Homework 3 Solutions

duong (ktd359) – Homework 3 – Spurlock – (44103) 1

This print-out should have 89 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 (part 1 of 2) 10.0 pointsA truck travels up a hill with a 13◦ incline.The truck has a constant speed of 17 m/s.What is the horizontal component of the

truck’s velocity?

Correct answer: 16.5643 m/s.

Explanation:

Let : v = 17 m/s and

θ = 13◦ .

vy

vx

17m/s

13◦

vx = v sin θ = (17 m/s) cos 13◦

= 16.5643 m/s .

002 (part 2 of 2) 10.0 pointsWhat is the vertical component of the truck’svelocity?


Explanation:

vy = v sin θ = (17 m/s) sin 13◦

= 3.82417 m/s .

003 10.0 pointsA hiker makes four straight-line walks

A 27 km at 277◦

B 33 km at 75◦

C 25 km at 164◦

D 23 km at 234◦

in random directions and lengths starting atposition (41 km, 41 km) ,

AB

C

D

How far from the starting point is the hikerafter these four legs of the hike? All anglesare measured in a counter-clockwise directionfrom the positive x-axis.

Correct answer: 26.5623 km.

Explanation:

Let : (x0, y0) = (41 km, 41 km) .

∆ax = (27 km) cos 277◦ = 3.29053 km ,

∆ay = (27 km) sin 277◦ = −26.7987 km ,

∆bx = (33 km) cos 75◦ = 8.54101 km ,

∆by = (33 km) sin 75◦ = 31.8756 km ,

∆cx = (25 km) cos 164◦ = −24.0316 km ,

∆cy = (25 km) sin 164◦ = 6.8909 km ,

∆dx = (23 km) cos 234◦ = −13.519 km ,

∆dy = (23 km) sin 234◦ = −18.6074 km ,


AB

C

D

E

Scale: 10 km =

∆x = ∆ax +∆bx +∆cx +∆dx

= 3.29053 km + (8.54101 km)

+ (−24.0316 km) + (−13.519 km)

= −25.719 km and

∆y = ∆ay +∆by +∆cy +∆dy

= −26.7987 km + (31.8756 km)

+ (6.8909 km) + (−18.6074 km)

= −6.63969 km ,

so the resultant is

E =√

(∆x)2 + (∆y)2

=√

(−25.719 km)2 + (−6.63969 km)2

= 26.5623 km .

004 (part 1 of 2) 10.0 pointsA particle undergoes three displacements.The first has a magnitude of 15 m and makesan angle of 37 ◦ with the positive x axis. Thesecond has a magnitude of 7.6 m and makesan angle of 151◦ with the positive x axis. (seethe figure below).After the third displacement the particle

returns to its initial position.

151◦

37◦

15m

7.6 m

Find the magnitude of the third displace-ment.

Correct answer: 13.7849 m.

Explanation:

Let : ‖~A‖ = 15 m ,

θa = 37◦ ,

‖~B‖ = 7.6 m , and

θB= 151◦ .

θC

θA

θB

A

B

C

−C

~A+ ~B + ~C = 0 , so ~C = −~A− ~B

Cx = −Ax −Bx

= −A cos θA−B cos θb

= −(15 m) cos 37◦ − (7.6 m) cos 151◦

= −5.33242 m and

Cy = −Ay −Bx

= −A sin θA−B sin θb


= −(15 m) sin 37◦ − (7.6 m) sin 151◦

= −12.7118 m ,

so the magnitude of ~C is

‖~C‖ =√

C2x + C2

y

=√

(−5.33242 m)2 + (−12.7118 m)2

= 13.7849 m .

005 (part 2 of 2) 10.0 pointsFind the angle of the third displacement (mea-sured from the positive x axis, with counter-clockwise positive within the limits of −180◦

to +180◦).

Correct answer: −112.757◦.

Explanation:

tan θC=

Cy

Cx

θC= arctan

(

Cy

Cx

)

= arctan

(−12.7118 m

−5.33242 m

)

= −112.757◦ .

006 (part 1 of 2) 10.0 pointsA particle undergoes two displacements, mea-sured from the positive x-axis, with counter-clockwise positive. The first has a magnitudeof 11 m and makes an angle of 81 ◦ with thepositive x axis. The resultant displacementhas a magnitude of 9.8 m directed at an angleof 139◦ from the positive x axis.

139◦81◦

11m

9.8m

Find the magnitude of the second displace-ment.


Explanation:

Let : ‖~A‖ = 11 m ,

α = 81◦ ,

‖~R‖ = 9.8 m , and

θ = 139◦ .

The resultant is the diagonal of the paral-lelogram defined by the two displacements:

θα

β

A

B

R

~R = ~A+ ~B~B = ~R − ~B .

The components of the second displace-ment ~B are

Bx = Rx −Ax = R cos θ − A cosα

= (9.8 m) cos 139◦ − (11 m) cos 81◦

= −9.11693 m and

By = Ry −Ax = R sin θ − A sinα

= (9.8 m) sin 139◦ − (11 m) sin 81◦

= −4.43519 m ,

so the magnitude of ~B is

‖~B‖ =√

B2x +B2

y

=√

(−9.11693 m)2 + (−4.43519 m)2

= 10.1385 m .


007 (part 2 of 2) 10.0 pointsFind the angle of the second displacement(measured from the positive x axis, with coun-terclockwise positive and within the limits of−180◦ to +180◦).


Explanation:

tanβ =By

Bx

β = arctan

(

By

Bx

)

= arctan

(−4.43519 m

−9.11693 m

)

= −154.058◦ .

008 (part 1 of 3) 10.0 pointsAparticle moves in the xy plane with constantacceleration. At time zero, the particle isat x = 1.5 m, y = 5.5 m, and has velocity~vo = (5 m/s) ı+(−3 m/s) . The accelerationis given by ~a = (2 m/s2) ı+ (8.5 m/s2) .What is the x component of velocity after

6.5 s?

Correct answer: 18 m/s.

Explanation:

Let : ax = 2 m/s2 ,

vxo = 5 m/s , and

t = 6.5 s .

After 6.5 s,

~vx = ~vxo + ~ax t

= (5 m/s) ı+ (2 m/s2) ı (6.5 s)

= (18 m/s) ı .

009 (part 2 of 3) 10.0 pointsWhat is the y component of velocity after6.5 s?


Explanation:

Let : ay = 8.5 m/s2 and

vyo = −3 m/s .

~vy = ~vyo + ~ay t

= (−3 m/s) + (8.5 m/s2) (6.5 s)

= (52.25 m/s) .

010 (part 3 of 3) 10.0 pointsWhat is the magnitude of the displacementfrom the origin (x = 0 m, y = 0 m) after6.5 s?


Explanation:

Let : do = (1.5 m, 5.5 m) ,

vo = (5 m/s,−3 m/s) , and

a = (2 m/s2, 8.5 m/s2) .

From the equation of motion,

~d = ~do + ~vo t+1

2a t2

=[

(1.5 m) ı+ (5.5 m) ]

+ [(5 m/s) ı+ (−3 m/s) ] (6.5 s)

+1

2

[

(2 m/s2) ı+ (8.5 m/s2) ]

(6.5 s)2

= (76.25 m) ı+ (165.562 m) , so

|~d| =√

d2x + d2y

=√

(76.25 m)2 + (165.562 m)2

= 182.277 m .

011 (part 1 of 2) 10.0 pointsA particle has an initial horizontal velocityof 2.5 m/s and an initial upward velocity of


4.7 m/s. It is then given a horizontal accelera-tion of 2.1 m/s2 and a downward accelerationof 1.8 m/s2.What is its speed after 4.1 s?


Explanation:Basic Concepts:The direction of the motion depends only

on the horizontal and vertical components ofthe velocity at any moment.Solution:For the horizontal motion,

vx = v0x + axt

= 2.5 m/s +(

2.1 m/s2)

(4.1 s)

= 11.11 m/s

For the vertical motion,

vy = v0y − ayt

= 4.7 m/s−(

1.8 m/s2)

(4.1 s)

= −2.68 m/s

A resultant negative velocity vy would indi-cate downward motion.A right triangle is formed by the compo-

nents, so

v =√

v2x + v2y

=√

(11.11 m/s)2 + (−2.68 m/s)2

= 11.4287 m/s

012 (part 2 of 2) 10.0 pointsWhat is the direction of its velocity at thistime with respect to the horizontal? Answerbetween −180◦ and +180◦.


Explanation:The vertical component vy is the side oppo-

site the angle θ and the horizontal componentvx is the side adjacent to the angle, so

tan θ =vyvx

θ = arctanvyvx

= arctan−2.68 m/s

11.11 m/s

= 11.4287 m/s

The angle must be in degrees and a positiveangle indicates upward motion while a nega-tive angle indicates downward motion.

013 10.0 pointsAssume: A 78 g basketball is launched at anangle of 40.1◦ and a distance of 13.8 m fromthe basketball goal. The ball is released at thesame height (ten feet) as the basketball goal’sheight.A basketball player tries to make a long

jump-shot as described above.The acceleration of gravity is 9.8 m/s2 .What speed must the player give the ball?


Explanation:Basic concepts: Horizontally,

voh = v cos θ

vh = voh

d = voh t

Vertically,vov = v sin θ

vv = vov − g t

h = vov t−1

2g t2 .

Solution: At the maximum range of the ball,vfv = −vov , so,

−vov = vov − g t

−2 vov = −g t

t = 2vovg

.

The maximum distance covered is

d = voh t =2 voh vov

g


d =2 v cos θ v sin θ

g

d =v2 (2 sin θ cos θ)

g=

v2 sin(2 θ)

g.

Thus the initial velocity is

v =

√

d g

sin[2 θ]

=

√

(13.8 m) (9.8 m/s2)

sin[2 (40.1◦)]

= 11.7151 m/s .

014 10.0 pointsA golf ball is hit at ground level. The ball isobserved to reach its maximum height aboveground level 7.9 s after being hit. 0.55 s afterreaching this maximum height, the ball isobserved to barely clear a fence that is 636 ftfrom where it was hit.The acceleration of gravity is 32 ft/s2 .How high is the fence?

Correct answer: 993.72 ft.

Explanation:Basic Concepts:

~v = ~v0 + ~a t

~s = ~s0 + ~v0 t+1

2~a t2

SolutionThe golf ball reaches its maximum height

after t1 seconds, so

vy = 0 = vy0 − g t1

vy0 = g t1

The ball barely clears the fence after t =(t1 + t2) seconds, so

h = v0yt−1

2g t2

= (g t1) (t1 + t2)−1

2g (t1 + t2)

2

= g t21 + g t1t2 −1

2g(

t21 + 2 t1t2 + t22)

= g t21 + g t1 t2 −1

2g t21 − g t1 t2 −

1

2g t22

=1

2g t21 −

1

2g t22

=1

2g[

t21 − t22]

=1

2(32 ft/s2)

[

(7.9 s)2 − (0.55 s)2]

= 993.72 ft .

Intuitive Reasoning: The time taken forthe ball to rise from the ground to the topequals the time taken for the ball to fall fromthe top to the ground. So the maximum

height is at y = h =1

2g t21. After reaching the

height h, the amount of the subsequent fall in

the time interval t2 is ∆h =1

2g t22 . So the y-

coordinate at t = t1+t2 is h−∆h = h−1

2g t22 .

015 (part 1 of 2) 10.0 pointsA car is parked on a cliff overlooking the

ocean on an incline that makes an angle of 16◦

below the horizontal. The negligent driverleaves the car in neutral, and the emergencybrakes are defective. The car rolls from restdown the incline with a constant accelerationof 2.7 m/s2 and travels 43 m to the edge ofthe cliff. The cliff is 34 m above the ocean.How long is the car in the air? The acceler-

ation of gravity is 9.81 m/s2 .

Correct answer: 2.23925 s.

Explanation:

Let : θ = −16◦ ,

a = 2.7 m/s2 ,

d = 43 m ,

∆y = −34 m , and

g = 9.81 m/s2 .

The car is allowed to accelerate at a rate of2.7 m/s2 for a distance of 43 m, so

v2b = v2t + 2 a d

vb =√2 a d

=√

2(2.7 m/s2)(43 m) = 15.2381 m/s .


The car leaves the cliff at the same angle asthe road, so

∣

∣vb,y∣

∣ = vb sin θ = (15.2381 m/s) sin(16◦)

= 4.20019 m/s and

vb,x = vb cos θ = (15.2381 m/s) cos(16◦)

= 14.6478 m/s .

The vertical motion defines the time:

∆y = −1

2g t2 − vb,y t

0 = g t2 + 2 vb,y t+ 2∆y

t =−2 vb,y ±

√

4 v2b,y − 8∆y g

2 g

=−vb,y +

√

v2b,y − 2∆y g

g.

Since

v2b,y − 2∆y g = (4.20019 m/s)2

− 2(−34 m)(9.81 m/s2)

= 684.722 m2/s2 ,

t =−4.20019 m/s +

√

684.722 m2/s2

9.81 m/s2

= 2.23925 s .

016 (part 2 of 2) 10.0 pointsWhat is the car’s position relative to the baseof the cliff when the car lands in the ocean?


Explanation:There is no acceleration in the horizontal

direction, so

x = vb,x t = (14.6478 m/s)(2.23925 s)

= 32.8 m .

017 10.0 pointsTwo men decide to use their cars to pull atruck stuck in mud. They attach ropes andone pulls with a force of 699 N at an angleof 26◦ with respect to the direction in whichthe truck is headed, while the other car pullswith a force of 1186 N at an angle of 17◦ withrespect to the same direction.

699N

26◦

1186 N

17◦

What is the net forward force exerted onthe truck in the direction it is headed?

Correct answer: 1762.43 N.

Explanation:

Let : F1 = 699 N ,

F2 = 1186 N ,

θ1 = 26◦ , and

θ2 = 17◦ .

For the first vehicle, the forward componentis

F1f = F1 cos θ1

= (699 N) cos 26◦

= 628.257 N .

Similarly, for the second vehicle,

F2f = F2 cos θ2

= (1186 N) cos 17◦

= 1134.18 N .

Thus the net forward force on the truck is

Ff = F1f + F2f

= 628.257 N + 1134.18 N

= 1762.43 N .

018 10.0 points


A 6.1 g bullet leaves the muzzle of a rifle witha speed of 327.6 m/s.What constant force is exerted on the bullet

while it is traveling down the 0.7 m length ofthe barrel of the rifle?


Explanation:Average acceleration can be found from

v2f = v2o + 2 a ℓ

Since vo = 0, we have

a =v2

2 ℓ

Thus

F = ma = mv2

2 ℓ

=(6.1 g)(327.6 m/s)2

2 (0.7 m)· 1 kg

1000 g

= 467.616 N .

019 10.0 pointsOne of the great dangers to mountain climbersis an avalanche, in which a large mass of snowand ice breaks loose and goes on an essentiallyfrictionless “ride” down a mountainside on acushion of compressed air.The acceleration of gravity is 9.8 m/s2 .

avala

nche

µ=0

413 m

35.4◦

If you were on a 35.4 ◦ slope and anavalanche started 413 m up the slope, howmuch time would you have to get out of theway?


Explanation:Consider the free body diagram for the

avalanche:

Mg s

inθ N =

Mg c

osθF f

=0

M g

Since the slope is essentially frictionless, theonly force with a non-zero component parallelto the surface is the weight, hence

F netx = mg sin θ

which gives us the downhill acceleration

a =1

mF netx = g sin θ = 5.67696 m/s2.

The avalance starts at zero initial speed. Atthe above acceleration, it would take time

t =

√

2s

a= 12.0624 s .

to cover distance s = 413 m.

020 10.0 pointsTake the mass of the Earth to be 5.98 ×1024 kg.If the Earth’s gravitational force causes a

falling 46 kg student to accelerate downwardat 9.8 m/s2, determine the upward accelera-tion of the Earth during the student’s fall.

Correct answer: 7.53846× 10−23 m/s2.

Explanation:By Newton’s third law, the force Fse ex-

erted on the student by the earth is equal inmagnitude and opposite in direction to theforce Fes exerted on the earth by the student.Thus

Fes = −Fse

me ae = −ms g

ae =−ms g

me

=−(46 kg)(−9.8 m/s2)

5.98× 1024 kg

= 7.53846× 10−23 m/s2


021 (part 1 of 2) 10.0 pointsConsider the 659 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 400 N andmakes an angle of 35◦ with the ceiling.The right-hand cable makes an angle of 35◦

with the ceiling and has a tension of 400 N .

659 N

T2

400 N

35◦

θ2

a) What is the tension T2 in the left-handcable slanted at an angle of θ2 with respect tothe wall?


Explanation:Observe the free-body diagram below.

F2

F1 θ1

θ2

Wg

Note: The sum of the x- andy-components of F1 , F2 , andWg are equal to zero.

Given : Wg = 659 N ,

F1 = 400 N ,

θ1 = 35◦ , 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 y2 −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)

= (400 N) cos 35◦

= 327.661 N , and

F y2 = F3 − F1 sin θ1 (2)

= 659 N− (400 N) sin 35◦

= 659 N− 229.431 N

= 429.569 N , so

F2 =√

(F x2 )

2 + (F y2 )

2

=√

(327.661 N)2 + (429.569 N)2

= 540.27 N .

022 (part 2 of 2) 10.0 pointsb) What is the angle θ2 which the left-handcable makes with respect to the ceiling?

Correct answer: 52.6649◦.

Explanation:Using Eq. 2, we have

θ2 = arctan

(

F y1

F x1

)

= arctan

(

229.431 N

327.661 N

)

= 52.6649◦ .

023 (part 1 of 2) 10.0 pointsA block with a mass of 4.1 kg is held inequilibrium on a frictionless incline of 33.0◦

by the horizontal force ~F , as shown.The acceleration of gravity is 9.81 m/s2 .


4.1kgF

33◦

33◦

What is the magnitude of F?


Explanation:

Let : m = 4.1 kg ,

g = 9.81 m/s2 , and

θ = 33◦ .

Consider the free body diagram for theblock

mg s

inθ N =

mg c

osθF

mg

Basic Concepts:

Fx,net = F cos θ − Fg‖ = 0

Fg‖ = mg sin θ = 0

Solution: Choose the coordinate axes withx along the incline and y perpendicular tothe incline. In equilibrium, the acceleration iszero. Hence from Newton’s second law in thex direction,

∑

Fx = F cos θ −mg sin θ = 0 .

Therefore

F cos θ = mg sin θ

F =mg sin θ

cos θ

F =(4.1 kg) (9.81 m/s2) sin 33◦

cos 33◦

= 26.1198 N .

024 (part 2 of 2) 10.0 pointsWhat is the magnitude of the normal force?



Fy,net = Fn − F sin θ − Fg⊥ = 0

Fg⊥ = mg cos θ

Solution:

Fn = F sin θ +mg cos θ

= (26.1198 N) sin 33◦

+ (4.1 kg) (9.81 m/s2) cos 33◦

= 47.958 N .

025 10.0 pointsA mass of 2 kg lies on a frictionless table,pulled by another mass of 3.3 kg under theinfluence of Earth’s gravity.The acceleration of gravity is 9.8 m/s2 .

2 kg

3.3 kg

What is the magnitude of the accelerationa of the two masses?

Correct answer: 6.10189 m/s2.

Explanation:

Given : m1 = 2 kg and

m2 = 3.3 kg .


m1 m2

a

T

N m1 g

a

T

m2 g

Let the direction of acceleration as indi-cated in the figure be positive. The net forceon the system is simply the weight of m2.

Fnet = m2 g = 32.34 N .

From Newton’s second law,

Fnet = m2 g = (m1 +m2) a .

Solving for a,

a =m2

m1 +m2

g

=3.3 kg

2 kg + 3.3 kg

= 6.10189 m/s2 .

026 10.0 pointsTwo blocks are arranged at the ends of a mass-less string as shown in the figure. The systemstarts from rest. When the 3.09 kg mass hasfallen through 0.442 m, its downward speed is1.35 m/s.The acceleration of gravity is 9.8 m/s2 .

3.09 kg

4.54 kg

µ

a

What is the frictional force between the4.54 kg mass and the table?


Explanation:

Given : m1 = 3.09 kg ,

m2 = 4.54 kg ,

v0 = 0 m/s , and

v = 1.35 m/s .

Basic Concept: Newton’s Second Law

F = M a

Solution: The acceleration ofm1 is obtainedfrom the equation

v2 − v20 = 2 a (s− s0)

a =v2 − v202h

=(1.35 m/s)2 − (0 m/s)2

2 (0.442 m)

= 2.06165 m/s2 .

Consider free body diagrams for the twomasses

T

m1 g

a

T

N

µN

a

m2 g

Because m1 and m2 are tied together withstring, they have same the speed and the sameacceleration, so the net force exerted on m2 is

F2 = m2 a

The net force on m1 is m1 a = m1 g − T , sothat T = m1 g −m1 a.Thus

F2 = T − fk ,

fk = T − F2

= m1 g − (m1 +m2) a

= (3.09 kg) (9.8 m/s2)

− (3.09 kg + 4.54 kg)

× (2.06165 m/s2)

= 14.5516 N .


027 (part 1 of 2) 10.0 pointsA car is traveling at 35.9 mi/h on a horizontalhighway.The acceleration of gravity is 9.8 m/s2 .If the coefficient of friction between road

and tires on a rainy day is 0.12, what is theminimum distance in which the car will stop?(1 mi = 1.609)


Explanation:Newton’s second law in the direction of

motion gives

−fk = −µkmg = ma

Solving for aa = −µkg (1)

The acceleration a may be found from thefollowing kinematics equation with vf = 0:

v2f = v20 + 2ax

a = − v202x

(2)

Combining equations (1) and (2), we obtain

v202x

= µkg

or solving for x,

x =v20

2µkg

Plugging in the appropriate values,

xwet =(16.0453 m/s)2

2(0.12) (9.8 m/s2)= 109.461 m

028 (part 2 of 2) 10.0 pointsWhat is the stopping distance when the sur-face is dry and µdry = 0.631?


Explanation:

xdry =(16.0453 m/s)2

2(0.631) (9.8 m/s2)= 20.8167 m

(A considerable difference)

029 (part 1 of 2) 0.0 pointsA descent vehicle landing on the moon hasa vertical velocity toward the surface of themoon of 29.9 m/s. At the same time, it has ahorizontal velocity of 55.7 m/s.At what speed does the vehicle move along

its descent path?


Explanation:

Let : vv = 29.9 m/s , and

vh = 55.7 m/s .

vh

vv v

θ

The speeds act at right angles to each other,so

v =√

v2h + v2v

=√

(55.7 m/s)2 + (29.9 m/s)2

= 63.2179 m/s .

030 (part 2 of 2) 0.0 pointsAt what angle with the vertical is its path?


Explanation:vh is the side opposite and vv is the side

adjacent to the angle, so

tan θ =vhvv

θ = arctan

(

vhvv

)

= arctan

(

55.7 m/s

29.9 m/s

)

= 61.7729◦ .


031 (part 1 of 2) 0.0 pointsA child rides a toboggan down a hill thatdescends at an angle of 19.1◦ to the horizontal.The hill is 20.4 m long.What is the horizontal component of the

child’s displacement?


Explanation:

Let : d = 20.4 m and

θ = −19.1◦ .

∆y

∆x

d

θ

∆x = d cos θ = (20.4 m) cos(−19.1◦)

= 19.277 m .

032 (part 2 of 2) 0.0 pointsWhat is the vertical component of the child’sdisplacement?

Correct answer: −6.67525 m.

Explanation:

∆y = d sin θ = (20.4 m) sin(−19.1◦)

= −6.67525 m .

033 (part 1 of 2) 0.0 pointsA roller coaster travels 31.8 m at an angle of18.0◦ above the horizontal.How far does it move horizontally?


Explanation:

Let : d = 31.8 m and

θ = 18.0◦ .

∆y

∆x

d

θ

∆x = d cos θ = (31.8 m) cos 18◦

= 30.2436 m .

034 (part 2 of 2) 0.0 pointsHow far does it move vertically?


Explanation:

∆y = d sin θ = (31.8 m) sin 18◦

= 9.82674 m .

035 0.0 pointsWhen the Sun is directly overhead, a hawkdives toward the ground at a speed of4.92 m/s.If the direction of his motion is at an angle

of 24.6◦ below the horizontal, calculate thespeed of his shadow along the ground.


Explanation:

Let : v = 4.92 m/s and

θ = 24.6◦ .

The velocity of the hawk’s shadow on theground will be equal to the horizontal projec-tion of the velocity of the hawk:


u = v cos θ = (4.92m/s) cos 24.6◦ = 4.47344 m/s .

036 0.0 pointsBob heads out into a lake at an angle of 29◦

with respect to the shore.If his boat is capable of a speed of 2 m/s,

how far from land will he be in 4 min and39 s ?


Explanation:Let v be the rate at which the boat is mov-

ing away from the shore.

2 m/s

v

29◦

Although the boat’s speed is vboat, it ismoving away from the shore at less thanthis speed. The boat’s speed vboat is thehypotenuse of a right triangle with v the sideopposite the angle θ, so

sin θ =v

vboatv = vboat sin θ and

(d = v t = vboat sin θ) t

= (2 m/s)(sin 29◦)(279 s)

= 270.524 m .

037 (part 1 of 2) 0.0 pointsA skier squats low and races down a 12 ◦

ski slope. During a 7 s interval, the skieraccelerates at 4.4 m/s2.What is the horizontal component of the

skier’s acceleration (perpendicular to the di-rection of free fall)?


Explanation:

Let : θ = −12◦ and

a = 4.4 m/s2 .

ay

ax

a

θ

ax = a cos θ = (4.4 m/s2) cos(−12◦)

= 4.30385 m/s2 .

038 (part 2 of 2) 0.0 pointsWhat is the vertical component of the skier’sacceleration?

Correct answer: −0.914812 m/s2.

Explanation:

ay = a sin θ = (4.4 m/s2) sin(−12◦)

= −0.914812 m/s2 .

039 0.0 pointsTwo airplanes leave an airport at the sametime. The velocity of the first airplane is660 m/h at a heading of 56.5◦. The velocityof the second is 570 m/h at a heading of 139◦.How far apart are they after 2.2 h?


Explanation:

Let : v1 = 660 m/h ,

θ1 = 56.5◦ ,

v2 = 570 m/h , and

θ2 = 139◦ .


Under constant velocity, the displacementfor each plane in the time t is

d = v t.

These displacements form two sides of a tri-angle with the angle

α = θ2 − θ1 = 82.5◦

between them. The law of cosines applies forSAS, so the distance between the planes is

d =√

d21 + d22 − 2 d1 d2 cosα .

Since

2 d1 d2 cosα = 2 (1452 m) (1254 m) cos 82.5◦

= 4.75324× 105 m2 , then

d =[

(1452 m)2 + (1254 m)2

−4.75324× 105 m2]1/2

= 1790.39 m .

040 (part 1 of 2) 0.0 pointsA car travels 11.8 km due north and then28.7 km in a direction φ = 71.7 ◦ west ofnorth.

β

θ

φ

A

B

R

x

y

N

E

S

W

Find the magnitude of the car’s resultantdisplacement.


Explanation:

Let : A = 11.8 km ,

B = 28.7 km , and

φ = 71.7◦ .

θ = 180◦ − φ = 180◦ − 71.7◦ = 108.3◦ ,

applying the Law of Cosines,

R2 = A2 +B2 − 2AB cos θ .

Since

−2AB cos θ = −2 (11.8 km)(28.7 km) cos 108.3◦

= 212.674 km2 , then

R =√

(11.8 km)2 + (28.7 km)2 + 212.674 km2

= 34.2871 km .

041 (part 2 of 2) 0.0 pointsCalculate the direction of the car’s resul-tant displacement, measured counterclock-wise from the northerly direction.

Correct answer: 52.6285 ◦.

Explanation:Applying the Law of Sines,

sinβ

B=

sin θ

R

sinβ =B

Rsin θ =

28.7 km

34.2871 kmsin 108.3◦

= 0.794716

β = arcsin 0.794716 = 52.6285 ◦ .

042 0.0 pointsA person walks 33 m East and then walks37 m at an angle 45◦ North of East.What is the magnitude of the total dis-

placement?


Explanation:


Let : Ra = 33 m ,

Rb = 37 m , and

θ = 45◦ .

The total displacement is the vector sum ofthe two displacements, so

Rx = Ra +Rb cos θ = 33 m+ (37 m) cos 45◦

= 59.163 m and

Ry = Rb sin θ = (37 m) sin 45◦ = 26.1629 m .

The final displacement is

R =√

R2x +R2

y

=√

(59.163 m)2 + (26.1629 m)2

= 64.6897 m .

043 0.0 pointsThree vectors ~A, ~B, and ~C have the followingx and y components: Ax = 2.3, Ay = −3.7;Bx = −3.4, By = 1.9; Cx = 1.6, Cy = 7.3.

What is the magnitude of ~A+ ~B + ~C ?

Correct answer: 5.52268.

Explanation:

Given : ~A = (2.3,−3.7) ,

~B = (−3.4, 1.9) , and

~C = (1.6, 7.3) .

Let ~D = ~A+ ~B + ~C. The x component o f~D is

Dx = Ax+Bx+Cx = 2.3+(−3.4)+1.6 = 0.5

and the y component of ~D is

Dy = Ay +By +Cy = −3.7+1.9+7.3 = 5.5 .

Thus the magnitude of ~A+ ~B + ~C is given by

D =√

D2x +D2

y

=√

0.52 + 5.52

= 5.52268 .

044 0.0 pointsVector ~A has a magnitude of 12 and pointsin the positive x-direction. Vector ~B has amagnitude of 17 and makes an angle of 29◦

with the positive x-axis.What is the magnitude of ~A− ~B?


Explanation:

A

A

B

−B

A−B

A−B

θ

θ

The component of vector ~B along the x-axisis

Bx = B cos θ = 17 cos 29◦ = 14.8685

and the component along the y-axis is

By = B sin θ = 17 sin 29◦ = 8.24176 .

Vector ~A points in the x-direction, so it hasno component along y-axis, so

(~A− ~B)x = Ax −Bx

= 12− 14.8685 = −2.86854 ,

(~A− ~B)y = Ay −By

= 0− 8.24176 = −8.24176 , and

‖~A− ~B‖ =√

(−2.86854)2 + (−8.24176)2

= 8.72669 .

045 (part 1 of 2) 0.0 pointsA particle has ~r(0) = (4 m) and ~v(0) =(2 m/s) ı.


If its acceleration is constant and given by~a = −(2 m/s2) (ı+ ), at what time t does theparticle first cross the x axis?

Correct answer: 2 s.

Explanation:

~r(t) = ~r(0) + ~v(0) t+1

2~a t2 , so

~r(t) = (4 m) + (2 m/s) ı t

+(−1 m/s2) (ı+ ) t2

= [(2 m/s) t− (1 m/s2) t2] ı

+[4 m− (1 m/s2) t2]

~r(t) will not have a y component when

4 m− (1 m/s2) t2 = 0

(1 m/s2) t2 = 4 m

t = 2 s .

046 (part 2 of 2) 0.0 pointsAt what time t is the particle moving parallelto the y axis; that is, in the direction?

Correct answer: 1 s.

Explanation:

~r(t) = ~r(0) + ~v(0) t+1

2~a t2

~v(t) =d~r

d t= ~v(0) + ~a t , so

~v(t) = (2 m/s) ı− (2 m/s2) (ı+ ) t

= [(2 m/s)− (2 m/s2) t] ı

−(2 m/s2) t

~v(t) will not have an x component when

2 m/s− (2 m/s2) t = 0

t = 1 s .

At this time ~v(t) = −(2 m/s) parallel tothe y-axis.

047 (part 1 of 2) 0.0 pointsA particle travels horizontally between twoparallel walls separated by 18.4 m. It movestoward the opposing wall at a constant rateof 7.4 m/s. Also, it has an acceleration in thedirection parallel to the walls of 3.9 m/s2 .

18.4 m

3.9 m/s2

7.4 m/s

What will be its speed when it hits theopposing wall?


Explanation:

Let : d = 18.4 m ,

vx = 7.4 m/s ,

a = 3.9 m/s2 .

d

b

a

vy

vx

v f θ

y

The horizontal motion will carry the parti-cle to the opposite wall, so

d = vx tf

tf =d

vx=

18.4 m

7.4 m/s= 2.48649 s


is the time for the particle to reach the oppo-site wall.Horizontally, the particle reaches the maxi-

mum parallel distance when it hits the oppo-

site wall at the time of t =d

vx, so the final

parallel velocity vy is

vy = a t =a d

vx=

(3.9 m/s2) (18.4 m)

7.4 m/s

= 9.6973 m/s .

The velocities act at right angles to eachother, so the resultant velocity is

vf =√

v2x + v2y

=√

(7.4 m/s)2 + (9.6973 m/s)2

= 12.1983 m/s .

048 (part 2 of 2) 0.0 pointsAt what angle with the wall will the particlestrike?


Explanation:When the particle strikes the wall, the ver-

tical component is the side adjacent and thehorizontal component is the side opposite theangle, so

tan θ =vxvy

θ = arctan

(

vxvy

)

= arctan

(

7.4 m/s

9.6973 m/s

)

= 37.3472◦ .

049 0.0 pointsInitially (at time t = 0) a particle is mov-ing vertically at 6.5 m/s and horizontally at0 m/s. Its horizontal acceleration is 1.8 m/s2 .At what time will the particle be traveling

at 36◦ with respect to the horizontal? Theacceleration due to gravity is 9.8 m/s2 .


Explanation:

Let : vy0 = 6.5 m/s ,

g = 9.8 m/s2 ,

vx0= 0 , and

a = 1.8 m/s2 , and

θ = 36◦ .

vxt

vytvt

36◦

The vertical velocity is

vyt = vy0 − g t

and the horizontal velocity is

vxt= vy0 + a t = a t .

The vertical component is the opposite sideand the horizontal component the adjacentside to the angle, so

tan θ =vytvxt

=vy0 − g t

a t

a t tan θ = vy0 − g t

a t tan θ + g t = vy0

t =vy0

a tan θ + g

=6.5 m/s

(1.8 m/s2) tan(36◦) + 9.8 m/s2

= 0.585176 s .

050 0.0 pointsA cannon fires a 0.499 kg shell with initialvelocity vi = 9.5 m/s in the direction θ = 50◦

above the horizontal.


∆x

∆h

9.5m/s

50◦

∆y

y

The shell’s trajectory curves downward be-cause of gravity, so at the time t = 0.343 sthe shell is below the straight line by somevertical distance ∆h.Find this distance ∆h in the absence of

air resistance. The acceleration of gravity is9.8 m/s2 .


Explanation:In the absence of gravity, the shell would fly

along the straight line at constant velocity:

x = t vi cos θ ,

y = t vi sin θ .

The gravity does not affect the x coordinateof the shell, but it does pull its y coordinatedownware at a constant acceleration ay = −g,so

x = t vi cos θ,

y = t vi sin θ −g t2

2.

Thus, x = x but y = y− 1

2gt2; in other words,

the shell deviates from the straight-line pathby the vertical distance

∆h = y − y =g t2

2.

Note: This result is completely indepen-dent of the initial velocity vi or angle θ of the

shell. It is a simple function of the flight timet.

∆h =g t2

2

=(9.8 m/s2) (0.343 s)2

2

= 0.57648 m .

051 (part 1 of 2) 0.0 pointsDuring World War I, the Germans had a guncalled Big Bertha that was used to shell Paris.The shell had an initial speed of 1.47 km/s atan initial inclination of 65.1◦ to the horizontal.The acceleration of gravity is 9.8 m/s2 .How far away did the shell hit?


Explanation:The range R is given by

R =v20g

sin(2 θ0)

=(1470 m/s)2

9.8 m/s2sin 130.2◦

= 168.417 km .

052 (part 2 of 2) 0.0 pointsHow long was it in the air?


Explanation:The time in the air is

t =R

v0x

=R

v0 cos θ0

=1.68417× 105 m

(1470 m/s) cos 65.1◦

= 272.113 s .

053 0.0 pointsA 1.8 tall basketball player attempts a goal

13.1 from the basket that is 3.05 high.


1.8 m3.05 m

13.1 m

v 0

41◦

If he shoots the ball at a 41◦ angle, at whatinitial speed must he throw the basketball sothat it goes through the hoop without strikingthe backboard? The acceleration of gravity is9.81 m/s2 .


Explanation:

Let : ∆x = 13.1 m ,

θ = 41◦ , and

g = 9.81 m/s2 .

∆y = 3.05 m− 1.8 m = 1.25 m .

1.8

m 3.05m

13.1 m

The horizontal motion defines the time:

∆x = vi cos θ∆t

∆t =∆x

vi cos θ.

Then vertically,

∆y = vi sin θ∆t− 1

2g (∆t)2

∆y = vi sin θ

(

∆x

vi cos θ

)

− 1

2g

(

∆x

vi cos θ

)2

= ∆x tan θ − g (∆x)2

2 v2i cos2 θ

g (∆x)2

2 v2i cos2 θ= ∆x tan θ −∆y

vi =

√

g (∆x)2

2 (cos θ)2 (∆x tan θ −∆y)

Since

∆x tan θ −∆y

= (13.1 m) tan 41◦ − 1.25 m

= 10.1377 m , then

vi =

√

g (∆x)2

2 cos2 θ (∆x tan θ −∆y)

=

√

(9.81 m/s2) (13.1 m)2

2 cos2 41◦ (10.1377 m)

= 12.0737 m/s .

054 0.0 pointsTo win the game, a place kicker must kick afootball from a point 15 m (16.404 yd) fromthe goal, and the ball must clear the crossbar,which is 3.05 m high. When kicked, the ballleaves the ground with a speed of 14 m/s atan angle of 56.9◦ from the horizontal.The acceleration of gravity is 9.8 m/s2 .By how much vertical distance does the ball

clear the crossbar?


Explanation:First, compute the components of the ini-

tial velocity.

vx = v cos θ

= (14 m/s) cos 56.9◦

= 7.64542 m/s ,


vy = v sin θ

= (14 m/s) sin 56.9◦

= 11.7281 m/s .

We can find the time required for the ballto reach the position of the crossbar fromx = vx t as

t =x

vx

=15 m

7.64542 m/s= 1.96196 s .

At this time the height of the football abovethe ground is

y = vy t−1

2g t2

= (11.7281 m/s)(1.96196 s)

− 1

2(9.8 m/s2)(1.96196 s)2

= 4.1485 m .

Thus, the ball clears the crossbar by

∆y = 4.1485 m− 3.05 m

= 1.0985 m .

055 (part 1 of 2) 0.0 pointsTwo lifeguards pull on ropes attached to araft. If they pull in the same direction, the raftexperiences a net external force of 327 N tothe right. If they pull in opposite directions,the raft experiences a net external force of130 N to the left.Draw a free body diagram for each situation

and find the magnitude of the larger of the twoindividual forces.


Explanation:

~Fnet = ~F1 + ~F2 ;

the larger force ~F2 must be applied to the leftin the second situation.

F2

F1

F2 F1

Let : F1 + F2 = 327 N and

F1 − F2 = −130 N .

Solve the simultaneous equations:

F1 + F2 = 327 N

−F1 + F2 = 130 N

2 F2 = 457 N

F2 =457 N

2= 228.5 N .

056 (part 2 of 2) 0.0 pointsWhat is the magnitude of the smaller of theindividual forces?


Explanation:

F1 = 327 N− F2

= 327 N− 228.5 N = 98.5 N .

057 0.0 pointsAn elevator starts from rest with a constantupward acceleration and moves 1 m in the first1.7 s. A passenger in the elevator is holding a3.7 kg bundle at the end of a vertical cord.What is the tension in the cord as the ele-

vator accelerates? The acceleration of gravityis 9.8 m/s2 .


Explanation:

h = 1 m ,

t = 1.7 s ,

m = 3.7 kg , andg = 9.8 m/s2 .

T

mg

aelevatorg


Let h be the distance traveled and a theacceleration of the elevator. Since the initialvelocity is zero,

h = v0 t+1

2a t2 =

1

2a t2

a =2h

t2.

The equation describing the forces acting onthe bundle is

Fnet = ma = T −mg

T = m (g + a) = m

(

g +2h

t2

)

= (3.7 kg)

[

9.8 m/s2 +2 (1 m)

(1.7 s)2

]

= 38.8206 N .

058 0.0 pointsA 6.1 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of 15.8 N.Find the speed of the block after it has

moved 4.5 m.


Explanation:

Let : m = 6.1 kg ,

F = 15.8 N , and

∆x = 4.5 m .

The acceleration of the block is

a =F

m=

15.8 N

6.1 kg

= 2.59016 m/s2 .

Since the force is constant, the acceleration isconstant. We find the final speed of the blockusing the equation

vf =√

v20 + 2 a (xf − x0)

=√

2 (2.59016 m/s2) (4.5 m)

= 4.8282 m/s .

059 0.0 pointsA 2.8 kg particle starts from rest and moves adistance of 3 m in 2.4 s under the action of asingle, constant force.Find the magnitude of the force.


Explanation:Basic Concept:

vf = vi + a t

v2f = v2i + 2 a x

Solution:Since vi = 0,

vf2 = 2 a x

a2 t2 = 2 a x

a2 t2 − 2 a x = 0

a(a t2 − 2 x) = 0

a =2 x

t2

The solution a = 0 must be ignored.From Newton’s second law, the force on the

particle is

F = ma

=2mx

t2

=2 (2.8 kg) (3 m)

2.4 s2

= 2.91667 N

060 0.0 pointsA 2.1 kg otter starts from rest at the top of amuddy incline 96.7 cm long and slides downto the bottom in 0.40 s.What net external force acts on the otter

along the incline?


Explanation:


Basic Concepts:

~Fnet = Σ ~F = m~a

∆x =1

2a(∆t)2

since vi = 0 m/s.Given:

m = 2.1 kg

∆x = 96.7 cm = 0.97 m

∆t = 0.40 s

Solution:

a =2∆x

(∆t)2

=2(0.967 m)

(0.4 s)2

= 12.0875 m/s2

down the incline, so

Fnet = (2.1 kg)(

12.0875 m/s2)

= 25.3837 N

down the incline.

061 0.0 pointsA 4.5 kg bucket of water is raised from a wellby a rope.The acceleration of gravity is 9.81 m/s2 .If the upward acceleration of the bucket is

3.6 m/s2, find the force exerted by the ropeon the bucket of water.


Explanation:

FT

Fg

3.6 m/s2

Note: Figure is not drawn to scale.Basic Concept:

Fnet = ma = FT − Fg

Given:

m = 4.5 kg

a = 3.6 m/s2

g = 9.81 m/s2

Solution:

FT = ma+ Fg

= ma+mg

= (4.5 kg)(

3.6 m/s2)

+ (4.5 kg)(

9.81 m/s2)

= 60.345 N

upward.

062 (part 1 of 3) 0.0 points

A 3 kg object is subjected to two forces, ~F1 =(1.8 N) ı + (−1.6 N) and ~F2 = (4.7 N) ı +(−11.6 N) . The object is at rest at the originat time t = 0.What is the magnitude of the object’s ac-

celeration?


Explanation:

Let : ~F1 = (1.8 N) ı+ (−1.6 N) ,

~F2 = (4.7 N) ı+ (−11.6 N) , and

~vo = 0m/s .

~Fnet = m~a = ~F1 + ~F2 , so

~a =~Fnet

m

=~F1 + ~F2

m

=(1.8 N + 4.7 N) ı+ (−1.6 N + (−11.6 N))

3 kg

=(

2.16667 m/s2)

ı+(

−4.4 m/s2)

.

‖~a‖ =

√

(2.16667 m/s2)2+ (−4.4 m/s2)

2

= 4.90453 m/s2 .

063 (part 2 of 3) 0.0 pointsWhat is the magnitude of the velocity at t =3.2 s ?



Explanation:

Let : t = 3.2 s .

~v = ~vo + ~a t = ~a t

=[

(

2.16667 m/s2)

ı+(

−4.4 m/s2)

]

(3.2 s)

= (6.93333 m/s) ı+ (−14.08 m/s) .

|~v| =√

(6.93333 m/s)2 + (−14.08 m/s)2

= 15.6945 m/s .

064 (part 3 of 3) 0.0 pointsWhat is the magnitude of the object’s positionat t = 3.2 s ?


Explanation:

~vavg =~vo + ~v

2=

~v

2

~r = ~vavg t

=1

2~v t

=1

2

[

(6.93333 m/s) ı+ (−14.08 m/s) ]

(3.2 s)

= (11.0933 m) ı+ (−22.528 m) .

|~r| =√

(11.0933 m)2 + (−22.528 m)2

= 25.1112 m .

065 0.0 pointsA person weighing 0.9 kN rides in an elevatorthat has a downward acceleration of 2.7 m/s2.The acceleration of gravity is 9.8 m/s2 .What is the magnitude of the force of the

elevator floor on the person?

Correct answer: 0.652041 kN.


∑

~F = m~a

~W = m~g

Solution: Since W = mg,

Fnet = ma = W − f

f = W −ma

= W(

1− a

g

)

= (0.9 kN)

(

1− 2.7 m/s2

9.8 m/s2

)

= 0.652041 kN .

066 (part 1 of 2) 0.0 pointsA child holds a sled on a frictionless, snow-covered hill, inclined at an angle of 31◦.

76N

F

31◦

If the sled weighs 76 N, find the force ex-erted on the rope by the child.


Explanation:

Given : W = 76 N and

θ = 31◦ .


mg s

inθ N =

mg c

osθ

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-nentsW sin θ acting down the hill andW cos θacting straight into the hill.The system is in equilibrium, so for forces

parallel to the hill,

Fnet = T −W sin θ = 0

=⇒ T = W sin θ

= (76 N) sin 31◦

= 39.1429 N

067 (part 2 of 2) 0.0 pointsWhat force is exerted on the sled by the hill?


Explanation:For forces perpendicular to the hill,

Fnet = N −W cos θ = 0

=⇒ N = W cos θ

= (76 N) cos 31◦

= 65.1447 N

068 (part 1 of 2) 0.0 pointsConsider the 48 N weight held by two cablesshown below. The left-hand cable is horizon-tal.

48 N

50◦

a) What is the tension in the cable slantedat an angle of 50◦?


Explanation:Observe the free-body diagram below.

40.2768 N40.2768 N

62.6595N

48N

48NScale: 10 N

50◦

Note: The sum of the x- andy-components of T1 , T2 , andWg are equal to zero.

Given : Wg = 48 N and

θ = 50◦ .

Basic Concept: Vertically, we have

Fy,net = F1 sin θ −Wg = 0

Solution:

F1(sin θ) = Wg

F1 =Wg

sin θ

=48 N

sin 50◦

= 62.6595 N


069 (part 2 of 2) 0.0 pointsb)What is the tension in the horizontal cable?


Explanation:Basic Concept: Horizontally,

Fx,net = F1 cos θ − F2 = 0

Solution:

F2 = F1 cos θ

=Wg cos θ

sin θ

=(48 N) cos 50◦

sin 50◦

= 40.2768 N

070 (part 1 of 2) 0.0 pointsThe 5.6 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 5.1 N force actsfrom above on the right at an angle of 57◦ withthe horizontal. The force 5.6 N acts straightdown.

F

5.1N

5.6 N

57◦

α

What is the magnitude of the force F?


Explanation:Standard angular measurements are from

the positive x-axis in a counter-clockwise di-rection.

Let : F1 = F ,

F2 = 5.1 N ,

α2 = 57◦ , and

F3 = −5.6 N .

Consider the free body diagram. The greenvectors are the components of the slantedforces.

F1

F 2

F3

α2

α

The weight is is equilibrium, so

∑

Fx = F1x + F2x + F3x = 0

F1x = −F2 cosα2 − 0

= −(5.1 N) cos 57◦

= −2.77766 N and

∑

Fy = F1y + F2y + F3y = 0

F1y = −F2 sinα2 − F3

= −(5.1 N) sin 57◦ − (−5.6 N)

= 1.32278 N , and

F1 =√

F 21x + F 2

1y

=√

(−2.77766 N)2 + (1.32278 N)2

= 3.07655 N .

071 (part 2 of 2) 0.0 pointsWhat is the angle α of the force F as shownin the figure?


Explanation:

α1 = arctan

(

F1y

F1x

)

= arctan

(

1.32278 N

−2.77766 N

)

= 154.535◦

measured from the positive x-axis, so

α = 180◦ − α1 = 180◦ − 154.535◦

= 25.4648◦ .


072 (part 1 of 2) 0.0 pointsAn object in equilibrium has three forces ex-erted on it. A(n) 25 N force acts at 71.6◦, anda(n) 54 N force acts at 60.2◦.What is the direction of the third force?

(Consider all angles to be measured counter-clockwise from the positive x-axis.)


Explanation:Basic Concepts:Choose a coordinate system with the pos-

itive x-axis representing 0◦ and the positivey-axis representing 90◦.Any force F has an x-component of Fx =

F cos θ and a y-component of Fy = F sin θwhere θ is measured from the x-axis.Solution:Since F1 and F2 act up and to the right, theforce F must act down and to the left, so Flies in the third quadrant.

θ

θθ

1

1

2

2

ref

F

F

F

For vertical equilibrium,

FyNet = ΣFup − ΣFdown = 0

F1y + F2y − Fy = 0

F1 sin θ1 + F2 sin θ2 − F sin θ = 0

F sin θref = F1 sin θ1 + F2 sin θ2 (1)

For horizontal equilibrium,

FxNet = ΣFright − ΣFleft = 0

F1x + F2x − Fx = 0

F1 cos θ1 + F2 cos θ2 − F cos θ = 0

F cos θref = F1 cos θ1 + F2 cos θ2 (2)

Dividing (1) by (2), we have

F sin θ

F cos θ=

F1 sin θ1 + F2 sin θ2F1 cos θ1 + F2 cos θ2

tan θ =F1 sin θ1 + F2 sin θ2F1 cos θ1 + F2 cos θ2

θref = arctan

(

F1 cos θ1 + F2θ2F1 cos θ1 + F2 cos θ2

)

Since the reference angle is in the thirdquadrant, and

F1 sin θ1 + F2 sin θ2

= (25 N) sin 71.6◦ + (54 N) sin 60.2◦

= 70.5813 N

and

F2 cos θ1 + F2 cos θ2

= (25 N) cos 71.6◦ + (54 N) cos 60.2◦

= 34.7278 N , then

θ = 180◦ + θref

= 180◦ + arctan

[

70.5813 N

34.7278 N

]

= 78.6621 N

073 (part 2 of 2) 0.0 pointsWhat is the magnitude of the third force?


Explanation:From equation (1),

F =

∣

∣

∣

∣

F1 cos θ1 + F2 cos θ2cos θ2

∣

∣

∣

∣

=

[

(25 N) cos 71.6◦ + (54 N) cos 60.2◦

cos θ2

]

= 243.802◦

074 (part 1 of 2) 0.0 pointsYour 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.


392 N 3◦

27 m

Find the force exerted by the rope on thecar when the angle is 3◦ and you are pullingwith a force of 392 N but the car does notmove.


Explanation:

Let : F = 392 N ,

θ = 3◦ , and

ℓ = 27 m .

Before the car starts moving,

∑

Fy = may ,

so

2T sin θ − F = 0

T =F

2 sin θ

=392 N

2 sin 3◦· kN

1000 N

= 3.74504 kN .

075 (part 2 of 2) 0.0 pointsHow strong must the rope be if it takes a forceof 608 N to move the car when θ is 3.1◦?


Explanation:

Let : θ = 3.1◦ .

T =F

2 sin θ

=608 N

2 sin 3.1◦· kN

1000 N

= 5.62143 kN .

076 (part 1 of 2) 0.0 pointsOn takeoff, the combined action of the enginesand wings of an airplane exerts a(n) 7210 Nforce on the plane, directed upward at anangle of 62.3◦ above the horizontal. The planerises with constant velocity in the verticaldirection while continuing to accelerate in thehorizontal direction.The acceleration of gravity is 9.8 m/s2 .What is the weight of the plane?


Explanation:The plane moves with constant velocity in

the vertical direction, so the sum of the verti-cal components of the forces equal zero.

Wplane = F sin θ

= (7210 N) sin 62.3◦

= 6383.69 N.

077 (part 2 of 2) 0.0 pointsWhat is its horizontal acceleration a?


Explanation:The mass of the plane is

m =Wplane

g

and the horizontal acceleration is

a =Fx

m

=F cos θWplane

g

=F g cos θ

Wplane

=(7210 N) (9.8 m/s2) cos 62.3◦

6383.69 N= 5.14511 m/s2.

078 (part 1 of 3) 0.0 pointsThree masses are connected by light stringsas shown in the figure.


m2

m3

m1

The string connecting the m1 and the m2

passes over a light frictionless pulley.Given m1 = 1.7 kg, m2 = 4.26 kg, m3 =

2.08 kg, and g = 9.8 m/s2. The accelerationof gravity is 9.8 m/s2 .Find the downward acceleration of m2

mass.


Explanation:Consider the free body diagrams:

m2

m3

m1

T1

a ↓

T2

a ↓

T1

↑ a

Applying Newton’s second law to each ofthese masses we get

m1 a = T1 −m1 g (1)

m2 a = T2 +m2 g − T1 (2)

m3 a = m3 g − T2 (3)

Adding these equations yields

(m1 +m2 +m3) a = (−m1 +m2 +m3) g ,

so

a =

(−m1 +m2 +m3

m1 +m2 +m3

)

g

=

(−1.7 kg + 4.26 kg + 2.08 kg

1.7 kg + 4.26 kg + 2.08 kg

)

× (9.8 m/s2)

= 5.65572 m/s2 .

079 (part 2 of 3) 0.0 pointsFind the tension in the string connecting them1 and the m2 masses.



T1 = m1 (a+ g)

= (1.7 kg)(

5.65572 m/s2 + 9.8 m/s2)

= 26.2747 N .

080 (part 3 of 3) 0.0 pointsFind the tension in the string connecting them2 and the m3 masses.



T2 = m3 (g − a)

= (2.08 kg)(

9.8 m/s2 − 5.65572 m/s2)

= 8.6201 N .

081 0.0 pointsAt a instant when a 5.7 kg object has anacceleration equal to ~a = (ax ı + ay ), where

ax = 6.8 m/s2, ay = 1.2 m/s2, one of thetwo forces acting on the object is known to

be ~f1 = (f1x ı + f1y ), where f1x = 19 N,f1y = 26 N.Determine the magnitude f2 of the other

force acting on the object.



∑

F = ma


Solution:

~Fnet = m~a = maxı+may

~f1 + ~f2 = ~Fnet, so

~f2 = ~Fnet − ~f1

= (max − f1x)ı+ (may − f1y)

Thus the magnitude of f2 is

f2 =

√

(max − f1x)2 + (max − f1y)

2

082 0.0 pointsThere is friction between the block and thetable.The suspended 2 kg mass on the left is

moving up, the 4 kg mass slides to the righton the table, and the suspended mass 4 kg onthe right is moving down.The acceleration of gravity is 9.8 m/s2 .

2 kg

4 kg

4 kg

µ = 0.13

What is the magnitude of the accelerationof the system?


Explanation:

m1

m2

m3

µ

a

Let : m1 = 2 kg ,

m2 = 4 kg ,

m3 = 4 kg , and

µ = 0.13 .

Basic Concepts: The acceleration a ofeach mass is the same, but the tensions in thetwo strings will be different.

Fnet = ma 6= 0

Solution: Let T1 be the tension in the leftstring and T2 be the tension in the right string.Consider the free body diagrams for each

mass

T1

m1 g

a T2

m3 g

a

T1 T2

N

µN

a

m2 g

For the mass m1, T1 acts up and the weightm1 g acts down, with the acceleration a di-rected upward, so

Fnet1 = m1 a = T1 −m1 g . (1)

For the mass on the table, a is directed tothe right, T2 acts to the right, T1 acts to theleft, and the motion is to the right so that thefrictional force µm2 g acts to the left and

Fnet2 = m2 a = T2 − T1 − µm2 g . (2)

For the mass m3, T2 acts up and the weightm3 g acts down, with the acceleration a di-rected downward, so

Fnet3 = m3 a = m3 g − T2 . (3)

Adding these equations yields

(m1 +m2 +m3) a = m3 g − µm2 g −m1 g

a =m3 − µm2 −m1

m1 +m2 +m3

g

=4 kg− (0.13) (4 kg)− 2 kg

2 kg + 4 kg + 4 kg

× (9.8 m/s2)

= 1.4504 m/s2 .


083 (part 1 of 2) 0.0 pointsA block weighing 9.3 N requires a force of3.9 N to push it along at constant velocity.What is the coefficient of friction for the

surface?


Explanation:Constant velocity implies the system is in

equilibrium. Thus

ΣFright = ΣFleft

The force necessary to overcome friction isF = µN . Friction always opposes the mo-tion, and acts in the opposite direction of themotion.Consider the free body diagram:

µW1

fW1

The normal force is W1, so

f − µW1 = 0

µ =f

W1

=3.9 N

9.3 N= 0.419355 .

084 (part 2 of 2) 0.0 pointsA weight W is now placed on the block and4.7 N is needed to push them both at constantvelocity.What is the weight W of the block?


Explanation:Consider the free body diagram:

µ1N

f1

W

W1

The normal force is W1 +W, so

Fnet = f1 − µ(W1 +W) = 0

f1 = µW1 + µW

W =f1 − µW1

µ

=4.7 N− (9.3 N)(0.419355)

0.419355

= 1.90769 N .

085 (part 1 of 2) 0.0 pointsTwo blocks are arranged at the ends of a mass-less cord over a frictionless massless pulley asshown in the figure. Assume the system startsfrom rest. When the masses have moved a dis-tance of 0.36 m, their speed is 1.33 m/s.The acceleration of gravity is 9.8 m/s2 .

4.2 kg

2.7 kg

µ

What is the coefficient of friction betweenm2 and the table?


Explanation:

Given : m1 = 2.7 kg ,

m2 = 4.2 kg ,

s = 0.36 m , and

v0 = 0 m/s .

Basic Concept: Newton’s Second Law

F = M a

Solution: The acceleration ofm1 is obtainedfrom the equation

v2 − v20 = 2 a (s − s0)

a =v2 − v202h

=(1.33 m/s)2 − (0 m/s)2

2 (0.36 m)

= 2.45681 m/s2 .


Consider free body diagrams for the twomasses

m2 m1

a

T

Nm2 g

a

T

m1 gµm2 g

which leads to∑

F1y : m1 a = m1 g − T (1)∑

F2x : m2 a = T − fk (2)∑

F2y : N = m2 g , (3)

where T is the tension in the cord and fromEq. 3, fk ≡ µN = µm2 g.Becausem1 andm2 are tied together with a

cord, they have same the speed and the sameacceleration. Adding Eqs. 1 & 2 we have

(m1 +m2) a = m1 g − fk = m1 g − µm2 g

so that

µm2 g = m1 g − (m1 +m2) a .

Thus

µ =m1

m2

− (m1 +m2)

m2

a

g

=2.7 kg

4.2 kg− 2.7 kg + 4.2 kg

4.2 kg

× 2.45681 m/s2

9.8 m/s2

= 0.231002 .

086 (part 2 of 2) 0.0 pointsWhat is the magnitude of the tension in thecord?


Explanation:Using Eq. 1 the tension T is

T = m1 (g − a)

= (2.7 kg) (9.8 m/s2 − 2.45681 m/s2)

= 19.8266 N

or, using Eq. 2 and µ from Part 1, the tensionT is

T = m2 [a+ µ g]

= (4.2 kg) [(2.45681 m/s2)

+ (0.231002) (9.8 m/s2)]

= 19.8266 N .

Since the T is the same using Eqs. 1 & 2: Part1, µ = 0.231002, is verified.

087 0.0 pointsA 33 kg box rests on the back of a truck. Thecoefficient of static friction between the boxand the truck bed is 0.351.The acceleration of gravity is 9.81 m/s2 .What maximum acceleration can the truck

have before the box slides backward?



Fs,max = matruck,max = µsmg

Given:

m = 33 kg

µs = 0.351

g = 9.81 m/s2

Solution: If Fs,max > ma, then the box staysin place, but if Fs,max < ma, then the box willslide to the back of the truck bed. Thus

matruck,max = µsmg

atruck,max = µsg

= (0.351)(

9.81 m/s2)

= 3.44331 m/s2

forward.

088 0.0 pointsA girl coasts down a hill on a sled, reachinglevel ground at the bottom with a speed of6.6 m/s. The coefficient of kinetic frictionbetween the sled’s runners and the hard, icy


snow is 0.060, and the girl and sled togetherweigh 768 N.The acceleration of gravity is 9.81 m/s2 .How far does the sled travel on the level

ground before coming to a rest?


Explanation:

768 NFn

Fkµk = 0.06

vi = 6.6 m/s

Note: Figure is not drawn to scale.Basic Concepts:

Fk = µkFn

Fx,net = max = −Fk

0 = v2i + 2a∆x

since vf = 0 m/s.

Given : vi = 6.6 m/s ,

µk = 0.060 ,

Fn = Fg = 768 N , and

g = 9.81 m/s2 .

Solution:

m =Fg

g=

768 N

9.81 m/s2= 78.2875 kg

ax = −µkFn

m

= −0.06(768 N)

78.2875 kg

= −0.5886 m/s2

∆x = − v2i2ax

= − (6.6 m/s)2

2 (−0.5886 m/s2)

= 37.0031 m .

089 0.0 pointsThe coefficient of static friction between the4.41 kg crate and the 35.4◦ incline is 0.287.The acceleration of gravity is 9.8 m/s2 .

4.41 k

g

µ k

F

35.4◦

What minimum force F must be applied tothe crate perpendicular to the incline to pre-vent the crate from sliding down the incline?


Explanation:

Given : m = 4.41 kg ,

w = mg = (4.41 kg) (9.8 m/s2) = 43.218 N ,

θ = 35.4◦ , and

µs = 0.287 .


mg s

inθ

N =mg c

osθ

µN

F

m g

Vertically,

∑

Fy = n−Wy − F = 0

so the normal force is

n = Wy + F = mg cos θ + F

Horizontally,

∑

Fx = mg sin θ − fs = 0 ,


so the force of static friction is

fs = mg sin θ .

But

fs ≤ µsN = µs (mg cos θ + F )

mg sin θ ≤ µsmg cos θ + µs F

Thus

F ≥ mg sin θ

µs−mg cos θ

=(4.41 kg)

(

9.8 m/s2)

sin 35.4◦

0.287− (4.41 kg)

(

9.8 m/s2)

cos 35.4◦

= 52.0031 N .

Homework 3 Solutions

Documents

free body

free body

net external

t1 t2

m1 m2

hand cable

cars resultant

body diagram