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TRANSCRIPT
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 1
This print-out should have 47 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.
Weight of a Boxer001 (part 1 of 4) 10.0 points
A(n) 98.3 kg boxer has his first match inthe Canal Zone with gravitational acceler-ation 9.782 m/s2 and his second match atthe North Pole with gravitational accelera-tion 9.832 m/s2.
a) What is his mass in the Canal Zone?Correct answer: 98.3 kg.
Explanation:An object’s mass is constant, regardless of
the gravitational acceleration.
002 (part 2 of 4) 10.0 pointsb) What is his weight in the Canal Zone?Correct answer: 961.571 N.
Explanation:An object’s weight varies with gravitational
position and is given by
W = mg = (98.3 kg)(9.782 m/s2)
= 98.3 kg
003 (part 3 of 4) 10.0 pointsc) What is his mass at the North Pole?Correct answer: 98.3 kg.
Explanation:His mass doesn’t vary.
004 (part 4 of 4) 10.0 pointsd) What is his weight at the North Pole?Correct answer: 966.486 N.
Explanation:
W = mg = (98.3 kg)(9.832 m/s2)
= 98.3 kg
Hewitt CP9 04 E15005 (part 1 of 4) 10.0 points
Gravity on the surface of the moon is only1
6as strong as gravity on the Earth.
What is the weight of a 16 kg object onthe Earth? The acceleration of gravity is10 m/s2 .Correct answer: 160 N.
Explanation:On the Earth the weight is
W = mg = (16 kg)(10 m/s2) = 160 N .
006 (part 2 of 4) 10.0 pointsWhat is the weight on the moon?Correct answer: 26.6667 N.
Explanation:On the moon the weight is
W =1
6mg =
1
6(16 kg)(10 m/s2)
= 26.6667 N .
007 (part 3 of 4) 10.0 pointsWhat is the mass on the earth?Correct answer: 16 kg.
Explanation:The mass would be 16 kg everywhere.
008 (part 4 of 4) 10.0 pointsWhat is the mass on the moon?Correct answer: 16 kg.
Explanation:The mass would be 16 kg everywhere.
Dragster Acceleration009 (part 1 of 3) 10.0 points
A dragster and driver together have mass930.6 kg . The dragster, starting from rest,attains a speed of 25 m/s in 0.57 s .
Find the average acceleration of the drag-ster during this time interval.Correct answer: 43.8596 m/s2.
Explanation:The average acceleration during time t is
a =∆v
t
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 2
=v − 0
t
=v
t
=(25 m/s)
(0.57 s)
= 43.8596 m/s2 .
010 (part 2 of 3) 10.0 pointsWhat is the size of the average force on thedragster during this time interval?Correct answer: 40815.8 N.
Explanation:The average force on the dragster is
fd = md a
= (930.6 kg) (43.8596 m/s2)
= 40815.8 N .
011 (part 3 of 3) 10.0 pointsAssume: The driver has a mass of 78.6 kg .
What horizontal force does the seat exerton the driver?Correct answer: 3447.37 N.
Explanation:The force on the driver is
F = ma
= (78.6 kg) (43.8596 m/s2)
= 3447.37 N .
Force on a Bullet 02012 10.0 points
A 9.7 g bullet leaves the muzzle of a rifle witha speed of 520.5 m/s.
What constant force is exerted on the bulletwhile it is traveling down the 0.9 m length ofthe barrel of the rifle?Correct answer: 1459.96 N.
Explanation:Average acceleration can be found from
v2
f = v2
o + 2 a ℓ
Since vo = 0, we have
a =v2
2 ℓ
Thus
F = ma = mv2
2 ℓ
=(9.7 g)(520.5 m/s)2
2 (0.9 m)·
1 kg
1000 g
= 1459.96 N .
Force on a Coasting Car013 10.0 points
A 1698.7 kg car is traveling at 35 m/s whenthe driver takes his foot off the gas pedal. Ittakes 5.1 s for the car to slow down to 20 m/s.
How large is the net force slowing the car?Correct answer: 4996.18 N.
Explanation:The acceleration of the car is given by
a =v − v0
t
=20 m/s − 35 m/s
5.1 s= −2.94118 m/s2 .
Applying Newton’s second law yields the forceslowing the car:
‖~F‖ = |ma|
=∣
∣(1698.7 kg)(−2.94118 m/s2)∣
∣
= 4996.18 N .
Force on a Lawn Spreader014 10.0 points
Joe pushes down the length of the handle ofa 14.3 kg lawn spreader. The handle makesan angle of 45.6 ◦ with the horizontal. Joewishes to accelerate the spreader from rest to1.35 m/s in 1.3 s.
What force must Joe apply to the handle?Correct answer: 21.2245 N.
Explanation:
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 3
The horizontal component of the force is
Fh = F cos θ.
Let v be the final velocity of the spreader.According to Newton’s second law,
Fh = mah
so
F cos θ = m∆vh
∆t
F =m∆vh
∆t cos θ
=mv
t cos θ
=(14.3 kg) (1.35 m/s)
(1.3 s) cos 45.6◦
= 21.2245 N
keywords:
Hanging Weight 01015 (part 1 of 2) 10.0 points
Consider the 666 N weight held by two cablesshown below. The left-hand cable had tensionT and makes an angle of θ with the wall. Theright-hand cable had tension 740 N and makesan angle of 33◦ with the ceiling.
666 N
T
740 N33
◦
θ
a) What is the tension T in the left-handcable slanted at an angle of θ with respect tothe wall?Correct answer: 674.03 N.
Explanation:Observe the free-body diagram below.
F2 F1 θ1θ2
Wg
Note: The sum of the x- andy-components of F1 , F2 , andWg are equal to zero.
Let : F2 ≡ T ,
F1 = 740 N ,
Wg = 666 N ,
θ1 = 33◦ , and
θ2 = 90◦ − θ .
Basic Concept: Vertically and Horizontally,we have
F xnet = F x
1 − F x2 = 0
= F1 cos θ1 − F2 cos θ2 = 0 (1)
F ynet = F y
1+ F y
2−Wg = 0
= F1 sin θ1 + F2 sin θ2 −Wg = 0 (2)
Solution: Using Eqs. 1 and 2, we have
F x2 = F1 cos θ1 (3)
= (740 N) cos 33◦
= 620.616 N , and
F y2
= F3 − F1 sin θ1 (4)
= (666 N) − (740 N) sin 33◦
= (666 N) − (403.033 N)
= 262.967 N
So
F2 =√
(F x2)2 + (F y
2)2
=√
(620.616 N)2 + (262.967 N)2
= 674.03 N .
016 (part 2 of 2) 10.0 pointsb) What is the angle θ which the left-handcable makes with respect to the wall?Correct answer: 67.0367◦.
Explanation:
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 4
Using Eqs. 3 & 4, we have
θ2 = arctan
(
F y2
F x2
)
= arctan
(
262.967 N
620.616 N
)
= 22.9633◦ , so
θ = 90◦ − θ2
= 90◦ − 22.9633◦
= 67.0367◦ .
Hanging Weight 04017 (part 1 of 2) 10.0 points
Consider the 671 N weight held by two cablesshown below. The left-hand cable had tensionT2 and makes an angle of θ2 with the ceiling.The right-hand cable had tension 460 N andmakes an angle of 43◦ with the ceiling.
The right-hand cable makes an angle of 43◦
with the ceiling and has a tension of 460 N .
671 N
T2
460N
43◦θ
2
a) What is the tension T2 in the left-handcable slanted at an angle of θ2 with respect tothe wall?Correct answer: 490.744 N.
Explanation:Observe the free-body diagram below.
F2 F 1 θ1
θ2
Wg
Note: The sum of the x- andy-components of F1 , F2 , andWg are equal to zero.
Given : Wg = 671 N ,
F1 = 460 N ,
θ1 = 43◦ , and
θ2 = 90◦ − θ .
Basic Concept: Vertically and Horizontally,we have
F xnet = F x
1 − F x2 = 0
= F1 cos θ1 − F2 cos θ2 = 0 (1)
F ynet = F y
1+ F y
2−Wg = 0
= F1 sin θ1 + F2 sin θ2 −Wg = 0 (2)
Solution: Using Eqs. 1 and 2, we have
F x2 = F1 cos θ1 (1)
= (460 N) cos 43◦
= 336.423 N , and
F y2
= F3 − F1 sin θ1 (2)
= 671 N − (460 N) sin 43◦
= 671 N − 313.719 N
= 357.281 N , so
F2 =√
(F x2)2 + (F y
2)2
=√
(336.423 N)2 + (357.281 N)2
= 490.744 N .
018 (part 2 of 2) 10.0 pointsb) What is the angle θ2 which the left-handcable makes with respect to the ceiling?Correct answer: 46.7222◦.
Explanation:Using Eq. 2, we have
θ2 = arctan
(
F y1
F x1
)
= arctan
(
313.719 N
336.423 N
)
= 46.7222◦ .
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 5
Hanging Weights 02019 (part 1 of 3) 10.0 points
In the figure below the left-hand cable has atension T1 and makes an angle of 55◦ withthe horizontal. The right-hand cable has atension T3 and makes an angle of θ3 with thehorizontal. A 61.4 N weight is on the left anda 53.1 N weight is on the right. The cableconnecting the two weights is horizontal.
61.4 N
53.1 N
T2
T1 T 3
θ355 ◦
a) Find the tension T1.Correct answer: 74.9556 N.
Explanation:
Given : W1 = M1 g ,
W2 = M2 g ,
θ1 = 55◦ ,
θ3 = 51.0044◦ , and
T2 = 42.9927 N .
T 3
T1 θ3
θ1
T3 cos θ3T1 cos θ1
W2W1
Note: T1 cos θ1 = T2 = T3 cos θ3
Basic Concepts:
∑
~F = m~a = 0
~W = m~g
Solution: Consider the point of attachmentof cable 1 and cable 2.
Vertically, W1 = M1 g acts down andT1 sin θ1 acts up, so
Fnet = W1 − T1 sin θ1 = 0
=⇒ T1 sin θ1 = W1 . (1)
Horizontally, T1 acts to the left and T1 cos θ1
acts to the right, so
Fnet = T1 − T1 cos θ1 = 0
=⇒ T1 cos θ1 = T2 . (2)
Using Eq. 1 we have
T1 =W1
sin θ1
=61.4 N
sin 55◦
= 74.9556 N .
020 (part 2 of 3) 10.0 pointsb) Find the tension T2.Correct answer: 42.9927 N.
Explanation:Dividing Eq. 1 by Eq. 2, we have
T2 =W1
tan θ1
=61.4 N
tan 55◦
= 42.9927 N .
021 (part 3 of 3) 10.0 pointsc) Find the angle θ3.Correct answer: 51.0044◦.
Explanation:Balancing forces at the point where the
right-hand weight is suspended, we have
T3 sin θ3 = W2 vertical (3)
T3 cos θ3 = T2 horizontal . (4)
dividing Eq. 3 by Eq. 4, we have
tan θ3 =W2
T2
θ = arctan
(
W2
T2
)
= arctan
(
53.1 N
42.9927 N
)
= 51.0044◦ .
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 6
The tension T3 =T2
cos θ3
= 68.3277 N is not
required in this problem.
Hanging Weights 01022 10.0 points
In the figure below the left-hand cable has atension T1 and makes an angle of 48◦ withthe horizontal. The right-hand cable has atension T3 and makes an angle of 54◦ withthe horizontal. A W1 weight is on the leftand a W2 weight is on the right. The cableconnecting the two weights has a tension 39 Nand is horizontal.
The acceleration of gravity is 9.8 m/s2 .
M1
M2
39 N
T1 T 3
54◦48 ◦
Determine the mass M2.Correct answer: 5.47744 kg.
Explanation:
Given : W1 = M1 g ,
W2 = M2 g ,
θ1 = 48◦ ,
θ3 = 54◦ , and
T2 = 39 N .
T 3T1 θ3θ
1
T3 cos θ3T1 cos θ1
W2
W1
Note: T1 cos θ1 = T2 = T3 cos θ3
Basic Concepts:∑
~F = m~a = 0
~W = m~g
Solution: Consider the point of attachmentof cable 2 and cable 3.
Vertically, W2 = M2 g acts down andT3 sin θ3 acts up, so
Fnet = W2 − T3 sin θ3 = 0
=⇒ T3 sin θ3 = W2 . (1)
Horizontally, T2 acts to the left and T3 cos θ3
acts to the right, so
Fnet = T2 − T3 cos θ3 = 0
=⇒ T3 cos θ3 = T2 . (2)
Dividing Eq. 1 by Eq. 2, we have
tan θ3 =W2
T2
.
W2 = T2 tan θ3
= (39 N) tan 54◦ = 66.3508 N
M2 =W2
g=
53.6789 N
9.8 m/s2= 5.47744 kg
and by symmetry, we have
T1 cos θ1 − T3 cos θ3 = 0 , so
W1 = T2 tan θ1
= (39 N) tan 48◦ = 58.2846 N
M1 =W1
g=
43.3139 N
9.8 m/s2= 4.41978 kg .
Holt SF 04Rev 68023 (part 1 of 2) 10.0 points
Consider the 44 N weight held by two cablesshown below. The left-hand cable is horizon-tal.
44 N
33◦
a) What is the tension in the cable slantedat an angle of 33◦?Correct answer: 80.7875 N.
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 7
Explanation:Observe the free-body diagram below.
67.7541 N67.7541 N
80.78
75N
44
N44
N
Scale: 10 N
33◦
Note: The sum of the x- andy-components of T1 , T2 , andWg are equal to zero.
Given : Wg = 44 N and
θ = 33◦ .
Basic Concept: Vertically, we have
Fy,net = F1 sin θ −Wg = 0
Solution:
F1(sin θ) = Wg
F1 =Wg
sin θ
=44 N
sin 33◦
= 80.7875 N
024 (part 2 of 2) 10.0 pointsb) What is the tension in the horizontal ca-ble?Correct answer: 67.7541 N.
Explanation:Basic Concept: Horizontally,
Fx,net = F1 cos θ − F2 = 0
Solution:
F2 = F1 cos θ
=Wg cos θ
sin θ
=(44 N) cos 33◦
sin 33◦
= 67.7541 N
Serway CP 04 16025 10.0 points
Consider the 90 N light fixture supported asin the figure.
35◦ 35◦
Find the tension in the supporting wires.Correct answer: 78.4551 N.
Explanation:
Given : W = 90 N and
θ = 35◦ .
W = 90 N
x
y
T2
θT1
θ
Horizontally,∑
Fx = 0 , so
T1 cos θ − T2 cos θ = 0
T1 = T2 .
Vertically,∑
Fy = 0 , so
2 (T1 sin θ) −W = 0
T1 = T2
=W
2 sin θ
=90 N
2 sin 35◦
= 78.4551 N .
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 8
Static Equilibrium 01026 (part 1 of 2) 10.0 points
The 7.1 N weight is in equilibrium under theinfluence of the three forces acting on it. TheF force acts from above on the left at an angleof α with the horizontal. The 5.8 N force actsfrom above on the right at an angle of 48◦ withthe horizontal. The force 7.1 N acts straightdown.
F5.8N
7.1 N
48◦α
Note: F, α are not to scale.What is the magnitude of the force F?
Correct answer: 4.7796 N.
Explanation:Note: Standard angular measurements are
from the positive x-axis in a counter-clockwisedirection.
Given : F1 = F ,
α1 = 180◦ − α ,
F2 = 5.8 N ,
α2 = 48◦ ,
F3 = 7.1 N , and
α3 = 270◦ ,
Basic Concepts:∑
F x = 0
F x1 + F x
2 + F x3 = 0
F x1 + F2 cos α2 + F3 cos α3 = 0 (1)
∑
F y = 0
F y1
+ F y2
+ F y3
= 0
F y1
+ F2 sin α2 + F3 sinα3 = 0 (2)
Solution: Using Eqs. 1 and 2, we have
F x1 = −F2 cos α2 − F3 cos α3 (1)
= −(5.8 N) cos 48◦ − (7.1 N) cos 270◦
= −3.88096 N , and
F y1
= −F2 sin α2 − F3 sin α3 (2)
= −(5.8 N) sin 48◦ − (7.1 N) sin 270◦
= 2.78976 N , so
F1 =√
(F x1)2 + (F y
1)2
=√
(−3.88096 N)2 + (2.78976 N)2
= 4.7796 N .
027 (part 2 of 2) 10.0 pointsWhat is the angle α of the force F as shownin the figure above?Correct answer: 35.7098◦.
Explanation:
α1 = arctan
(
F y1
F x1
)
= arctan
(
2.78976 N
−3.88096 N
)
= 144.29◦ , from the positive x−axis
α = 180◦ − α1
= 180◦ − 144.29◦
= 35.7098◦ .
Observe the free-body diagram below.
4.78 N 5.8N
7.1 N
48◦
35.7◦
The vertical dashed vector is the sumof F1 sin α1 = 2.78976 N and F2 sinα1 =4.31024 N and is equal in length to F y
3=
7.1 N .
Static Equilibrium 04028 (part 1 of 2) 10.0 points
The knot at the junction is in equilibriumunder the influence of four forces acting onit. The F force acts from above on the left atan angle of α with the horizontal. The 5.6 Nforce acts from above on the right at an angle
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 9
of 17◦ with the horizontal. The 6.2 N forceacts from below on the left at an angle of 79◦
with the horizontal.
5.6 N
6.2N
F
knot17◦
79 ◦
α
Note: F, α are not to scale.What is the magnitude of the force F?
Correct answer: 6.09917 N.
Explanation:Note: Standard angular measurements are
from the positive x-axis in a counter-clockwisedirection.
Given : F1 = 5.6 N ,
α1 = 180◦ − 17◦ = 163◦ ,
F2 = 6.2 N ,
α2 = 180◦ + 79◦ = 281◦ ,
F3 = F , and
α3 = α .
Observe the free-body diagram belowwhere the vectors are decomposed into com-ponents along the x- and y-axes.
5.6 N
6.2N
6.1N
17◦
79 ◦
46.8
Note: The vectors along the x-and y-coordinates add to zero.
Basic Concepts:
∑
F x = 0
F x1 + F x
2 + F x3 = 0
F1 cos α1 + F2 cos α2 + F x3 = 0 (1)
∑
F y = 0
F y1
+ F y2
+ F y3
= 0
F1 sin α1 + F2 sinα2 + F y3
= 0 (2)
Solution: Using Eqs. 1 and 2, we have
F x3 = −F1 cos α1 − F2 cos α2 (1)
= −(5.6 N) cos 163◦ − (6.2 N) cos 281◦
= 4.17229 N , and
F y3
= −F1 sin α1 − F2 sinα2 (2)
= −(5.6 N) sin 163◦ − (6.2 N) sin 281◦
= 4.44881 N , so
F3 =√
(F x3)2 + (F y
3)2
=√
(4.17229 N)2 + (4.44881 N)2
= 6.09917 N .
029 (part 2 of 2) 10.0 pointsWhat is the angle α of the force F as shownin the figure above?Correct answer: 46.8371◦.
Explanation:
α3 = arctan
(
F y3
F x3
)
= arctan
(
4.44881 N
4.17229 N
)
= 46.8371◦ , from the positive x−axis
α = 46.8371◦ .
Tipler PSE5 04 45030 (part 1 of 4) 10.0 points
A 0.62 kg block is suspended from the middleof a 1.68 m long string. The ends of the stringare attached to the ceiling at points separatedby 1 m, and the block can slip along the longstring.
The acceleration of gravity is 9.81 m/s2 .
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 10
1 m
0.84 m
T1
0.84 m
T2
0.62 kg
What angle does the string make with theceiling?Correct answer: 53.4704◦.
Explanation:
Given : ℓ = 1.68 m ,
d = 1 m .
Consider the physical distances
cos θ =
1
2d
1
2ℓ
=d
ℓ
θ = cos−1d
ℓ
= cos−11 m
1.68 m
= 53.4704◦ .
031 (part 2 of 4) 10.0 pointsWhat is the tension in the string?Correct answer: 3.78458 N.
Explanation:
Given : m = 0.62 kg .
Applying∑
Fy = may = 0 to the mass,
2 T sin θ − mg = 0
T =mg
2 sin θ
=(0.62 kg)
(
9.81 m/s2)
2 sin 53.4704◦
= 3.78458 N .
032 (part 3 of 4) 10.0 pointsThe 0.62 kg block is removed and two 0.31 kgblocks are attached to the string such that thelengths of the three string segments are equal.
1 m
0.56 mT1
0.56 m
T2
0.56 mT3
0.31 kg 0.31 kg
What is the tension in the string segmentattached to the ceiling on the right ?Correct answer: 3.30698 N.
Explanation:
Let m = 0.31 kg .
The length of each segment is
ℓ′ =ℓ
3=
1.68 m
3= 0.56 m .
Find the distance d′
d′ =d − ℓ′
2
=(1 m) − (0.56 m)
2= 0.22 m .
cos θ′ =d′
ℓ′
θ′ = cos−1
(
d′
ℓ′
)
= cos−1
(
0.22 m
0.56 m
)
= 66.8676◦ .
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 11
Applying∑
Fy = may = 0 to the 0.31 kg
block
T sin θ′ − mg = 0
T3 =mg
sin θ′
=(0.31 kg)
(
9.81 m/s2)
sin 66.8676◦
= 3.30698 N .
033 (part 4 of 4) 10.0 pointsWhat is the tension in the horizontal seg-ment?Correct answer: 1.29917 N.
Explanation:
Applying∑
Fx = max = 0 to the block,
T3 cos θ′ − T2 = 0
T2 = T3 cos θ′
= (3.30698 N) cos 66.8676◦
= 1.29917 N .
Tipler PSE5 04 53034 (part 1 of 2) 10.0 points
Your car is stuck in a mud hole. You are alone,but you have a long, strong rope. Havingstudied physics, you tie the rope tautly to atelephone pole and pull on it sideways at themidpoint, as shown.
404 N 3.3◦
22 m
Find the force exerted by the rope on thecar when the angle is 3.3◦ and you are pullingwith a force of 404 N but the car does notmove.Correct answer: 3.50914 kN.
Explanation:
Let : F = 404 N ,
θ = 3.3◦ , and
ℓ = 22 m .
Before the car starts moving,
∑
Fy = may ,
so
2 T sin θ − F = 0
T =F
2 sin θ
=404 N
2 sin 3.3◦·
kN
1000 N
= 3.50914 kN .
035 (part 2 of 2) 10.0 pointsHow strong must the rope be if it takes a forceof 599 N to move the car when θ is 3.1◦?Correct answer: 5.53821 kN.
Explanation:
Let : θ = 3.1◦ .
T =F
2 sin θ
=599 N
2 sin 3.1◦·
kN
1000 N
= 5.53821 kN .
Forces on a Sled036 (part 1 of 2) 10.0 points
A child holds a sled on a frictionless, snow-covered hill, inclined at an angle of 31◦.
62N
F
31◦
If the sled weighs 62 N, find the force ex-erted on the rope by the child.Correct answer: 31.9324 N.
Explanation:
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 12
Given : W = 62 N and
θ = 31◦ .
Consider the free body diagram for theblock
mg sin
θ N=
mg cos
θ
F
W
Basic Concepts: If we “tilt” our world,and consider the forces parallel to the hill,
Fnet =∑
Fup −∑
Fdown = 0
then the forces perpendicular to the hill,
Fnet =∑
Fout −∑
Fin = 0
Solution: Consider the free body diagramfor the sled: The weight of the sled has compo-nents W sin θ acting down the hill and W cos θacting straight into the hill.
The system is in equilibrium, so for forcesparallel to the hill,
Fnet = T −W sin θ = 0
=⇒ T = W sin θ
= (62 N) sin 31◦
= 31.9324 N
037 (part 2 of 2) 10.0 pointsWhat force is exerted on the sled by the hill?Correct answer: 53.1444 N.
Explanation:For forces perpendicular to the hill,
Fnet = N −W cos θ = 0
=⇒ N = W cos θ
= (62 N) cos 31◦
= 53.1444 N
Forces on a Skier038 (part 1 of 2) 10.0 points
A skier of mass 109 kg comes down a slope ofconstant angle 25◦ with the horizontal.
The acceleration of gravity is 9.8 m/s2 .What is the force on the skier parallel to
the slope?Correct answer: 451.441 N.
Explanation:
a
N W θ
The weight of the skier acting verticallydownward can be split into two components.The sine component acts along the slope.
Wparallel = W sin θ
= mg sin θ
= (109 kg) (9.8 m/s2) sin 25◦
= 451.441 N.
039 (part 2 of 2) 10.0 pointsWhat force normal to the slope is exerted bythe skis?Correct answer: 968.118 N.
Explanation:The cosine component acts perpendicular
to the slope.
Wnormal = W cos θ
= mg cos θ
= (109 kg) (9.8 m/s2) cos 25◦
= 968.118 N.
Tipler PSE5 04 61040 10.0 points
A 65 kg student weighs himself by standingon a scale mounted on a skateboard that isrolling down an incline, as shown. Assumethere is no friction so that the force exerted
Version PREVIEW – Chapter 4 Newton’s 1st 2nd Laws – Johnson – (84248) 13
by the incline on the skateboard is normal tothe incline.
The acceleration of gravity is 9.81 m/s2 .
28◦
What is the reading on the scale if the angleof the slope is 28◦?Correct answer: 563.012 N.
Explanation:
Let : θ = 28◦ ,
m = 65 kg , and
g = 9.81 m/s2 .
Consider the forces acting on the scale:
y x
N
W
28◦
Let the positive x-axis be parallel to anddown the plane, and the positive y-axis in thedirection of the normal force.
Applying∑
Fy = may = 0 ,
N −W cos θ = 0
N = W cos θ
= mg cos θ
= (65 kg)(
9.81 m/s2)
cos 28◦
= 563.012 N .
Up a Slope
041 (part 1 of 2) 10.0 pointsTwo blocks are connected by an extensiblemassless cord on an inclined plane as shownin the figure below.
The acceleration of gravity is 9.8 m/s2 .
41 kg
T 68 kg
F
3.8m/s
2
20◦
What is the force F pulling both theblocks?Correct answer: 779.546 N.
Explanation:
Let : θ = 20◦ ,
M1 = 41 kg ,
M2 = 68 kg ,
g = 9.8 m/s2 , and
a = 3.8 m/s2 .
F
N2
W2‖
W2⊥W2
N1
W1⊥
W1‖
W1
Tθ
The resulting acceleration is due to the ap-plied force acting against the component ofthe weights along the inclined surface. Thissystem is equivalent to a combined block ofmass M1 + M2 being accelerated up the slope
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by a force F , and can be described by theequation
F − (M1 + M2) g sin θ = (M1 + M2) a . (1)
Solving for F we obtain
F = (M1 + M2)(a + g sin θ)
= (41 kg + 68 kg)
×(
3.8 m/s2 + 9.8 m/s2)
sin 20◦
= 779.546 N .
042 (part 2 of 2) 10.0 pointsWhat is the tension in the cord pulling thelower block?Correct answer: 293.224 N.
Explanation:Using Eq. (1) from Part 1 for the lower
block, we have
T − M1 g sin θ = M1 a .
Solving for T we obtain
T = M1 (a + g sin θ)
= (41 kg)
×(
3.8 m/s2 + 9.8 m/s2)
sin 20◦
= 293.224 N .
Pulleys 01043 (part 1 of 3) 10.0 points
Assume all pulleys are massless and fric-tionless, and the systems are in equilibrium.
The acceleration due to gravity is 9.8 m/s2.
44 N
T
Find the tension T .Correct answer: 22 N.
Explanation:
Let : W = 44 N .
A weight hanging at the end of a stringdefines the tension in that string, which is thesame throughout the string.
When a pulley system is in equilibrium, thesum of the upward forces will equal the sumof the downward forces.
2 T = W
T =W
2=
44 N
2= 22 N .
044 (part 2 of 3) 10.0 pointsThe suspended weight is 43 N.
43 N
T
Find the tension T .Correct answer: 43 N.
Explanation:
Let : W2 = 43 N .
The weight W defines the tension:
T = W2 = 43 N .
045 (part 3 of 3) 10.0 pointsThe suspended weight is 35 N.
35 N
T
What is the tension T?Correct answer: 70 N.
Explanation:
Let : W3 = 35 N .
The pulley is in equilibrium, so
T = 2W3 = 2 (35 N) = 70 N .
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Pulleys 06046 10.0 points
In the pulley system, all pulleys are mass-less and frictionless.
The acceleration of gravity is 9.8 m/s2 .
40 N
T
Find the tension T .Correct answer: 20 N.
Explanation:
Let : W = 40 N .
W
1
2
3
T1 T1
T1 T1
T
T2 T3
At pulleys 1 and 3,
2 T1 = T2 and 2 T1 = T3 .
At the weight W,
W = T2 + T3 = 4 T1
T1 =W
4.
At pulley 2,
T = 2 T1 =W
2=
40 N
2= 20 N .
Pulleys 12047 10.0 points
The system is in equilibrium and the pulleysare weightless and frictionless. The suspendedweight is 19 N.
The acceleration of gravity is 9.8 m/s2 .
19 N
T
Find the tension T .Correct answer: 4.75 N.
Explanation:
Let : W = 19 N .
W
2
1
T1
T
T1
T
T
Consider equilibrium at pulley 1, we have
2 T1 = W
T1 =W
2.
At pulley 2,
2 T = T1 so
T =T1
2
=W
4
=19 N
4
= 4.75 N .