Download ExamView - S15--Physics Q4 Torque.tst

Name: ________________________ Class: ___________________ Date: __________
ID: A
S15--Physics Q3 Torque
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1. A ball is whirled on a string, then the string breaks. What causes the ball to move off in a straight line?
a. centripetal acceleration b. inertia c. centrifugal force d. centripetal force
2. When an object is moving with uniform circular motion, the centripetal acceleration of the object
a. is circular. b. is directed toward the center of motion. c. is perpendicular to the plane of motion. d. is
zero.
3. A meter stick is balanced at the 50.0-cm mark. You tie a 10.0-N weight at the 15.0-cm mark. Where should a
30.0-N weight be placed so the meter stick will again be balanced?
a. 10.0-cm mark b. 30.0-cm mark c. 15.0-cm mark d. 25.0-cm mark e. 61.7-cm mark
4. What are the SI units for momentum?
a. kg•m/s 2 b. J c. N•m d. kg•m/s
5. A heavy bank-vault door is opened by the application of a force of 3.0 × 10 2 N directed perpendicular to the plane
of the door at a distance of 0.80 m from the hinges. What is the torque?
a. 240 N•m b. 360 N•m c. 300 N•m d. 120 N•m
6. Where should a force be applied on a lever arm to produce the most torque?
a. in the middle of the lever arm b. closest to the axis of rotation c. It doesn’t matter where the force is
applied. d. farthest from the axis of rotation
7. If you cannot exert enough force to loosen a bolt with a wrench, which of the following should you do?
a. Use a wrench with a longer handle. b. You should exert a force on the wrench closer to the bolt. c. Tie a
rope to the end of the wrench and pull on the rope. d. Use a wrench with a shorter handle.
8. Suppose a doorknob is placed at the center of a door. Compared with a door whose knob is located at the edge,
what amount of force must be applied to this door to produce the torque exerted on the other door?
a. four times as much b. one-fourth as much c. two times as much d. one-half as much
9. Suppose you try loosening a nut with a wrench, and the nut doesn't give at all. You increase your chance of
success if you _____.
a. be sure to exert force perpendicular to the lever arm b. have a friend help you pull on the wrench c. extend
the lever arm d. exert a larger force e. all of the above
10. Torque is defined as _____.
a. momentum times radius b. mass times acceleration c. force times time d. force times lever arm e. mass
times velocity
11. If you want to open a swinging door with the least amount of force, where should you push on the door?
a. close to the hinges b. as far from the hinges as possible c. It does not matter where you push. d. in the
middle
12. Two people sit on a balanced seesaw. When one person leans toward the center of the seesaw, that person's end of
the seesaw will _____.
a. fall b. stay at the same level c. rise
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Name: ________________________
ID: A
13. Which of the following equations can be used to directly calculate an object’s momentum, p?
m
a. p = F∆t b. ∆p = F∆t c. p = mv d. p =
v
Problem
14. Suppose the torque produced by a wrench on a stubborn nut is 30 N·m. If the lever arm is doubled with no change
in force, what is the resulting torque?
15. A 40-kg boy sits on a seesaw 2.0 m from the fulcrum. What distance from the fulcrum should a 30-kg girl sit in
order to balance the seesaw?
16. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a
velocity of 15 m/s to the west?
17. A 4.0 m board with a mass of 19 kg is pivoted at its center of gravity. A helium balloon attached 0.23 m from the
left end of the board produces an upward force of 7.0 N. A 3.5 kg book is placed 0.73 m from the left end of the
board, and another book of 1.3 kg is placed 0.75 m from the right end of the board. Find the torque on the board
and the direction of rotation.
18. A bucket filled with water has a mass of 23 kg and is attached to a rope wound around a cylinder with a radius of
0.050 m at the top of a well. What torque does the weight of the water and bucket produce on the cylinder? (g =
9.81 m/s 2 )
19. A force of 4.0 N is applied to a door at an angle of 60.0° and a distance of 0.30 m from the hinge. What is the
torque produced?
20. A child with a weight of 4.50 × 10 2 N sits on a seesaw 0.60 m from the axis of rotation. How far from the axis of
rotation on the other side should a child with a weight of 6.00 × 10 2 N sit so the seesaw will remain balanced?
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ID: A
S15--Physics Q3 Torque
Answer Section
MULTIPLE CHOICE
1. ANS: B
2. ANS: B
3. ANS: E
OBJ: 11.2
4. ANS: D
5. ANS: A
PTS:
PTS:
PTS:
STA:
PTS:
1
1
1
Ph.1.k | Ph.1.l
1
DIF:
DIF:
DIF:
I
I
3
OBJ: 7-1.3
OBJ: 7-1.1
REF: p. 153
DIF:
I
OBJ: 6-1.1
OBJ:
DIF:
DIF:
DIF:
DIF:
7-4.2
I
II
II
2
OBJ:
OBJ:
OBJ:
REF:
DIF:
1
REF: p. 151
DIF:
DIF:
I
2
OBJ: 7-4.1
REF: p. 152
DIF:
I
OBJ: 6-1.1
Given
F = 3.0 × 10 2 N
d = 0.80 m
Solution
Ê
Ë
ˆ
¯
τ = Fd = ÁÁ 3.0 × 10 2 N ˜˜ (0.80 m) = 2.4 × 10 2 Nm
6.
7.
8.
9.
10.
11.
12.
13.
PTS:
ANS:
ANS:
ANS:
ANS:
OBJ:
ANS:
OBJ:
ANS:
ANS:
OBJ:
ANS:
1
D
A
C
E
11.1
D
11.1
B
C
11.2
C
DIF:
PTS:
PTS:
PTS:
PTS:
STA:
PTS:
STA:
PTS:
PTS:
STA:
PTS:
IIIA
1
1
1
1
Ph.1.l
1
Ph.1.l
1
1
Ph.1.k | Ph.1.l
1
DIF:
3
REF: p. 151
OBJ: 11.1
PTS: 1
DIF:
STA: Ph.1.k | Ph.1.l
3
REF: p. 153
OBJ: 11.2
7-4.1
7-4.1
7-4.2
p. 152
PROBLEM
14. ANS:
60 N·m
PTS: 1
STA: Ph.1.l
15. ANS:
2.7 m
1
ID: A
16. ANS:
30 m/s to the west
Given
m1 = 2680 kg
v 1 = 15 m/s to the west
m2 = 1340 kg
Solution
m 1 v 1 = m2 v 2
v2 =
m1 v 1
m2
ÁÊÁ 2.68 × 10 3 kg ˜ˆ˜ (15 m/s west)
Ë
¯
=
= 3.0 × 10 1 m/s west
ÁÊÁ 1.34 × 103 kg ˜ˆ˜
Ë
¯
PTS: 1
DIF:
17. ANS:
15 N•m counterclockwise
IIIB
OBJ: 6-1.1
Given
F 1 = 7.0 N
d 1 = −2.0 m + 0.23 m = −1.8 m
m2 = 3.5 kg
d 2 = −2.0 m + 0.73 m = −1.3 m
m3 = 1.3 kg
d 3 = 2.0 m − 0.75 m = 1.2 m
g = −9.81 m/s 2
Solution
τ net = τ 1 + τ 2 + τ 3
τ 1 = F 1 d 1 = (7.0 N) (−1.8 m) = −13 Nm
Ê
ˆ
F 2 = m2 g = ÊÁË 3.5 kg ˆ˜¯ ÁÁ −9.81 m/s 2 ˜˜ = −34 N
Ë
¯
τ 2 = F 2 d 2 = (−34 N) (−1.3 m) = 44 Nm
Ê
ˆ
F 3 = m3 g = ÊÁË 1.3 kg ˆ˜¯ ÁÁ −9.81 m/s 2 ˜˜ = −13 N
Ë
¯
τ 3 = F 3 d 3 = (−13 N) (1.2 m) = −16 Nm
τ net = (−13 Nm) + (44 Nm) + (−16 Nm) = 15 Nm
τ net = 15 Nm counterclockwise
PTS: 1
DIF:
IIIC
OBJ: 7-4.2
2
ID: A
18. ANS:
11 N•m
Given
m = 23 kg
d = 0.050 m
g = 9.81 m/s 2
Solution
Ê
Ë
ˆ
¯
τ = Fd = mgd = ÊÁË 23 kg ˆ˜¯ ÁÁ 9.81 m/s 2 ˜˜ (0.050 m) = 11 Nm
PTS: 1
19. ANS:
1.0 N•m
DIF:
IIIA
OBJ: 7-4.2
Given
F = 4.0 N
d = 0.30 m
θ = 60.0°
Solution
τ = Fd sin θ = (4.0 N) (0.30 m) (sin60.0°) = 1.0 N •m
PTS: 1
DIF:
IIIB
OBJ: 7-4.2
3
ID: A
20. ANS:
0.45 m
Given
F 1 = −4.50 × 10 2 N
d 1 = −0.60 m
F 2 = −6.00 × 10 2 N
Solution
τ net = τ 1 + τ 2 = 0
τ 1 = −τ 2
F 1 d 1 = −F 2 d 2
d2 = −
F1 d 1
F2
ÊÁ
ˆ
Á −4.50 × 10 2 N ˜˜ (−0.60 m)
Ë
¯
d2 = −
= 0.45 m
ÊÁ
ˆ
Á −6.00 × 10 2 N ˜˜
Ë
¯
PTS: 1
DIF:
IIIB
OBJ: 7-4.2
4