for 5.0 s exerts a force on the cardboard of 2.0 N. What is the skater’s initial acceleration once the wind begins to blow?

26. Studies show that a male lion (170 kg) accelerates toward prey at about 10 m/s2, which is about the same rate a human sprinter can achieve (compare

that to 3.8 m/s2 for a Porsche). How much force must the lion exert on the ground during such a charge?

m = 170 kga = 10 m/s2

F = ?

30.Suppose a car stopped on the road is hit from behind by a bus so that itaccelerates up to 4.47 m/s in 0.10 s. If the driver of the car has a mass of 50kg and her front-seat passenger has a mass of 80 kg, what average force must the seat exert on them?

vi = 0 m/svf = 4.47 m/s t = 0.10 smtotal = 130 kgFave = ?

33. A bullet fired into wet clay will decelerate fairly uniformly. If a 10 g bullet hits a block of clay at 200 m/s and comes to rest in 20 cm, what average force does it exert on the block?

30.Suppose a car stopped on the road is hit from behind by a bus so that itaccelerates up to 4.47 m/s in 0.10 s. If the driver of the car has a mass of 50kg and her front-seat passenger has a mass of 80 kg, what average force must the seat exert on them? Fnet = ma

vi = 0 m/svf = 4.47 m/s t = 0.10 smtotal = 130 kgFave = ?

N 581.1

1.0

047.4)130(

=

−=

−⋅=

kg

t

vvmF

if

net

33. A bullet fired into wet clay will decelerate fairly uniformly. If a 10 g bullet hits a block of clay at 200 m/s and comes to rest in 20 cm, what average force does it exert on the block?

N 581.1 =

Ch4 Lab4.1 HW12. 500N

200N 300N

100N300N 200N

16. F1 = 200N

F2 =100N

F3 = 100N19.Net force on net:

Top Side view:view:

Ch4 Lab4.1 HW12. 500N

200N 300NFnet = 400N

100N300N 200N

16. F1 = 200N

F2 =100N

F3 = 100N19.Net force on net:

Top Side view:view:

Ch4 Lab4.1 HW12. 500N

200N 300NFnet = 400N

100N300N 200N

16. F1 = 200N

Top view: Side view:

Fnetx = 0 N Each person exerts: Fy = F.sin30°= 100 N

Fnety = 4 x 100 N = 400N

Ch4.3 – Apparent Weight

Ex1) A 50kg person stands on a bathroom scale on earth. Label the forces present. What is the person’s weight?

Ex2) A 50kg person stands on a bathroom scale on earth, in an elevator. The cable slips, and the elevator accl downward at 5m/s2. Label the forces present. What is the person’s apparent weight?

Ex3) elevator accl downward at 10m/s2….

Ex4) elevator accl upwards at 5m/s2….

Ex4) elevator accl upwards at 10m/s2….

Free FallAt Earth’s surface, all objects falling thru a vacuum

accelerate at the same constant rate,

regardless of weight, shape, etc.

Most people believe falling bodies descend at a rate

proportional to their weight.

Galileo started our correct thinking vi = 0 a = 9.8 m/s2

FgFg

Air Drag (contributes to our incorrect thinking)

As an object falls faster, breaks thru more air,

creating a larger resistance force.

Eventually reach the point where

force resisting = force pulling

Fair = Fg

If the net force = 0 the object will continue with

a constant velocity

Terminal speed – depends on

shape, surface area, and weight of object.

vi = 0 a = 9.8 m/s2

Fg


vi = 0 a = 9.8 m/s2

Fg

v a = lessFr






Fair = Fg


a constant velocity


shape, surface area, and weight of object. v a = less

Fg


vi = 0 a = 9.8 m/s2

Fg

v a = lessFr






Fair = Fg


a constant velocity


shape, surface area, and weight of object. v a = less

Fg

vmax a = 0

Fg

Fr


Ch4 HW#3 1 – 4

Ch4 HW#3 1 – 4

A 100kg person stands on a bathroom scale in an elevator.

Draw and label the forces present, set up an Fnet equation, and solve for

what the bathroom scale reads in each picture.

1) a = 0 m/s2 2) a = 2 m/s2 3) 4)

Ch4 HW#3 1 – 4

A 100kg person stands on a bathroom scale in an elevator.

Draw and label the forces present, set up an Fnet equation, and solve for

what the bathroom scale reads in each picture.

1) a = 0 m/s2 2) a = 2 m/s2 3) a = 10 m/s2 4) a = 2 m/s2

Fnet = Fg – FN Fnet = Fg – FN

m.a = 1000N – FN m.a = 1000N – FN

(100)(0) = 1000 – FN (100)(+2) = 1000 – FN

FN = 1000N FN = 800N

Ch4.4 – Incline Planes

Inclined Planes

Ex p105) A 50kg skier coasts along the frictionless snow,

tilted at 30°.

a. Compute the normal force.

b. Compute the force down the incline.

c. Compute her accl.

Ex2) A 15 kg chunk of ice slides down a glacier with an angle of incline of 11°. What is its acceleration?

Ch4 HW #4 1 – 5

Ch4 HW#4 Inclined Planes 1 – 5

1. Find Fg|| and Fg⊥ for a 10 kg mass on incline of 20°.

2. What is the normal force on a 15 kg box set on an incline of 45°?

Ch4 HW#4 Inclined Planes

1. Find Fg|| and Fg⊥ for a 10 kg mass on incline of 20°.

Fg|| = Fg.sinθ

= 100N.sin20°

= 33.5 N

Fg⊥ = Fg.cosθ

= 100N.cos20°

= 92.1 N

2. What is the normal force on a 15 kg box set on an incline of 45°?

Fnet = Fg⊥ – FN

3. What is the acceleration of a 7.5 kg bowling ball down a 40° incline?

4. What is the tension of the spring?


Fnet = Fg||

m.a = Fg.sinθ

a = 10m/s2.sin40°

=6.3 m/s2


Fnet = Fg|| - FTFnet = Fg|| - FT


Fnet = Fg||

m.a = Fg.sinθ

a = 10m/s2.sin40°

=6.3 m/s2


Fnet = Fg|| - FTFnet = Fg|| - FT

0 = Fg.sinθ - FT

FT = 50N.sin30°

FT = 24.5N

5. In which diagram is there a better chance of the board breaking?

25° 15°

5. In which diagram is there a better chance of the board breaking?

25° 15°

Fg⊥ breaks the boardFg⊥ breaks the board

Fg⊥ = Fg.cosθ Fg⊥ = Fg

.cosθ

= Fg.cos25° = Fg

.cos15°

= .91.Fg = .97.Fg

Ch4.7 – Coupled Motion

1. What is the acceleration of this system?

20kg

10kg

2. What is the acceleration of the system? (No friction)

10kg

5kg

3. What is the acceleration of the system? (no friction)

4kg

2kg

2kg

4. What is the tension in each rope?

20kg

10kg

Ch4 HW#5 1 – 5

20kg

Ch4 HW#5 Coupled Motion 1 – 5


8kg

6kg

Ch4 HW#5 Coupled Motion


Fnet = Fg8 – FT + FT – Fg6

mTa = 80N – 60N

(14kg)a = 20N 8kg

6kg

a = 1.4 m/s2


6kg

2kg


Fnet = Fg6 – FT + FT

6kg

2kg

mTa = 60N

(8kg)a = 20N

a = 7.4 m/s2


3kg

2kg

1kg


F = F – F + F + F + F – F

3kg

2kg

1kg

Fnet = Fg3 – FT1 + FT1 + FT2 + FT2 – Fg1

mTa = 30N – 10N

(6kg)a = 20N

a = 3.3 m/s2


5kg 5kg F=25N


F = 25N

Fnet = F – FT1 + FT1 – FT2 + FT2

m a = 25N

5kg 5kg

mTa = 25N

(10kg)a = 25N

a = 2.5 m/s2


30kg

30kg


Fnet = Fg30 – FT1 + FT1 + Fg30 – FT2

mTa = 300N + 300N – FT2

0 = 600N – FT2

FT2 = 600N

F = F – F

30kg

Fnet = Fg30 – FT1

mTa = 300N – FT1

0 = 300N – FT1

FT1 = 300N

30kg

Ch 4.7 – Coupled Motion cont.

Ex 5) What is the accl of the system?

What is the tension in the rope?

20N

5N

Ex6) What is the accl?

What is the tension in both ropes?

4kg 6kg

Ex7) Accl? Tension?

5kg

3kg

10N

10N

5N

Ex8) Accl? Tension?

Ex p106) 50kg Mary and 70kg Don are tied together by a rope of negligible mass. Mary stands on a frictionless sheet of ice, when Don falls..

a. What is the tension in the rope?

b. What is the accl?

c. What would happen if she cut the rope?

Ch4 HW#6

p131 38,45,50,61

Lab4.2 – Mass of a Key

- lab due @ end of period

- go over HW#6 first

Ch4 HW#6 p131, 38,45,50,61

38. What is the smallest force needed to lift a 0.50-kg bullfrog up from the ground, and under what circumstances would that answer be applicable?

45. Someone of mass 100 kg is standing on the top of a steep cliff with only an old 45. Someone of mass 100 kg is standing on the top of a steep cliff with only an old rope that he knows will only

hold 500 N. He plans to slide down the rope using friction to keep him from falling freely. At what minimum rate can he accl down the rope and not break it?

45. Someone of mass 100 kg is standing on the top of a steep cliff

with only an old rope that he knows will only hold 500 N.

He plans to slide down the rope using friction to keep him

from falling freely. At what minimum rate can he accl down

the rope and not break it?

Fnet = Fg – FT

m.a = 1000N – 500N

(100kg)a = 500N

a = 5 m/s2a = 5 m/s2

50. A rescue helicopter lifts 2 people from the sea on an essentially weightless rope. Jamey (100 kg) hangs

15 m below Amy (50.0 kg), who is 5 m below the aircraft. What is the tension:

a. On the top part of the rope? b. Middle?

c. Repeat with the helicopter accl at 9.81 m/s2 upward.

50. A rescue helicopter lifts 2 people from the sea on an

essentially weightless rope. Jamey (100 kg) hangs

15 m below Amy (50.0 kg), who is 5 m below the aircraft.

What is the tension:

a. On the top part of the rope? b. Middle?

c. Repeat with the helicopter accl at 9.81 m/s2 upward.

FgA Fnet = FgJ – FT1 + FT1 + FgA – FT2 Fnet = FgJ – FT1

FT1 0 = 1000N + 500N – FT2 0 = 1000N – FT1

FT2 = 1500N FT1 = 1000NFT2 = 1500N FT1 = 1000N

FgJ Fnet = FgJ – FT1 + FT1 + FgA – FT2 Fnet = FgJ – FT1

FT1 mTa = 1000N + 500N – FT2 mTa = 1000N – FT1

(150).(-9.81) = 1500N – FT2 (100)(-9.81) = 1000N – FT1

FT2 = 3000N FT1 = 2000N

61. The pulley in fig 4.21a is weightless and frictionless.

If m1 = 10.0 kg and m2 = 98.1 N, what is the tension

of the rope and the accl of m1?

Ch4 Lab4.2 HW p131+ 40, 62,64,7140. The gravitational acceleration on the surface of Mercury is

0.38 times the accl on Earth. What is the weight of a 1kg mass?

62. What is tension on right rope if system accl at 0.50 m/s2?

1kg

20kg 10kg F = ?20kg 10kg FT = ?


0.38 times the accl on Earth. What is the weight of a 1kg mass?FgM = m.gM

= (1kg)(.38x9.81m/s2)

= 3.8N


1kg

20kg 10kg20kg 10kg


0.38 times the accl on Earth. What is the weight of a 1kg mass?FgM = m.gM

= (1kg)(.38x9.81m/s2)

= 3.8N


1kg

20kg 10kg

Fnet =FT2 Fnet =FT1 – FT2

mTa = FT2 (30)(.5) = FT1 – 10N (20)(.5) = FT2 FT1 = 25NFT2 = 10N

20kg 10kg

64. Someone pulls down on mass 1 with a force that causes an acclof 9.81 m/s2, what is the tension?

20kg

10kg

71. Tension, Accl

2kg

2kg 4kg

Ch4 Mid Chapter Review p128 18,27,31,49, HW: 68,69,70,7118. 3 people pulling down on ropes, each with 200N, on a 20m tallpost, at a distance of 20m from the base. If the lines are 120° apart,what is the net force on the post?

(top view) (side view)

Ch4 Mid Chapter Review18. 3 people pulling down on ropes, each with 200N, on a 20m tallpost, at a distance of 20m from the base. If the lines are 120° apart,what is the net force on the post?

(top view) (side view)

(cancel out) Fy = F.sinθ= 200N.sin45°= 141N x 3= 424N

27. A 100kg gentleman standing on slippery grass is pulled by his two rather unruly children. One tugs him with a force of 50N north, whilethe other hauls with 120N east. If friction is negligible, what is the man’s accl?

27. A 100kg gentleman standing on slippery grass is pulled by his two rather unruly children. One tugs him with a force of 50N north, whilethe other hauls with 120N east. If friction is negligible, what is the man’s accl?

50N120N

120N50N

Fnet = 130N

Fnet = m.a130N = (100kg).a

a = 1.3 m/s2

31. When a golf club strikes a 0.046kg ball, the ball will attaina speed of 70m/s during the 50 ms collision. Find the averageforce exerted by the club on the ball.

t = .050svi = 0m/s vf = 70m/s

31. When a golf club strikes a 0.046kg ball, the ball will attaina speed of 70m/s during the 50 ms collision. Find the averageforce exerted by the club on the ball.

t = .050svi = 0m/s vf = 70m/s

Fnet = m.aFnet = m.a

N

s

smsmkg

t

vvmF

if

net

4.64

050.

)/0/70()046(.

)(

=

−=

−=

49. A 2000kg car at the top of a 20° incline driveway rolls freelydownhill. At what speed will it hit the garage door 20m down?

Fnet = Fg||

49. A 2000kg car at the top of a 20° incline driveway rolls freelydownhill. At what speed will it hit the garage door 20m down?

Fnet = Fg||

= Fg.sinθ

m.a = 20000N.sin20°(2000kg).a = 6703N

a = 3.35 m/s2

vf2 = vi

2 + 2asvf = vi + 2asvf = 11.6 m/s

68. (Modified) Compute the accl of the system, and tension.

1300N 1800N1300N 1800N

68. (Modified) Compute the accl of the system, and tension.

Fnet = Fg18 – FT + FT – Fg13

mTa = 1800N – 1300N

(310kg)a = 500N

a = 1.6 m/s2

Fnet = Fg18 – FT

(180kg).a = 1800N – FT

F = 1512N1300N 1800N

FT = 1512N1300N 1800N

69. An 80.0 kg man inside a 40.0 kg dumb-waiter pulls down on the rope.

At the moment the scale on which he is standing reads 200 N.

Determine the elevator’s acceleration.

Forces acting on man:

Forces acting on elevator:

69. An 80.0 kg man inside a 40.0 kg dumb-waiter pulls down on the rope.

At the moment the scale on which he is standing reads 200 N.

Determine the elevator’s acceleration.

Forces acting on man:

Fnet = FN – Fg + FT

mTa = 200N – 800N + FT

(80kg)a = – 600N + FT

Forces acting on elevator:

Fnet = FT – Fgtotal

mTa = FT – 1200N(120kg)a = FT – 1200N Substitute!

Ch4 Test #1

- Projectile Motion Review

- Force Vectors

- Fnet eqns

- Incline Planes

- Coupled Motion

Ch4.8 - Friction

- Force that opposes motion

Kinetic Friction (Ff,k) – moving friction

Ch4.8 - Friction



Static Friction (Ff,s) – non-moving friction (prevents object from moving)

(Ff,s) varies between 0 and a max value

- It will equal your force until your force exceeds the max value.

Ch4.8 - Friction



Static Friction (Ff,s) – non-moving friction (prevents object from moving)

(Ff,s) varies between 0 and a max value

- It will equal your force until your force exceeds the max value.

Ff depends on:

1. Types of materials

2. Force pressing surfaces together2. Force pressing surfaces together

Friction doesn’t care about the area of contact or speed!

Formula: FF= µ·FN

HW#73) A dog weighing 300N harnessed to a sled

can exert a max horizontal force of 160N

without slipping. What is the coefficient of

static friction, µs for the dog’s pads on snow?

Ex p113) Someone wants to push a 100kg box full of books along the floor by exerting a constant horizon force of 608N. If µs = 0.6 and µk = 0.1, can the crate move, and, if so, what is its accl?

HW#81) A crate in transported on a flatbed truck. The µs = 0.50 between the truck and crate. If the truck is traveling at 14 m/s, what minimum stopping distance is requited so that the deceleration is small enough and the crate doesn’t move.

Ch 4 HW #7 pg 133 + 73,74,76,81,84,85

Ch4 HW#7 p133, 73,74,76,81,84,8574. Assume your mass is 70.0 kg and that you are wearing leather soled

shoes on a wood floor. Now walk over to a wall and push horizontally on it. How much force can you exert before sliding away?

76. Mass m sits on top of mass m , which is pulled along at a constant 76. Mass m1 sits on top of mass m2, which is pulled along at a constant speed by a horizontal force F.

5kg

10kg

Ch4 HW#7 p133, 73,74,76,81,84,85

74. Assume your mass is 70.0 kg and that you are wearing leather soled shoes on a wood floor. Now walk over to a wall and push horizontally on it. How much force can you exert before sliding away?

Fnet = Fw pushing p – Ff s

0 = F – µ.FN

F = (.3)(700N)

F = 205N

76. Mass m1 sits on top of mass m2, which is pulled along at a constant 76. Mass m1 sits on top of mass m2, which is pulled along at a constant speed by a horizontal force F. µk = .3

5kg

10kg

Ch4 HW#7 p133, 73,74,76,81,84,85

74. Assume your mass is 70.0 kg and that you are wearing leather soled shoes on a wood floor. Now walk over to a wall and push horizontally on it. How much force can you exert before sliding away?

Fnet = Fw pushing p – Ff s

0 = F – µ.FN

F = (.3)(700N)

F = 205N

76. Mass m1 sits on top of mass m2, which is pulled along at a constant speed by a horizontal force F. µk = .3

Fnet = F– Ff k

0 = F – µ.FN

F = (.3)(150N)

F = 45N

5kg

10kg

84. (Mod) Place a block flat on a table and press down on it with your hand. Now suppose coefficient of kinetic friction between the table and block is 0.40. The block’s mass is 1.0 kg and your downward force push on it is 10N. How much horizontal force is needed to pull the block at a constant speed if your hand stays on the block while it is moving?

84. (Exactly) Place a block flat on a table and press down on it with your hand. Slide the block on the table, µk = 0.40 still. Suppose the block slides under the hand while hand remains stationary with respect to the table, and µk = 0.50 between block and the hand. The block’s mass is 1.0 kg and your downward force push on it is 10N. How much horizontal force is needed to keep the block moving at a constant speed?


Fnet = F– Ff k

0 = F – µ.FN1

F = (.4)(20N) F = 8N

84. (Exactly) Place a block flat on a table and press down on it with your hand. 84. (Exactly) Place a block flat on a table and press down on it with your hand. Slide the block on the table, µk = 0.40 still. Suppose the block slides under the hand while hand remains stationary with respect to the table, and µk = 0.50 between block and the hand. The block’s mass is 1.0 kg and your downward force push on it is 10N. How much horizontal force is needed to keep the block moving at a constant speed?


Fnet = F– Ff k

0 = F – µ.FN1

F = (.4)(20N) F = 8N

84. (Exactly) Place a block flat on a table and press down on it with your hand. 84. (Exactly) Place a block flat on a table and press down on it with your hand. Slide the block on the table, µk = 0.40 still. Suppose the block slides under the hand while hand remains stationary with respect to the table, and µk = 0.50 between block and the hand. The block’s mass is 1.0 kg and your downward force push on it is 10N. How much horizontal force is needed to keep the block moving at a constant speed?

Fnet = F– Ff k1 – Ff k2

0 = F – µ.FN1 – µ.FN2

F = (.4)(20N) + (.5)(10N) F = 13N

Ch4.8b - Friction on Incline Planes

Ex 1) A block slides down an incline plane angled at 30° at a constant speed. What is the coefficient of kinetic friction between the block

and the plane?

Ch4.8b - Friction on Incline Planes

Ex 2) A 5kg iron block is placed on a wooden board. The coefficient of static friction between iron and wood is known to be 0.55. At what angle will the block begin to slide?

Ex 3) A 2kg block is placed on a 30° incline plane.

If it is sliding and µk = 0.25, what is its accl?

Ch4 HW#8 Friction Problems

Ch4 HW#8 Friction on Incline Planes

1. A block slides down a carpeted incline plane angled at 45° at a constant speed. What is the coefficient of kinetic friction between the block and the plane?


1. A block slides down a carpeted incline plane angled at 45° at a constant speed. What is the coefficient of kinetic friction between the block and the plane?

Fnet = Fg|| – Ff k

0 = Fg.sinθ – µ.FN

0 = Fg.sinθ – µFg

.cosθ

0 = sinθ – µ.cosθ

µ = 1.0


2. A 7.5 kg iron block is placed on a iron board. µs = 0.6.

a. At what angle will the block begin to slide?

b. Once sliding, if µk = 0.5, what is its acceleration?

c. If the ramp is 2 m long, how fast will the block be sliding at the bottom?

d. If instead you wanted the block to move at constant speed after it had started sliding, what angle would you lower the board to?







a. Fnet = Fg|| – Ff k



.cosθ

0 = sinθ – 0.6.cosθ

θ = 31°

b.







a. Fnet = Fg|| – Ff snet g|| f s



.cosθ


θ = 31°

b. Fnet = Fg|| – Ff k

m.a = Fg.sinθ – µ.FN

m.a = Fg.sinθ – µFg

.cosθ

a = g.sin31° – (0.5)g.cos31°

a = 0.9 m/s2










.cosθ


θ = 31°

b. Fnet = Fg|| – Ff k c. vf2 = vi

2 + 2as d. Fnet = Fg|| – Ff k

m.a = Fg.sinθ – µ.FN vf = 1.9 m/s 0 = mgsinθ - µmgcosθ


.cosθ

a = g.sin31° – (0.5)g.cos31°

a = 0.9 m/s2










.cosθ


θ = 31°

b. Fnet = Fg|| – Ff k c. vf2 = vi

2 + 2as d. Fnet = Fg|| – Ff k

m.a = Fg.sinθ – µ.FN vf = 1.9 m/s 0 = mgsinθ - µmgcosθ


.cosθ θ = 27°

a = g.sin31° – (0.5)g.cos31°

a = 0.9 m/s2

3. A 13 kg block is placed on a 10° incline plane.

If µs = 0.12, µk = 0.06, will the block slide, and if so,

what is its acceleration?

4. A skier on a 4.0°inclined, snow covered run skied downhill at a constant speed. Compute the coefficient of friction between the waxed skies and the snow on that day when the temperature was around 0°C. Why is it that if the temp were to drop to –10°C, µkwould rise to around 0.22? Neglect air friction.

5. A youngster shoots a bottle cap up a 20°inclined board at 2.0 m/s. The cap slides in a strait line, slowing to 1.0 m/s after traveling some distance.

If µk = 0.4, find that distance.

5. A youngster shoots a bottle cap up a 20°inclined board at 2.0 m/s. The cap slides in a strait line, slowing to 1.0 m/s after traveling some distance.

If µk = 0.4, find that distance.

Fnet = – Fg|| – Ff k

m.a = – Fg.sinθ – µ.FN

m.a = – Fg.sinθ – µFg

.cosθa = – g.sin20° – (0.4)g.cos20°a = – g sin20° – (0.4)g cos20°

a = – 7.17 m/s2

vf2 = vi

2 + 2as

0 = 4 + 2(-7.17)s

s = .21 m

Ch4 HW#9 Lab4.3 HW1. A 50 kg box slides down a 30° incline plane at constant speed.

What is the net force acting on the box? What is the net accl of the box? What is µk for the box?

2. A 100 kg trunk loaded with old books is to be slid across a floor by a young woman who exerts a force of 300 N down and forward at 30°with the horizontal. If µk = 0.4 and µs = 0.5, compute accl.

2. A 100 kg trunk loaded with old books is to be slid across a floor by a young woman who exerts a force of 300 N down and forward at 30°with the horizontal. If µk = 0.4 and µs = 0.5, compute accl.

Fnet = Fx – Ff snet x f s

= F.cosθ – µ.FN

= 300N.cos30° – (.5)(1000N+300N.sin30°)= 260N – 575N

No!

3. Suppose the woman in #2 puts aside some of the books so that the mass of the load is 50 kg.

a. Pushing with the same force, what’s the accl now?

b. A bit annoyed, she squirts some oil under the trunk so that

µs= 0.4 and µk= 0.3. What now?

3. Suppose the woman in #2 puts aside some of the books so that the mass of the load is 50 kg.

a. Pushing with the same force, what’s the accl now?

b. A bit annoyed, she squirts some oil under the trunk so that

µs= 0.4 and µk= 0.3. What now?

a. Fnet = Fx – Ff s

= F.cosθ – µ.FN

= 300N.cos30° – (.5)(500N+300N.sin30°)= 260N – 338N= 260N – 338N

No!

b. Fnet = Fx – Ff k

ma = F.cosθ – µk.FN

= 260N – (.3)(500N+300N.sin30°)50a = 65N

a = 1.3 m/s2

4. Determine the force (F) needed to keep the blocks moving at a constant speed: m1 (5N) to the right and m2

(10N) to the left. µk = .40 for all surfaces.

m2

m1

4. Determine the force (F) needed to keep the blocks moving at a constant speed: m1 (5N) to the right and m2

(10N) to the left. µk = .40 for all surfaces.

m2

m1

Fnet = F – Ff k2 – FT + FT – Ff k1

0 = F – µk.FN2 – µk

.FN1

0 = F – (.4)(15N) – (.4)(5N)

F = 8N

Ch4.8c – Friction and Coupled Motion

Ex1) What is the accl of the system? µk = .10 between m2 and table.

What is the tension in the rope?

15

10

Ex2) What is the accl of the system? µk = .22 between m2 and table,

m1 = 5 kg, m2 = 2 kg, m3 = 1 kg. What are the tensions in the ropes?

m3 m1

m2

Ch4 HW#10 1 – 3

Ch4 HW#10 1 – 3

1. A 50g mass hangs over the edge of a table. It is attached to a 25g mass,

on the table, with µk = .5 coefficient of kinetic friction.

What is the accl of the system, what is the tension in the rope?

25

50

Ch4 HW#10 1 – 3

1. A 50g mass hangs over the edge of a table. It is attached to a 25g mass,

on the table, with µk = .5 coefficient of kinetic friction.

What is the accl of the system, what is the tension in the rope?

25

50

Fnet = Fg50 – FT1 + FT1 – Ff k

mT.a = 0.50N – µ.FN Fnet = Fg50 – FT1

(.075kg)a = 0.50N – (.5)(.25N) ma = 0.50N – FT1

a = 5 m/s2 (.050)(5) = 0.50N – FT1

FT1 = 0.25N

Fg50 > FT1 > Ffk

0.50N > 0.25N > 0.125N

2. What is the accl of the system? µk = .15 between m2 and table,

m1 = 10kg, m2 = 3kg, m3 = 3kg. What are the tensions in the ropes?

m3 m1

m2

m3 m1

2. What is the accl of the system? µk = .15 between m2 and table,

m1 = 10kg, m2 = 3kg, m3 = 3kg. What are the tensions in the ropes?

Fnet = FT2 – Fg3

ma = FT2 – 30N

(3)(4) = FT2 – 30FT2 = 42N

m3 m1

m2

Fnet = Fg1 – FT1 + FT1 – Ff k + FT2 + FT2 – Fg3

mT.a = 100N – µ.FN – 30N Fnet = Fg10 – FT1

(16kg)a = 70N – (.15)(30N) ma = 100N – FT1

a = 4 m/s2 (10)(4) = 100 – FT1

FT1 = 60N

m3 m1

3. A 15g mass hangs over the edge of a table. It is attached to a 25 g mass, on the table, with an unknown coefficient of kinetic friction. If the system is moving at constant speed, what is the value of the coefficient?

25

15

Ch4.9d – More Friction and Coupled Motion

Ex1) m1 = 20 kg, m2 = 3 kg, µk = .14, find the accl, tension.

m2

Ex2) Write an expression for µ in terms of the given masses, angles, and tension (T).

θ

M

Ch4 HW#11 1 – 3

Ch4 HW#11 1 – 3

1. m1 = 10 kg, m2 = 4 kg, µk = .2, θ = 30°, find the accl, tension.

m2

Ch4 HW#11 1 – 3

1. m1 = 10 kg, m2 = 4 kg, µk = .2, θ = 30°, find the accl, tension.

m2

Fnet = Fg||1 – Ffk1 – FT1 + FT1 – Fg2

mT.a = m1gsinθ1 – µ.FN1 – m2g

14a = 50N – (.2)87N – 40N

14a = - 7.4N

Whaa, Whaa, no movement FT = ______

2. m1 = 10 kg, m2 = 4 kg, grease the incline µk = .02, θ = 30°,

find the accl, tension.

m2

2. m1 = 10 kg, m2 = 4 kg, grease the incline µk = .02, θ = 30°,

find the accl, tension.

Fnet = Fg||1 – Ffk1 – FT1 + FT1 – Fg2

m .a = m gsinθ – µ.F – m g

m2

mT.a = m1gsinθ1 – µ.FN1 – m2g

14a = 50N – (.02)87N – 40N

14a = 8.3N a = 0.6m/s2

Fnet = FT1 – Fg2

ma = FT1 – 40N

(4kg)(0.6m/s2) = FT1 – 40N FT1 = 42.4N


m1 = 25kg, m2 = 10kg, µk = .25

30° 30°


m1 = 25kg, m2 = 10kg, µk = .25

Fnet = Fg||1 – Ffk1 – FT1 + FT1 – Ff k2 – Fg||2

mT.a = m1gsinθ1 – µ.FN1 – m2gsinθ2 – µ.FN2

35a = 125N – (.25)217N – 50N – (.25)(87N)

a = 1.17 m/s2

Ch4.9 - Statics

Net Force = 0 (Object either at rest or moving at vcon)

Ex p119) Engine weighs 800N. θ = 20°, compute:

A) Tension in each rope.

B) Horizontal force pulling on pins.

Ex p121) Determine the angle in arrangement. (Fig 4.38)

θ 20°

scale = 150N

200N

Ex p122) A 400 N sign is suspended from the end of a 2.0 m long, negligibly light bar, pivoting on wall. The bar is held up by a rope, at 50° with the bar, with 0.70kN of tension. The bar is 2.00m long, and the rope is attached 0.50m from its end. Find the reaction force on the pivot on the wall.

50°

Ch4 HW#11

400N

Ch4 HW#11 p134 95,97,101,107,109

95. What are the tension forces acting on each of the weightless ropes?

10kg

10kg

Ch4 HW#11 p134 95,97,101,107,109

95. What are the tension forces acting on each of the weightless ropes? FT2

FnetTop = Fg1 – FT1 + FT1 + Fg2 – FT2 FT1 Fg2

0 = 100N + 100N – FT2 FT1

FT2 = 200N

F

10kg

10kg

Fg1

FnetBottom = Fg1 – FT1

0 = 100N – FT1

FT1 = 100N

97. The 10kg block is held at rest by the four ropes shown. If the tension in the ropes on the right and top are each 98N. What is the tension in the remaining two ropes?

97. The 10kg block is held at rest by the four ropes shown. If the tension in the ropes on the right and top are each 98N. What is the tension in the remaining two ropes?

FT = 98N

FR = 98N

Fnetx = FL – FR Fnety = Fg + FTbottom – FTtop

10kg

Fnetx = FL – FR Fnety = Fg + FTbottom – FTtop

0 = FL – 98N 0 = 98N + FTB – 98N

FL = 98N FTB = 0N

99. Determine the scale reading in the right arm of the suspension and the angle θ.

30N

20N 30° θ F = ?

99. Determine the scale reading in the right arm of the suspension and the angle θ.

20N 30° θ F = ?

F = F – F F = F – F – F

30N

Fnetx = FTLX – FTRX Fnety = Fg – FTLY – FTRY

0 = 20.sin30°– FTRX 0 = 30N – 20N.cos30°– FTRY

FTRX = 10N FTRY = 12.7N

θ = tan-1 (10/12.7) = 38°

101. In the static arrangement, the pulleys and ropes are essentially weightless. Determine the value of m3. Check that the system is in equilibrium.

15N 31Nm3

45°20°

101. In the static arrangement, the pulleys and ropes are essentially weightless. Determine the value of m3.

45° 20°

15N 31Nm3

FnetY = Fg3 – FT15Y – FT31Y

0 = Fg3 – (15N)(cos45) – (31N)(cos20)

Fg3 = 39.7N

107. Determine the weight of the mass m . Assume the pulleys and ropes are all essentially weightless. Both ropes make 20° angles to the horizontal.

1kg 1kg

m

107. Determine the weight of the mass m . Assume the pulleys and ropes are all essentially weightless.

(FT = Fg1 = 10N)

Fnet = Fgm – 2FTY FTY FTY

0 = Fgm – 2(10N.sin20°) FT

Fgm = 6.7N Fgm

Fg1

1kg 1kg

m

Fg1

109. A 533.8-N tightrope walker dances out to the middle of a 20-m-long wire stretched parallel to the ground between two buildings. She is wearing a pink tutu, of negligible weight; the wire sags, making a 5.0° angle on both sides of her feet with the horizontal. Find the tension.

109. A 533.8-N tightrope walker dances out to the middle of a 20-m-long wire stretched parallel to the ground between two buildings. She is wearing a pink tutu, of negligible weight; the wire sags, making a 5.0° angle on both sides of her feet with the horizontal. Find the tension.

5°

Fnet = Fg – 2FTY

0 = 533.8N – 2(FTsin5)

FT = 3062N

Ch4.9 cont.

Ex1) What is the tension in the rope?

What is force on each wall attachment?

400N

35°

HW#104) What is the tension in each length of rope, if the object is at rest?

HW#108) This apparatus is set up, and when the 100N mass is attached

to the rope, the rope sags 10cm. Determine the spring scale reading.

Ch4 HW#13 p135+ 103,104,108,110,111,113,114,116

Lab4.4 – Statics

- due tomorrow

- Ch4 Rev#2 due tomorrow

- Go over Ch4 HW#13 103,110,111,113 today

(114,116 with review)(114,116 with review)

Ch4 HW#13 p135+ 103,104,108,110,111,113,114,116

103) If the hand pulls down so that the 10N weight remains at rest,

what is the tension in the long rope?

What force is exerted down on the right-most ceiling bracket?

110) If a force of 100N holds the weight motionless, what is its mass?

What is the tension in rope 1?

111) Determine both the angle at which the pulley hangs

and the tension in the hook supporting it?

113) Figure P113 depicts a rod of negligible mass pinned to a wall at one end and supported by a rope at the other end. The load is 2.00kN and the scale reads 1.00kN. Compute the horizontal and vertical reaction forces on the rod at the wall.

114) A very light-weight beam is attached to a wall via a pinned bracket

if the load is 2.00kN, determine the reading of the spring scale

and the force exerted by the wall on the beam.

116) The three bodies are at rest. If a fly lands on the 80.0 kg mass and the system then begins to move what is the static coefficient of friction for the box and surface? The pulleys are weightless and frictionless.

Ch4 HW#13 p128+ 15,25,35,78

15. What is the net force acting on the ring?

300N 400N

50° 40°

Ch4 HW#13 p128+ 15,25,35,78

15.What is the net force acting on the ring?

Elegant way:

300N 400N Fnet = 500N

50° 40° θ = 40° + tan-1(300/400)

= 77°

Not so elegant way:

Fnetx = F400X – F300X FnetY = F400Y + F300Y

= 400N.cos40° – 300N.cos50° = 400N.sin40° – 300N.sin50°

Fnet = FnetX + FnetY (pythag)

θ = tan-1 (FnetY/FnetX)

25. A kid on an old dirt bike is being pushed along at constant speed

by her father w/ 10N force downward at 30°.

Combined mass of kid+bike is 30kg, find the total retarding force.

30kg30°30°

F = 10N

25. A kid on an old dirt bike is being pushed along at constant speed

by her father w/ 10N force downward at 30°.

Combined mass of kid+bike is 30kg, find the total retarding force.

FN

(FX = Ff)

(FN = Fg + FY)

Ff FX

30°

30kg30°

FY F=10N

Fg

Fnet = FX – Ff

0 = F.cos30°– Ff

Ff = 8.7N

78. A 30kg youngster is dragged around the living room

at a constant speed via a 60N force. Find the coeff

of friction.

30kg F = 60N

78. A 30kg youngster is dragged around the living room

at a constant speed via a 60N force. Find the coeff

of friction. FN

Ff k F = 60N

30kg

Fg Fnet = F – Ff

0 = 60N – µ.FN

0 = 60N – µ(300N)

µ = .2

87. A 100kg bale of hay falls from a truck travelling at 24.4m/s.

It skids for 100m before coming to a stop. Find µ.

100kg

87. A 100kg bale of hay falls from a truck travelling at 24.4m/s.

It skids for 100m before coming to a stop. Find µ.

vi = 24.4m/s vf = 0

s = 100m

vf2 = vi

2 + 2as a = – 3 m/s2

100kg

Fnet = – Ff k

ma = – µ.FN

(100kg)(–3m/s2) = – µ(1000N)

µ = 0.3

Bonus#1) A snowboarder starts at rest, and slides down a 30°

incline with a coefficient of friction = 0.22.

How fast is he going after 10s?

Bonus #2) Find accl of system, tension in rope,

(µ = 0.3)

10g

50g

Bonus #3) What is the reading of scale 2?

50kg

400N 20°40° F = ?

A single fixed pulley doesn’t change the force, only the direction!

100N 100N 100N 100N 100N 100N

100N 100N

35. A 4265 lb Jag requires 164ft stopping distance from a speed of 60mph.

Compute ave stopping force.

Jag

35. A 4265 lb Jag requires 164ft stopping distance from a speed of 60mph.

Compute ave stopping force.

Ff vi = 60 s = 164

vf = 0

vf2 = vi

2 + 2as a = – 11

Fnet = – Ff

ma = – F

Jag

ma = – Ff

(4265)(–11) = – Ff Ff = 46,800N

F1 = 25NF2 = 20N

F = 20N

o50o

20

20g

F3 = 20N

5kg

4kg

1kg

1.0kg

40°

F = 20N

M

2M

6.05kg

50N 35° 42° F = ?

15N 40° θ 17N

7kg | | | | | | |

2M

M

Documents

AP Phys1 Ch4