AP PhysicsChapter 6 Review

€

Fc =mv2

R=

80Kg( ) 40 msec( )2

200m= 640N

1. A race car travels 40 around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of the resultant force on the 80kg driver of this car?

€

msec

2

€

Fc =mv 2

R=

70Kg( ) 140 msec( )2

1000m=1372N

3

2. An airplane moves 140 as it travels around a vertical circular loop which has a 1.0-km radius. What is the magnitude of the resultant force on the 70-Kg pilot of this plane at the bottom of this loop?

€

msec

€

Fc =4π2Rm

T 2 =4π2 20m( ) 30Kg( )

22sec( )2

Fc = 48.94N

4

3. A 30-Kg child rides on a circus Ferris wheel that takes her around a vertical circular path with a radius of 20 meters every 22 seconds. What is the magnitude of the resultant force on the child at the highest point on this trajectory?

4. An amusement ride consists of a car moving in a vertical circle on the end of a rigid boom. The radius of the circle is 10 m. The combined weight of the car and riders is 5.0kN. At the top of the circle the car has a speed of 5.0 which is not changing at that instant. What is the force of the boom on the car at the top of the circle?

€

msec

N

mg

€

Fc = mg − N

N = mg − Fc = mg − mv2

R

N = m g− v 2

R

= 510.2Kg 9.8 m

sec2 −5 m

sec( )2

10m

N = 3725N, up

6

N

mg

€

m =12000N

9.8 NKg

=1225Kg

€

f = Fc =mv 2

R=

1225Kg( ) 24 msec( )2

140m= 5038N

5. A unbanked highway curve has a radius of 0.14km. A car weighing 12kN goes around the curve at a speed of 24 without slipping. What is the magnitude of the horizontal force of the road on the car?

€

msec

7

€

Fc =mv 2

R=

4Kg( ) 2 msec( )2

0.8m= 20N

6. A 4.0-Kg mass on the end of a string rotates in a circular motion on a horizontal frictionless table. The mass has a constant speed of 2.0 and the radius of the circle is 0.80 meters. What is the magnitude of the resultant force acting on the mass?

€

msec

8

8. A stunt pilot weighing 0.70 kN performs a vertical circular dive of radius 0.80 km. At the bottom of the dive, the pilot has a speed of 0.20 which at that instant is not changing. What force does the plane exert on the pilot?

€

Kmsec

N

mg

€

m =Wg

=700N9.8 N

kg

= 71.4kg

9

#8

€

Fc = N −mg

N = Fc + mg =mv2

R+ mg

N = m v 2

R+ g

= 71.4Kg

200 msec( )2

800m+ 9.8 m

sec2

N = 4270N, up

N

mg

10

€

tanθ =v2

Rg

θ = tan−1 25 msec( )2

150m( ) 9.8 msec2( )

θ = 23°

11

9. A car travels around a banked highway curve (radius 0.15km) at a constant speed of 25 . What is the angle of the bank if the 1600N car and driver do not need friction to negotiate the curve?

€

msec

10. A 0.50kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m). When the mass is at the lowest point on the circle, the speed of the mass is 12 . What is the force of the string on the mass at this position?

€

msec

T

mg

€

Fc = T −mg

T = Fc + mg =mv2

R+ mg

T = m v 2

R+ g

= 0.5Kg

12 msec( )2

2m+ 9.8 m

sec2

T = 40.9N, up12

11. A roller-coaster car has a mass of 500kg when fully loaded with passengers. The car passes over a hill of radius 15 m, as shown. At the top of the hill, the car has a speed of 8.0 . What is the force of the track on the car at the top of the hill?

€

msec

€

8 msec

€

Fc = mg −N

N = mg− Fc = mg− mv2

R

N = m g − v 2

R

= 500Kg 9.8 m

sec2 −8 m

sec( )2

15m

N = 2766.6N , up€

mg€

N

13

€

Fc = T + mg

T = Fc −mg =mv 2

R−mg

T = m v 2

R− g

= 0.2Kg

4.5 msec( )2

0.8m−9.8 m

sec2

T = 3.1N

mg

14

12. A 0.20Kg object attached to the end of a string swings in a vertical circle (radius = 80cm). At the top of the circle the speed of the object is 4.5 . What is the magnitude of the tension in the string at this position?

T€

msec

15

13. A roller-coaster car has a mass of 500Kg when fully loaded with passengers. At the bottom of a circular dip of radius 40m (as shown in the figure) the car has a speed of 16 . What is the force of the track on the car at the bottom of the dip?

€

msec

€

16 msec →

€

Fc = N −mg

N = Fc + mg =mv 2

R+ mg

N = m v 2

R+ g

= 500Kg

16 msec( )2

40m+ 9.8 m

sec2

N = 8100N, up€

mg€

N

€

Fc =mv 2

R= 7.71N

€

T = mv2

R+ gcosθ

T = 0.3kg6 m

sec( )2

1.4m+ 9.8 m

sec2( )cos120°

T = 6.24N

14. A 0.30Kg mass attached to the end of a string swings in a vertical circle (R = 1.4 meters), as shown. At an instant when θ = 30°, the speed of the mass is 6.0 . What is the magnitude of the Tension on the mass at this instant?

€

msec

€

θ

€

R

€

m

€

g

16

€

Fc = T −mgcosθ

Fc = 8N − 0.3Kg( ) 9.8 NKg( ) cos50°( )

Fc = 6.11N

15. A 0.30Kg mass attached to the end of a string swings in a vertical circle (R = 1.6 meters), as shown. At an instant when θ = 50° with the vertical, the tension in the string is 8.0N. What is the magnitude of the resultant force on the mass at this instant?

€

v

€

R

€

θ

€

g

17

€

T = m v2

R+ gcosθ

m =T

v 2

R+ gcosθ

=20N

5 msec( )2

1.2m+ gcos240°

m =1.26Kg

18

16. An object attached to the end of a string swings in a vertical circle (R = 1.2 m), as shown. At an instant when θ = 30°, the speed of the object is 5.0 and the tension in the string has a magnitude of 20 N. What is the mass of the object?

€

msec

€

θ

€

R

€

v

€

Fc = mg− N

N = mg− Fc = mg− mv 2

R

N = m g − v 2

R

= 50Kg 9.8 m

sec2 −6.28 m

sec( )2

10m

N = 293N, up€

v =2πR

T

v =2π 10m( )10sec

v = 6.28 msec

19

17. A 50-Kg child riding a Ferris wheel ( R =10 m) travels in a vertical circle. The wheel completes one revolution every 10 seconds. What is the force on the child by the seat at the highest point on the circular path?

€

N

€

mg

18. A 0.40-Kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40° below the horizontal, the speed of the mass is 5.0 . What is the tension in the string at this instant?

€

msec

20

T

€

mgcosθ€

50°

€

Fc = T − mgcosθT = Fc + mgcosθ

T =mv 2

R+ mgcosθ

T = m v 2

R+ gcosθ

T = 0.4Kg5 m

sec( )2

1.8m+ 9.8 m

sec2( )cos50°

T = 8.07N

T

€

mgcosθ€

50°

21

€

Fc = T − mgcosθ but, cos90° = 0

Fc = T =mv2

R=

0.5Kg( ) 8 msec( )2

2m=16N

22

19. A 0.50-Kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m). When the string is horizontal, the speed of the mass is 8.0 . What is the magnitude of the force of the string on the mass at this position?

€

msec

€

T = m v2

R+ gcosθ

T = 4Kg5 m

sec( )2

2m+ 9.8 m

sec2( )cos35°

T = 82N

T

€

mgcosθ€

35°

23

20. A 4.0Kg mass attached to the end of a string swings in a vertical circle of radius 2.0m. When the string makes an angle of 35° with the vertical as shown, the speed of the mass is 5.0 . At this instant what is the magnitude of the force on the mass by the string?

€

msec

€

Fc = mg−N

N = mg − Fc = mg − mv 2

R

N = m g − v 2

R

=

128lbs32 ft

sec2

32 ftsec2 −

20 ftsec( )

2

40 ft

N = 88 lbs, up

21. A 128-lb girl, riding a Ferris Wheel, moves at a constant speed of 20 around a vertical, circular path 80 feet in diameter. What force does the seat exert on her at the top of the circle?

ftsec

€

N

€

mg

24

30 lbs

75 lbs

ƒ

NG

NW

€

Fx = 0 Fy = 0∑∑

€

τCW =τCCW

25

22. A 30 pound, uniform ladder 15 ft long is leaning against a frictionless wall at an angle 53° above the horizontal. If a 75 pound boy climbs 6.0 feet up the ladder, what is the magnitude of the friction force exerted on the ladder by the floor?

€

53°

€

τCW = τCCW

75lbs( ) 6 ft( )cos53° + 30lbs 7.5 ft( )cos53° = f 15 ft( )sin53°

75lbs( ) 6 ft( ) 35

+ 30lbs 7.5 ft( ) 35

= f 15 ft( ) 45

270 lbs +135 lbs12

= f = 33.75 lbs

#22.2

26

27

€

f = Fc

µmg =mv 2

RµRg = v

v = 1.1( ) 400m( ) 9.8 msec2( )

v = 65.7 msec