Ch2 p50 23,2723. Figure P23 is a plot of the speed of a cat

versus time. How far did the cat travel during the third second of its journey? What were its maximum and minimum speeds? When, if ever, was its speed a nonzero constant?

27. Figure P27 shows the distance traveled versus time for a toy car. What was the toy car’s average speed during the time interval from 2.0 s to 8.0 s?

6

5

speed 4

(m/s) 3

2

1

1 2 3 4 5 6 7 time (sec)

7

6

5

dist 4

(m) 3

2

1

1 2 3 4 5 6 7 8 time (sec)

Ch2 p50 23,2723. Figure P23 is a plot of the speed of a cat

versus time. How far did the cat travel during the third second of its journey? What were its maximum and minimum speeds? When, if ever, was its speed a nonzero constant?Max: 4-5secMin: 0,7sec

Const @: 1.5-2, 2-3, 4-5,

27. Figure P27 shows the distance traveled versus time for a toy car. What was the toy car’s average speed during the time interval from 2.0 s to 8.0 s?

6

5

speed 4

(m/s) 3

2

1

1 2 3 4 5 6 7 time (sec)

7

6

5

dist 4

(m) 3

2

1

1 2 3 4 5 6 7 8 time (sec)

smm

t

xv

total

totalave /66.

sec6

4

Ch2 p52+ 71,7871. A toy electric train runs along

a straight length of track. Its displacement vs time curve is shown. Is the train’s velocity ever constand if so, when? What is its inst vel at t = 2.0 s and at t = 6.5 s? Did it change direction if so, when?

78. A trolley runs along a straight run of track the graph is the plot of its vel vs time. Approx how far did it travel in the first 3.0 s of its journey? How far from starting point is it at t = 6.0 s?

7

6

5

dist 4

(m) 3

2

1

1 2 3 4 5 6 7 8 time (sec)

6

5

speed 4

(m/s) 3

2

1

-1

-2

1 2 3 4 5 6 7 time (sec)

Ch2 p52+ 71,7871. A toy electric train runs along

a straight length of track. Its displacement vs time curve is shown. Is the train’s velocity ever constand if so, when? What is its inst vel at t = 2.0 s and at t = 6.5 s? Did it change direction if so, when?

78. A trolley runs along a straight run of track the graph is the plot of its vel vs time. Approx how far did it travel in the first 3.0 s of its journey? How far from starting point is it at t = 6.0 s?

7

6

5

dist 4

(m) 3

2

1

1 2 3 4 5 6 7 8 time (sec)

6

5

speed 4

(m/s) 3

2

1

-1

-2

1 2 3 4 5 6 7 time (sec)

vcon @ 0-4, 5-6v2 = 1 m/s, v6.5 = -2 m/sChange dir @ 4.5 sec

s3 = (2 . 4) = 8 m

s6 = 8 – (1 . 2) = 6 m

18. Fig P18 is a vel-time graph for a test car on a straight track. The test car initially moved backward in the negative x-direction at 20 m/s. It slowed, came to a stop, and then moved off in the positive x-direction at t = 2.0 s. What was its ave accl during each of the

time intervals 0 to 0.5 s, 1.5 to 2.0 s, and 2.0 to 2.5 s?

What was its instantaneous acceleration at t = 2.25 s?

40

30

20

vel 10

(m/s) 0

-10

-20

1 2 3 4 t (sec)

18. Fig P18 is a vel-time graph for a test car on a straight track. The test car initially moved backward in the negative x-direction at 20 m/s. It slowed, came to a stop, and then moved off in the positive x-direction at t = 2.0 s. What was its ave accl during each of the

time intervals 0 to 0.5 s, 1.5 to 2.0 s, and 2.0 to 2.5 s?

What was its instantaneous acceleration at t = 2.25 s?

40

30

20

vel 10

(m/s) 0

-10

-20

1 2 3 4 t (sec)

0 to 0.5 s 0, 1.5 to 2.0 s 0, 2.0 to 2.5 s 0

Inst accl at t = 2.25 s (slope) = 40/0.5 = 80 m/s2

Ch3 p81 4747. The length of a straight tunnel through

a mountain is 25.0 m. A cyclist heads directly toward it, accelerating at a constant rate of 0.20 m/s2. If at the instant he enters the tunnel he is traveling at a speed of 5.00 m/s, how fast will he be moving as he emerges?

vi = 5 m/s vf = ? a = 0.20 m/s2

s = 25 m

Ch3 p81 4747. The length of a straight tunnel through

a mountain is 25.0 m. A cyclist heads directly toward it, accelerating at a constant rate of 0.20 m/s2. If at the instant he enters the tunnel he is traveling at a speed of 5.00 m/s, how fast will he be moving as he emerges?

vi = 5 m/s vf = ? a = 0.20 m/s2

s = 25 m

vf2 = vi

2 + 2as = 52 + 2(0.20)(25) = 5.9 m/s

vf = 98.1 m/s

a = - 9.8 m/s2

t = 10 sec vf = ? s = ?

a) vf = vi + at

Ch3 p82+ 82#82) An arrow is launched vertically upward from a crossbow at 98.1 m/s. a) What is the inst speed at the end of 10 sec? b) What is its ave speed to that moment? c) How high has it risen? d) What is the inst accl 4.10 sec into flight?

b) vave = ½(vi + vf)

c) s = vave.t d) a =

vf = 98.1 m/s

a = - 9.8 m/s2

t = 10 sec vf = ? s = ?

a) vf = vi + at

= 98.1 + (-9.8)(10)

= 0 m/s

(at the top of its path)

Ch3 p82+ 82#82) An arrow is launched vertically upward from a crossbow at 98.1 m/s. a) What is the inst speed at the end of 10 sec? b) What is its ave speed to that moment? c) How high has it risen? d) What is the inst accl 4.10 sec into flight?

b) vave = ½(vi + vf)

= ½(98.1 + 0)

= 49 m/s

c) s = vave.t

= 49 m/s . 10 sec

= 490 m

d) a = - 9.8 m/s2

93. A raw egg is thrown horizontally straight out of the open window of a fraternity house. If its initial speed is 20 m/s and it hits ground 2.0 s later, at what height was it launched? How far out from the house does it hit?

vix = 20 m/s

t = 2 sec

sy = ?

sy =

sx =

93. A raw egg is thrown horizontally straight out of the open window of a fraternity house. If its initial speed is 20 m/s and it hits ground 2.0 s later, at what height was it launched? How far out from the house does it hit?

vix = 20 m/s

t = 2 sec

sy = ?

sy = viyt + ½at2

= 0 + 4.9(2)2

= 19.6 m

sx = vx.t

= (20m/s)(2s) = 40m

Ch425. 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.

30° F=10N

Fnet =

30kg

Ch425. 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°FY F=10N

Fg

Fnet = FX – Ff 0 = F.cos30°– Ff Ff = 8.7N

30kg

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

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

120N 50N

Fnet = 130N

Fnet = m.a130N = (100kg).a a = 1.3 m/s2

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 forwhat the bathroom scale reads in each picture.

2) a = 2 m/s2 4) a = 2 m/s2

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 forwhat the bathroom scale reads in each picture.

2) a = 2 m/s2 4) a = 2 m/s2

Fnet = Fg – FN Fnet = Fg – FN m.a = 1000N – FN m.a = 1000N – FN

(100)(+2) = 1000 – FN (100)(–2) = 1000 – FN FN = 800N FN = 1200N

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 =

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 = 11.6 m/s

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 = 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

71. Tension, Accl

2kg

2kg 4kg

71. Tension, Accl

Fnet = Fg4 – FT1 + FT1 + FT2 + FT2 – Fg2 mTa = 40N – 20N (8kg)a = 20N a = 2.5 m/s2

2kg

2kg 4kg

Fnet = Fg40 – FT1 mTa = 40N – FT1 (4)(2.5) = 40N – FT1 FT1 = 30N

Fnet = FT2 – Fg2 mTa = FT2 – 20N (2)(2.5) = FT2 – 20N

FT2 = 25N

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 =

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

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

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?

Fnet = Fg15 – FT1 + FT1 – Ff k mT

.a = 0.15N – μ.FN (0.04kg)(0) = 0.15N – (μ)(0.25N) μ = 0.60

25

15

Ch61. A force is gradually placed on an object, uniformly increased to 20 N over a distance of 5

m. It then stays constant for the next 5 m. How much work was done on this object?

10

20

F (N)

5 10s (m)

A = ½.b.h = 50 NmA = b.h = 100 NmAtotal = 150 Nm

MEi = MEf 6.4 m

o30

74. An inclined plane at 30.0º is 6.40 m long. A book, which has a kinetic coefficient of friction with the incline of 0.20, is placed at the top and immediately begins to slide. Using energy considerations, how long will it take for the book to reach the bottom of the incline?

MEi = MEf PEi = KEf + Wfric

mgh = ½mv2 + Ff.s

mgh = ½mv2 + µFN.s

mgh = ½mv2 + µFg┴.s

mgh = ½mv2 + µmgcosθ .s

gh = ½v2 + µgcosθ .s

v = 6.4 m/s

s = ½(vi + vf)t t = 2 sec

6.4 miPE

fKEo30

fricW

74. An inclined plane at 30.0º is 6.40 m long. A book, which has a kinetic coefficient of friction with the incline of 0.20, is placed at the top and immediately begins to slide. Using energy considerations, how long will it take for the book to reach the bottom of the incline?

78. How much power does it take to raise an object weighing 100N a distance of 20.0m in 50s?

13. A 47 g golf ball is hit in the air at 60 m/s. It lands in sand at the same elevation and comes to rest in 10 ms. Ignoring air friction, what was the average force the sand exerted on the ball?

vi = 60m/s vf = 0m/s

ptF

47. A 1263 kg Triumph TR-8 sports car traveling south at 40 km/h (11.6 m/s) crashes head-on into a 1742 kg Checker cab moving north at 90 km/h (25 m/s). If the two cars remain tangled together but free to coast, describe their motion immediately after collision. What is their final speed? How much kinetic energy is lost?

pi = pf

m1v1i + m2v2i = (m1 + m2)vf

KEi KEf ½m1v1i

2 + ½m2v2i2 ½mTvf

2

54. Two identical blocks, each of mass 10.0 kg, are to be used in an experiment on a frictionless surface. The first is held at rest on a 20.0° inclined plane 10.0m from the second, which is at rest at the foot of the plane. a. What is speed of m1 at bottom of incline?b. m1 slams into and sticks to m2 what is their speed immediately after impact.

c. In exp, they don’t stick, m2 = 6.1 m/s. m1 speed? a)

b)

h = 10m.sinθ = 3.42m c)

h = 3.4m

10 mm1

o20 m2

54. Two identical blocks, each of mass 10.0 kg, are to be used in an experiment on a frictionless surface. The first is held at rest on a 20.0° inclined plane 10.0m from the second, which is at rest at the foot of the plane. a. What is speed of m1 at bottom of incline?b. m1 slams into and sticks to m2 what is their speed immediately after impact.

c. In exp, they don’t stick, m2 = 6.1 m/s. m1 speed? a) PEi = KEf

mgh = ½mv2

vf = 8.3 m/s

b) pi = pf

m1v1i + m2v2i = (m1 + m2)vf

(10)(8.3) + 0 = (20)vf vf = 4.15m/s

h = 10m.sinθ = 3.42m c) pi = pf

m1v1i + m2v2i = m1v1f + m2v2f (10)(8.3) + 0 = (10)v + (10)

(6.1) v1f = 2.2m/s

h = 3.4m

10 mm1

o20 m2

5. 700kg flat track car rounds a corner of radius 30m at a speed of 20m/s, without slipping. What is the minimum coefficient of static friction that can accomplish this?

Fnet = Ff,s

m.ac = μ.Fn

37. The acceleration due to gravity on the surface of mars is 3.7 m/s2. If the planet’s diameter is 6.8x106 m, determine the mass of the planet and compare it to Earth.

Fnet = FG

26

11

2

)104.3(

)1067.6(7.3

x

Mx

R

mGMam

m

m

m

Mm = 6.4x1023 kg

v = 40 m/s

Fnet = FN – Fg

mac = FN – 800N

NFN 2080800100

)40)(80( 2

v = 40 m/s

3. A scale is fitted in the seat of a roller coaster car and a person weighing 800 N sits down on it. The car then descends along a path that has the shape of a 100.0 m radius vertical circle with its lowest point at the bottom where the car reaches its greatest speed of 40.0 m/s. What is the maximum reading of the scale?

FN

Fg

49. A wheel is revolving at 20 rad/s when a brake is engaged and the wheel is brought to a stop in 15.92 rev. How much time elapsed and what was the angular deceleration?

ωi = 20 rad/s ωf = 0 rad/s θ = 15.92 rev 100 rad α = ? t = ?

68. The bridge has a uniform weight of 20.0 kN. Calculate forces a, b. FgB = ?

15m 35m 15m 20m

Pick a fulcrum, I picked A

Fg8000 Fg8000

Fg20000

Tnet = Fg8000.r + FgB

.r + (-Fg20000.r) + (-Fg8000

.r) 0 = (8000N)(15m) + FgB

.(70m) – (20000)(35) – (8000N)(50m)

FgB = 14,000NFgA = 36,000N – 14,000N = 22,000N

a b

Ch8 HW#5 p282 69,72, +2 Bonus Questions69. The beam is of negligible mass, what value does the scale read?

2 1

9. A 10kg solid steel cylinder with a 10cm radius is mounted on bearings so that it rotates freely about a horizontal axis. Around the cylinder is wound a number of turns of a fine gold thread. A 1.0kg monkey named Fred holds on to the loose end and descends on the unwinding thread as the cylinder turns. Compute Fred’s acceleration.

126. A solid cylinder of mass 2.0 kg rolls without slipping down a long curved track from a height of 10.0 m. Calculate the linear speed with which it leaves the track.

PE = KEt + KEr mgh = ½mv2 + I.ω2