solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 13

COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Chapter 13, Solution 1.
Given: Weight of satellite, 1000 lbW =
Speed of satellite, 14,000 mi/hv =
Find: Kinetic energy, T
( )( )
h
14,000 mi/h 5280 ft/mi 20,533 ft/s
3600 s
v
 
= = 
 
( )
( )
2
2
1000 lb
Mass of satellite 31.0559 lbs /ft
32.2 ft/s
= =
( )( )22 91 1
31.0559 20,533 6.5466 10 lb ft
2 2
T mv= = = × ⋅
9
6.55 10 lb ftT = × ⋅
Note: Acceleration of gravity has no effect on the mass of the satellite.
9
6.55 10 lb ftT = × ⋅ !

Circumference
Time
v
 
  
=
( )( )( )
( )( ) ( )( )
2 6370 km 35,800 km 1000 m/km
3075.2 m/s
23 hr 3600 s/hr 56 min 60 s/hr
v
π +
= =
+
3075.2 m/sv =
( )( )221 1
Kinetic energy, 500 kg 3075.2 m/s
2 2
T mv= =
2.36 GJT = !

Given: Mass of stone, 2 kgm =
Velocity of stone, 24 m/sv =
Acceleration of gravity on the moon, 2
1.62 m/smg =
Find:
(a) Kinetic energy, T
Height h, from which the stone was dropped
(b) T and h on the Moon
(a) On the Earth ( )( )221 1
2 kg 24 m/s 576 N m
2 2
T mv= = = ⋅ 576 JT = !
( )( )2
2 kg 9.81 m/s 19.62 NW mg= = =
1 1 2 2 1 1 2 20 576 JT U T T U Wh T− −= = = = =
( )
( )
2
2
576 N m
29.36 m
19.62 N
T
Wh T h
W
⋅
= = = =
29.4 mh = !
(b) On the Moon
Mass is unchanged. 2 kgm =
Thus T is unchanged. 576 JT = !
Weight on the moon is, ( )( )2
2 kg 1.62 m/sm mW mg= =
N24.3=mW
( )576 N m
177.8 m
3.24 N
m
m
T
h
W
⋅
= = =
177.8 mmh = !

2
1 lb 1
1.6203
16 oz 32.2 ft/s
m
  
=   
   
(a) 2 21 1 1.62
(160 ft/s)
2 2 16(32.2)
T mv
 
= =  
 
40.2 ft-lbT = !
At maximum height,
(160 ft/s) cos25xv v= = °
(b) 21
(160cos25 )
2
T m= °
33.1 ft-lbT = !

Use work and energy with position 1 at A and position 2 at C.
At 1
0yFΣ =
1
1
cos30 0
cos30
N mg
N mg
⇒ − °=
= °
0yFΣ =
2
2
= 0N mg
N mg
⇒ −
=
Work and energy
1 1 2 2T V T→+ = (1)
Where
2 2
1 1
1 1
(4ft/s) 8
2 2
T mv m m+ = =
1 2 1 2 (20) ( sin30 )k kV N d N mg dµ µ→ = − − + °
2 2
2 2
1 1
(8) 32
2 2
T mv m m= = =
Into (1)
8 cos30 (20) sin30 32k km mgd mg mgd mµ µ− ° − + °=
Solve for
32 8 20 32 8 (0.25)(32.2)(20)
20.3 ft
cos30 sin30 ( 0.25) (32.2)(0.866 32.2(0.5))
k
k
g
d
g g
µ
µ
− + − +
= = =
− °+ ° − +
!
At 2

(a) Use work and energy from A to B. 1 1 2 2T V T→+ =
2 2
1 1
1 1 50
(40) 1242.24 ft lb
2 2 32.2
T mv
 
= = = ⋅ 
 
2 0T = (Stops at top)
1 2 sin 20U Nx mg xµ→ = − − °
N is needed
y 0FΣ = cos20 (50 lb)cos20 46.985 lbN W⇒ = °= °=
So
1 2 0.15(46.985) 50sin 20
24.149
U x x
x
→ = − − °
= −
Substitute
1242.24 24.149 0x− =
51.44 ftx = 51.4 ftx = !
(b) Package returns to A – use work and energy from B to A
2 2 3 3T U T→+ =
Where 2 0T = (At B)
2 3 sin 20 kU W x Nxµ→ = ° −
(50) sin 20 (51.44) 0.15(46.985)(51.44)= ° −
517.13 ft lb= ⋅

2 2 2
3 3 3 3
1 1 50
0.7764
2 2 32.2
T mv v v
 
= = = 
 
Substitute
2
30 517.13 0.7764v+ =
3 25.81 ft/sv = 3 25.8 ft/sv = 20°!
(c) Energy dissipated is equal to change of kinetic energy
2 2
1 3 1 2
1 1
2 2
T T mv mv− = −
2 21 50
(40 25.81 )
2 32.2
 
= − 
 
725 ft lb= ⋅
Energy dissipated 725 ft lb= ⋅ !

Given: Automobile Weight W = mg = (2000 kg) (9.81)
19,620 NW =
Initial Velocity A, 0m/sAv =
Incline Angle, 6α = °
Vehicle brakes at impending slip for 20 m from B to C
0Cv =
Find; speed of automobile at point B, vB
Coefficient of static friction, µ
(a) (19620 N) (150 m)sin6A B A BU Wh→ →= − °
3
307.63 10 N m= × ⋅
21
0
2
A B B AU T T mv→ = − = −
3 21
307.63 10 N m (2000kg) 0
2
Bv× ⋅ = −
17.54 m/sBv = !
(b) 0A C A C B C C AU Wh Fd T T→ → →= − = − =
20 mB Cd → = NF µ=
Where coefficient of static frictionµ =
(19620N)(sin6 )(170m) (20m)A CU F→ = ° −
(19620N) cos6F µ= °
(19620N)(sin6 )(170m) (19620N)(cos6 ) (20m) 0µ° − ° =
170
tan6 0.893
20
µ = °= 0.893µ = !

Given: Automobile weight, (2000)(9.81) 19620 NW = =
Initial velocity at A, 0 m/sAv =
Incline Angle, 6α = °
Vehicle costs 150 m from A to B
Vehicle skids 20 m from B to C
Dynamic friction coefficient, 0.75µ =
Find: Work done on automobile by air resistance and rolling resistance between points A and C.
(20 m) 0A C R A C C AU U Wh F T T→ →= + − = − =
N 0.75 (19620 N) cos6F µ= = °
UR = Resistance work
0.75 (19620 N)cos6 (20 m) (19620 N)sin6 (170 m)= ° − °
3
56.0 10 N mRU = − × ⋅
3
56.0 10 Jor − × !

90cos20 sin50 0NF N P= − °− °=∑
90cos20 sin50N P= °+ °
1 2 [ cos50 90sin 20 0.35 N](3 ft)U P→ = °− °−
2
2 2
1 90 lb
(2 ft/s)
2 32.2 ft/s
T
 
=  
 
2( cos50 ) 3 (90sin 20 ) (3) 0.35 (90cos20 sin50 )3P P T° − ° − °+ ° =
21 90
(3 cos50 0.35(3) sin50 ) 90sin 20 (3) 0.35 (90cos20 )(3) (2)
2 32.2
P
 
°− ° = ° + ° +  
 
186.736
166.1 lb
1.12402
P = = !

(a) First stage: 1 2 2(5.5 3)(50) 125 lb ft =U T→ = − = ⋅
2
2 2
1 3
2 32.2
T v
 
=  
 
2 51.8 ft/sv = !
(b) At the top: 2 3 3( 50) 0 125U h→ = − − = −
275
,
3
h∴ = 91.7 fth = !
(c) At the return: 2
3 4 4 4
1 3
3(91.6667)
2 32.2
U T v→
 
= + = =   
4 76.8 ft/sv = !

WA = 7(9.81) = 68.67 N
Given: Block A is released from rest and moves up incline 0.6 m.
Friction and other masses are neglected
Find: Velocity of the block after 0.6 m, v
From the Law of Cosines
2 2 2
(1.2) (0.6) 2(1.2)(0.6)cos 15d = + − °
2 2
0.4091 md =
0.63958 md =
C CU W= (Distance pulley C lowered)
1
140 N (1.2 0.63958) m 39.229 N m
2
 
= − = ⋅ 
 
68.67 N (sin15 )(0.6 m) 10.6639 N mAU = − ° = − ⋅
2 1 C AU T T U U= − = −
21
0
2
A C Am v U U− = −
21
(7kg) (39.229 10.6639) N m
2
v = − ⋅
2
8.1615v =
2.857 m/sv = 2.86m/sv = 15°!

WA = 7(9.81) = 68.67 N
Given: Block A is released at the position shown at a velocity of
1.5 m/s up.
After moving 0.6 m the velocity is 3 m/s.
Find: work done by friction force on the block, Vf J
From the Law of Cosines
2 2 2
(1.2) (0.6) 2(1.2)(0.6)(cos 15 )d = + − °
2 2
0.4091md =
0.63958md =
UC
1
140N (1.2 0.63958) m 39.229 N m
2
 
= − = ⋅ 
 
68.67 N(sin15 )(0.6 m) 10.664 N mAU = − ° = − ⋅
2 2
2 1 2 1
1
[ ]
2
C A friction AU U U T T m v v+ − = − = −
2 2 21
39.229 10.664 (7 kg)[(3) (1.5) ] m
2
frictionU− − = −
39.229 10.664 23.625frictionU− = − + +
4.94 J= −
4 94 JfrictionU .= − !

Given: At A, 0v v=
For AB, 0.40kµ =
At B, 2 m/sv =
Find: 0v
( )22 2
0
1 1 1
2 m/s
2 2 2
A B BT mv T mv m= = =
2 mBT =
( )( )sin15 N 6 mA B kU W µ− = ° −
0 cos15 0F N WΣ = − ° =
cos15N W= °
( )( )sin15 0.40cos15 6 mA BU W− = ° − °
( )0.76531 0.76531A BU W mg− = − = −
A A B BT U T−+ =
2
0
1
0.76531 2 m
2
mv mg− =
( ) ( )( )( )2 2
0 2 2 0.76531 9.81 m/sv = +
2
0 19.0154v =
0 4.36 m/sv = !

Given: At A, 0v v=
At B, 0v =
For AB, 0.40kµ =
Find: 0v
2
0
1
0
2
A BT mv T= =
( )( )sin15 6 mA B kU W Nµ− = ° −
0 cos15 0F N WΣ = − ° =
cos15N W= °
( )( )sin15 0.40cos15 6 mA BU W− = ° − °
( )0.76531 0.76531A BU W mg− = − = −
A A B BT U T−+ =
2
0
1
0.76531 0
2
mv mg− =
( )( )( )2 2
0 2 0.76531 9.81 m/sv =
2
0 15.015v =
0 3.87 m/sv = !
Down to the left.

Car C 3 2 3
0 0 (35 10 kg) (9.81 m/s ) 343.35 10 Ny C C CF N M g NΣ − ⇒ − = ⇒ = × = ×
So, 3 3
(0.35)(343.35 10 ) 120.173 10 NCF = × = ×
Car B 3 2 3
0 0 (45 10 kg)(9.81m/s ) 441.45 10 Ny B B BF N M g NΣ = ⇒ − = ⇒ = × = ×
So, 3 3
0.35(441.45 10 ) 154.508 10 NBF = × = ×
Also,
1 (54 km/h)(1h/3600s)(1000 m/km) 15 m/sv = =
(a) Work and energy for the train
1 1 2 2T U T→+ =
3 3 3 2 3 31
(35 10 45 10 35 10 )(15) (120.173 10 154.508 10 ) 0
2
x× + × + × − × + × =
47.10 mx =
47.1 mx = !

(b) Force in each coupling
Car A
Car C
1 1 2 2T U T→+ =
( )( ) ( )231
35 10 15 47.10 0
2
ABF× − =
3
83.599 10 NABF = ×
83.6 kNABF = !
Tension
1 1 2xT U T→+ =
( )( ) ( )( )23 31
35 10 15 120.173 10 47.10 0
2
BCF× + − × =
Solve for FBC
3
36.6 10 NBCF = ×
36.6 kNBCF = !
Tension

0 0y A AF N M gΣ = ⇒ − =
( )( )3 3
35 10 9.81 343.35 10 NAN = × = ×
so,
( )( )3 3
0.35 343.35 10 120.173 10 NAF = × = ×
( )1 54 km/h 15 m/sv = =
(a) Work - energy for the entire train
1 1 2 2T U T→+ =
( ) ( ) ( ) ( )23 3 3 31
35 10 45 10 35 10 15 120.173 10 0
2
x × + × + × − × =
 
107.66 mx =
107.7 mx = !
(b) Force in each coupling
Car A
1 1 2 2T U T→+ =
( )( ) ( )( )23 31
35 10 15 120.173 10 107.66 0
2
ABF× − + × =
3
83.60 10 NABF = − ×
83.6 kNABF = !
Compression
1 1 2 2T U T−+ =
( )( ) ( )231
35 10 15 107.66 0
2
BCF× + =
3
36.57 10 NBCF = − ×
36.6 kNBCF = !
Compression
Car A
Car C

Car B:
Car A:
Given: Car B towing car A uphill at A constant speed of 30 ft/s
Car B skids to a stop. µk = 0.9
Car A strikes rear of car B.
Find: Speed of car A before collision, vA
Let d = Distance traveled by car B after braking.
1 2 2 1U T T− = − ( )21
sin5
2
Bm v mg F d− = − °−
( )21
30
2
sin5 0.9 cos5
m
d
mg mg
= −
°+ °
( ) ( )( )
450 450
32.2 sin5 0.9cos5 32.2 0.9837
d = =
°+ °
14.206 ft traveled byd B=
For car A, travel to contact
2 2
1 1 1
1 1
2 2
C A AU T T mv mv→ = − = −
( )( ) ( )221 1
sin5 15 30
2 2
Amg d mv m− ° + = −
( )( )21
450 32.2sin5 14.206 15
2
Av − = − ° +
21
368.036
2
Av =
27.13Av = 27.1 ft/sAv = !

F = 0.8 NA
Given: Car B tows car A at 30 ft/s uphill.
Car A brakes for 4 wheels skid µk = 0.8
Car B continues in same gear and throttle setting.
Find: (a) distance, d, traveled to stop
(b) tension in cable
(a) 1 Traction force (from equilibrium)F =
( ) ( )1 3000 sin5 2500 sin5F = ° + °
5500sin5= °
For system A + B
( )1 2 1 3000 sin5 2500sin5U F F d→  = − °− ° − 
( )22
2 1
1 1 5500
0 30
2 2 32.2
A BT T m v+
 
− = − = −   
Since ( )1 3000 sin5 2500 sin5 0F − ° − ° =
( )0.8 3000cos5 76863 ft lbFd d− = − ° = − ⋅
32.1 ftd = !
(b) cable tension, T
( )( )1 2 2 10.8 sin5 32.149A AU T N W T T→ = − − ° = −
( )( )( ) ( )
( )
( )
2
3000 30
0.8 3000 cos5 3000 sin5 32.149
2 32.2
T − ° − ° =
( )2652.3 1304T − = −
1348 lb=
1348 lbT = !
NA = 3000 cos 5°
NB = 2500 cos 5°

Given: Blocks A, B released from rest and friction and masses of pulleys
neglected.
Find: (a) Velocity of block A, vA, after moving down dA = 1.5 ft.
(b) The tension in the cable.
(a) constraint 3 0A Bv v+ =
1
3
B Av v=
Also,
1
3
B Ad d=
( ) ( )1 2 sin30 sin30A A B BU W d W d→ = ° − °
( )( ) ( )
1.5
20 sin30 1.5 16 sin30
3
 
= ° − °  
 
11 ft lb= ⋅
2 2
1 2
1
0,
2 2
A A B BT T m v m v
1
= = +
2
2 21 20 1 16
0.33816
2 32.2 2 32.2 3
A
A A
v
v v
     
= + =     
     
2
1 2 2 1 ; 11 0.33816 AU T T v→ = − =
5.703Av = 5.70 ft/sAv = 30° !
(b) For A alone
( ) ( ) ( )2
1 2
1
sin30
2
A A A A AU W d T d m v→ = ° − =
( )( ) ( ) ( )21 20
20 0.5 1.5 1.5 5.703 10.102
2 32.2
T
 
− = = 
 
3.265 ft lbT = ⋅
3.27 ft lbT = ⋅ !

Given: System at rest when 500 N force is applied to collar A. No
friction. Ignore pulleys mass.
Find: (a) Velocity, Av of A just before it hits C.
(b) Av If counter weight B is replaced by a 98.1 N
downward force.
Kinematics
2B AX X=
2B Av v=
(a) Blocks A and B
2 2
1 2
1 1
0
2 2
B B A AT T m v m v= = +
( )( ) ( )( )2 2
2
1 1
10 kg 2 20 kg
2 2
A AT v v= +
( )( )2
2 30 kg AT v=
( ) ( )( ) ( )( )1 2 500 A A A B BU X W X W X− = + −
( )( ) ( )( )2
1 2 500 N 0.6 m 20 kg 9.81 m/s 0.6 mU − = + ×
( )( )2
10 kg 9.81 m/s 1.2 m− ×
1 2 300 117.72 117.72 300 JU − = + − =

( ) 2
1 1 2 2 0 300 J 30 kg AT U T v−+ = + =
2
10Av =
3.16 m/sAv = !
(b) Since the 10 kg mass at B is replaced by a 98.1 N force, kinetic
energy at 2 is,
( )2 2
2 1
1 1
20 kg 0
2 2
A A AT m v v T= = =
The work done is the same as in part (a)
1 2 300 JU − =
( ) 2
1 1 2 2 0 300 J 10 kg AT U T v−+ = + =
2
30Av =
5.48 m/sAv = !

PROBLEM 13.21 CONTINUED
( ) 2
1 1 2 2 0 300 J 30 kg AT U T v−+ = + =
2
10Av =
3.16 m/sAv = !
(b) Since the 10 kg mass at B is replaced by a 98.1 N force, kinetic
energy at 2 is,
( )2 2
2 1
1 1
20 kg 0
2 2
A A AT m v v T= = =
The work done is the same as in part (a)
1 2 300 JU − =
( ) 2
1 1 2 2 0 300 J 10 kg AT U T v−+ = + =
2
30Av =
5.48 m/sAv = !

Given: 10 kg; 4 kg; 0.5 mA Bm m h= = =
System released from rest.
Block A hits the ground without rebound.
Block B reaches a height of 1.18 m.
Find: (a) Av just before block A hits the ground.
(b) Energy, ,PE dissipated by the pulley friction.
(a) Bv at 2 Av= at 2 just before impact.
from 2 to 3; Block B
( )2 2 2
3 2
1 1
0 4 2
2 2
B B B BT T m v v v= = = =
Tension in the cord is zero, thus
( )( )( )2
2 3 4 kg 9.81 m/s 0.18 m 7.0632 JU − = − = −
2 2
2 2 3 3; 2 7.0632; 3.5316B BT U T v v−+ = = =
2 2
3.5316 1.8793B A B Av v v v= = = = 1.879 m/sAv = !
(b) From 1 to 2 Blocks A and B,
( ) 2
1 2 2
1
0
2
A BT T m m v= = +
Just before impact 2 1.793 m/sB Av v v= = =
( )( )2
2
1
10 4 1.8793 24.722 J
2
T = + =
( ) ( )1 2 0.5 0.5 ;A B PU W W E− = − −
Energy dissipated by pulleyPE =
( )( ) ( )2
1 2 9.81 m/s 10 4 kg 0.5 m 29.43P PU E E−
 = − − = −
 
1 1 2 2; 0 29.43 24.722PT U T E−+ = + − =
4.708PE = 4.71 JPE = !

Given: 8 kg; 10 kg; 6 kgA B Cm m m= = =
System released from rest.
Collar C removed after blocks move 1.8 m.
Find: Av , just before it strikes the ground.
Position 1 to position 2
1 10 0v T= =
At 2, before C is removed from the system
( ) 2
2 2
1
2
A B CT m m m v= + +
( ) 2 2
2 2 2
1
24 kg 12
2
T v v= =
( ) ( )1 2 1.8 mA C BU m m m g− = + −
( ) ( )1 2 8 6 10 g 1.8 m 70.632 JU − = + − =
2
1 1 2 2 2; 0 70.632 12T U T v−+ = + =
2
2 5.886v =
Position 2 to position 3
( ) ( )2
2 2
1 18
5.886 52.974
2 2
A BT m m v′ = + = =
( ) 2 2
3 3 3
1
9
2
A BT m m v v= + =
( ) ( ) ( )( )( )2
2 3 2 0.6 2 kg 9.81 m/s 1.4 mA BU m m g′− = − − = −
2 3 27.468 JU ′− = −
2
2 2 3 3 352.974 27.468 9T U T v′−′ + = = − =
2
3 32.834 1.68345v v= = 1.683 m/sAv = !

Given: Conveyor is disengaged, packages held by friction and system is released from rest. Neglect mass of
belt and rollers. Package 1 leaves the belt as package 4 comes onto the belt.
Find: (a) Velocity of package 2 as it leaves the belt at A.
(b) Velocity of package 3 as it leaves the belt at A.
(a) Package 1 falls off the belt, and 2, 3, 4 move down.
2.4
0.8 m
3
=
2
2 2
1
3
2
T mv
 
=  
 
( ) 2
2 2
3
3 kg
2
T v=
2
2 24.5T v=
( )( )( ) ( )( ) ( )2
1 2 3 0.8 3 3 kg 9.81 m/s 0.8U W− = = ×
1 2 70.632 JU − =
2
1 1 2 2 20 70.632 4.5T U T v−+ = + =
2
2 15.696v =
2 3.9618v = 2 3.96 m/sv = !

Work and energy 1 1 2 2T U T→+ = (1)
Where 1 20; 0T T= =
Work
Outer spring ( )22
1 2 1
1 1 N
3000 1.5 m
2 2 m
V k x−
 
= − = −  
 
33.75 J= −
Inner spring ( )22
1 2 2
1 1 N
10,000 0.06 m
2 2 m
U k x−
 
= − = −  
 
18 J= −
Gravity ( )1 2 0.15U mg h− = +
( )( )( )8 9.81 0.15 78.48 11.722h h= + = +
Total work 1 2 33.75 18 78.48 11.772U h− = − − + +
39.978 78.48 h= − +
Substituting into (1)
0 39.978 78.48 0h− + =
0.5094 mh =
509 mmh = !

Work and energy
Assume 0.09 mx >
1 1 2 2T U T→+ = (1)
Where 1 20; 0T T= =
Work
Outer spring ( )1
2 2 2
1 2 1
1 1
3000 1500
2 2
U k x x x→ = − = = −
Inner spring ( ) ( )2
2 2
1 2 2
1
0.09 5000 0.09
2
U k x x→ = − − = − −
Gravity ( )1 2 0.6U mg x→ = +
( )( )( )8 9.81 0.6 78.48 47.09x x= + = +
Total work
( )22
1 2 1500 5000 0.09 78.48 47.09U x x x→ = − − − + +
Substitute into (1)
2
6500 978.48 6.588 0x x− + + =
Solve
0.1570 mx = or –0.00646
Reject negative solution 157.0 mmx = !

(a)
Given: A 0.7 lb block rests on a 0.5 lb block which is not attached to a
spring of constant 9 lb/ft; upper block is suddenly removed.
Find: (a) maxv of 0.5 lb block
(b) maximum height reached by the 0.5 lb block
At the initial position (1), the force in the spring equals the weight of both
blocks, i.e., 1.2 lb.
Thus at a distance x, the force in the spring is,
1.2sF kx= −
1.2 9sF x= −
Max velocity of the 0.5 lb block occurs while the spring is still in contact
with the block.
2 2 2
1 2
1 1 0.5 0.25
0
2 2
T T mv v v
g g
 
= = = = 
 
( ) 2
1 2 0
9
1.2 9 0.5 0.7
2
x
U x dx x x x− = − − = −∫
2 2
1 1 2 2
9 0.25
0.7
2
T U T x x v
g
−+ = = − =
2 29
4 0.7
2
v g x x
 
= − 
 
max when 0 0.7 9 0.077778 ft
dv
V x x
dx
= = − ⇒ =
( ) ( )
22
max
9
4 0.7 0.077778 0.077778
2
v g
 
= − 
 
2
max 3.5063v =
max 1.87249v =
max 1.872 ft/sv = !

(b)
0 Initial compressionx =
0
1.2 lb
0.133333 ft
9 lb/ft
x = =
1.2 9sF x= −
1 30, 0T T= =
0
1 3 0
0.5
x
sU F dx h− = −∫
( )0
1 3 0
1.2 9 0.5
x
U x dx h− = − −∫
2
0 0
9
1.2 0.5
2
x x h= − −
( ) ( )29
1.2 0.133333 0.133333 0.5
2
h= − −
0.08 0.5h= −
( )1 1 3 3: 0 0.08 0.5 0T U T h−+ = + − =
0.16 fth =
1.920 in.h = !

(a) Same as 13.25 solution for Part (a) max 1.872 ft/sv = !
(b) With 0.5 lb block attached to the spring, refer to figure in (b) of Problem 13.27.
( )1 3 1 3 0
0 0 1.2 9 0.5
h
T T U x dx h−= = = − −∫
Since the spring remains attached to the 0.5 lb block,
the integration must be carried out for the total distance, h.
2
1 1 3 3
9
0 0.7 0
2
T U T h h−+ = + − =
( )
2
0.7 lb 0.155556 ft
9 lb/ft
h
 
= = 
 
1.86667 in.h =
1.867 in.h = !

Position 1, initial condition
Position 2, spring deflected 5 inches
Position 3, initial contact of spring with collar
( )
2
1 2
18 5 1 5 18 5
60 + 7.5 sin30
12 2 12 12
(Friction) (Spring) (Gravity)
U F→
+ +     
= − − °     
     
1 2 1 20, 0T T U −= = ∴ =
( ) ( ) ( ) ( )
2
23 1 5 23
0 7.5 0.866 60 7.5 0.5
12 2 12 12
µ
     
= − − +          
(a) 0.1590µ = !
(b) Max speed occurs just before contact with the spring
( ) ( ) ( ) 2
1 3 3 max
18 18 1 7.5
7.5 0.866 7.5 0.5
12 12 2 32.2
U T vµ→
     
= − + = =          
max 5.92 ft/sv = !

(a) W = Weight of the block ( )10 9.81 98.1 N= =
1
2
B Ax x=
( ) ( ) ( )2 2
1 2
1 1
2 2
A A A B BU W x k x k x− = − −
(Gravity) (Spring A) (Spring B)
( )( ) ( )( )2
1 2
1
98.1 N 0.05 m 2000 N/m 0.05 m
2
U − = −
( ) ( )21
2000 N/m 0.025 m
2
−
( ) ( )2 2
1 2
1 1
10 kg
2 2
U m v v− = =
( ) 21
4.905 2.5 0.625 10
2
v− − =
0.597 m/sv = !
(b) Let Distance moved down by the 10 kg blockx =
( ) ( ) ( )
2
2 2
1 2
1 1 1
2 2 2 2
A B
x
U W x k x k m v−
 
= − − =  
( ) ( ) ( )21
0 2
2 8
B
A
d k
m v W k x x
dx
 
= = − − 
 

( ) ( ) ( )
2000
0 98.1 2000 2 98.1 2000 250
8
x x x= − − = − +
( )0.0436 m 43.6 mmx =
For ( ) 21
0.0436, 4.2772 1.9010 0.4752 10
2
x U v= = − − =
max 0.6166 m/sv =
max 0.617 m/sv = !

(a) W = Weight of the block = (10)(9.81) = 98.1 N
1
2
B Ax x=
Block moves down at release after spring A is stretched 25 mm
2 2
1 2
1 1
( ) ( ) ( )
2 2
A A A B BU W x k x k x− = + −
(Gravity) (Spring A) (Spring B)
2
1 2
1
(98.1 N)(0.025 m) (2000 N/m)(0.025 m)
2
U − = +
21
(2000 N/m)(0.0125 m)
2
−
2 21 1
( ) (10 kg)
2 2
m v v= =
2
1 2
1
2.4525 0.625 0.15625 (10)
2
U v− = + − =
0.764 m/sv = 0.764 m/sv = !
(b) Let x = Distance moved down by the 10 kg block
(for x > 25 mm)
2
1 2
1 1
(0.025) ( 0.025)
2 2
A AU Wx k k x− = + − −
2
21 1
( )
2 2 2
B
x
k M v
 
− = 
 
2
2 2
1 2
1 1 1
98.1 (2000)(0.025) (2000)( 0.025) (2000)
2 2 2 2
x
U x x−
 
= + − − −  
 
21
(10)
2
v=

21
0 (10) 98.1 2000( 0.025) 2000
2 4
d x
v x
dx
   
= = − − −   
   
98.1 2000 50 500x x= − + −
148.1
0.05924 m ( 59.24 mm)
2500
x = = =
For 0.05924 mx =
2
1 2 98.1(0.05924) 0.625 1000(0.03424) 0.87734U − = + − −
21
5.8114 0.625 1.1724 0.87734 (10)
2
v+ − − =
max 0.937 m/sv =
max 0.937 m/sv = !

(a)
Assume auto stops in 1.5 4 m.d< <
( )( )22
1 1 1
1 1
27.778 m/s 1000 kg 27.778 m/s
2 2
v T mv= = =
385809 J 385.81 kJ= =
2 20 0T v= =
( )( ) ( )( )1 2 80 kN 1.5 m 120 kN 1.5 120 120 180 120 60U d d d− = + − = + − = −
1 1 2 2 385.81 120 60 3.715 mT U T d d−+ = = − =
3.72 md = !
Assumption that 4 md < is O.K.
(b) Maximum deceleration occurs when F is largest.
For 3.3401 m, 120 kN, thus Dd F F ma= = =
( ) ( )( )120,000 N 1000 kg Da=
2
120 m/sDa =
2
120.0 m/sDa = !

Pressures vary inversely as the volume
L
L
P Aa Pa
P
P Ax x
= =
( ) ( )2 2
R
R
P Aa Pa
P
P A a x a x
= =
− −
Initially at 1 0
2
a
v x= =
1 0T =
At 2, 2
2
1
,
2
x a T mv= =
( )2 2
1 2
1 1
2
a a
a a
L Ru P P Adx PaA dx
x a x
−
 
= − = − − 
∫ ∫
( )1 2
2
ln ln 2
a
au paA x a x−  = + − 
1 2
3
ln ln ln ln
2 2
a a
u paA a a−
    
= + − −    
    
2
2
1 2
3 4
ln ln ln
4 3
a
u paA a paA−
   
= − =   
  
2
1 1 2 2
4 1
0 ln
3 2
T U T paA mv−
 
+ = + = 
 
2
4
2 ln
3
0.5754
paA
paA
v
m m
 
 
 = = 0.759
paA
v
m
= !

( ) ( )
2
2 2
/
1
E E
h
h
R
GM m GM m R
F mg
h R
= = =
+ +
At earths’ surface ( )0h = 02
EGM m
mg
R
=
( )
2
02 2
1
EGM
E R
h
h
R
GM
g g
R
= =
+
Thus
( )
0
2
1
h
h
R
g
g =
+
6370 kmR =
At altitude h, “true” weight h TF mg W= =
Assume weight 0 0W mg=
0 0 0
0 0 0
Error T h hW W mg mg g g
E
W mg g
− − −
= = = =
( )
( )
( )
0
20
1
0
2 2
0
1
1
1 1
h
R
g
h
h h
R R
g
g
g E
g
+
−  
 = = = −
 + +  
(a)
( )
2
1
6370
1
1 km: 100 100 1
1
h P E
 
 = = = −
 +  
0.0314%P = !
(b)
( )
2
1000
6370
1
1000 km: 100 100 1
1
h P E
 
 = = = −
 +  
25.3%P = !

Newtons law of gravitation
2 2
1 0 2
1 1
2 2
T mv T mv= =
( )
2
1 2 2
m n
m
R h m m
n nR
mg R
u F dr F
r
+
− = − =∫
2
1 2 2
m n
m
R h
m m R
dr
u mg R
r
+
− = − ∫
2
1 2
1 1
m m
m m n
u mg R
R R h
−
 
= − 
+ 
1 1 2 2T U T−+ =
2 2
0
1 1
2 2
m
m m
m n
R
mv mg R mv
R h
 
+ − = 
+ 
( )
( )2 2
0
2 2
0
2
2
m
n
v v
m
m gm
v v R
h
g
R
−
 
−  
=  
 −  
(1)
Uniform gravitational field
2 2
1 0 2
1
2
T mv T mv= =
( ) ( )1 2
m n
m
R h
u m m u m uR
u F dr mg R h R mgh
+
− = − = − + − = −∫
2 2
1 1 2 2 0
1 1
2 2
m uT u T mv mg h mv−+ = − =
( )2 2
0
2
u
m
v v
h
g
−
= (2)
Divide (1) by (2)
( )
( )
2 2
0
2
1
1
m m
n
v v
u
g R
h
h −
=
−
!

6
(3960 mi)(5280 ft/mi) 20.9088 10 ftR = = ×
1 2 2 1
2 1
GMm GMm
U T T
r r
− = − = −
Since 1r is very large,
1
1
0 thus 0
GMm
T
r
≈ ≈
2 2
2
2 2
1
,
2 2
GMm v R
mv g
r r
= =
( )( )
2
2 62
2
2 2
2 32.2 ft/s 20.9088 10 ft2gR
v
r r
×
= =
(a) For 6
2 20.9088 10 ft ((620 mi)(5280 ft/mi))r = × +
6
24.1824 10 ft= ×
34,121 ft/sv = 6.46 mi/sv = !
(b) For 6
2 20.9088 10 ft ((30 mi)(5280 ft/mi))r = × +
6
21.0672 10 ft= ×
36,557 ft/sv = 6.92 mi/sv = !
(c) For 6
2 20.9088 10 ft/sr = ×
36,695 ft/sv = 6.95 mi/sv = !

0 620 mi 3960 mi 4580 mi 24,182,400 ftr = + = =
3960 mi 5200 mi 9160 mi = 48,364,800 ftBr = + =
Parabola: 2
y Kx=
At B: ( )22
0 24,182,400 ft 48,364,800 ftBr Kr K= =
9 1
10.3381 10 ftK − −
= ×
At A: ( ) ( )2
0sin 45 , cos45A A A A Ax r y Kx r r= ° = = − °
( ) ( )22
0 cos45 sin 45A Ar r Kr− ° = °
2
0 0A AKx x r+ − =
0 (6.5)(5280) 34,320 ft/sv = =
( )0
1
1 1 4
2
Ax Kr
K
= − + +
6 6
20.0334 10 ft, 14.1657 10 ft.A Ax r= × = ×
(a) 2 2
0 0
0
1 1
2 2
A A
A
GMm GMm
U mv mv
r r
→ = − = −
( )
2
6 2 2
0
0
1 1
32.2 20.9088 10 , 2A
A
GM v v GM
r r
 
= × = + − 
 
( )22
6 6
1 1
34320 2
14.1657 10 24.1824 10
Av GM
 
= + − 
× × 
44734 ft/sAv = 8.47 mi/sAv = !
(b) 2 2
0
0
1 1
2B
B
v v GM
r r
 
= + − 
 
( )22
6 6
1 1
34320 2
48.3648 10 28.1824 10
Bv GM
 
= + − 
× × 
24408 ft/sBv = 4.62 mi/sBv = !

(a) 2
1.62 m/sg = Use work and energy 1 1 2 2T U T→+ = (1)
where 2 2
1 1
1 1
(600) 180,000
2 2
T mv m m= = =
2 0T = (maximum elevation)
1 2U mgh→ = −
(1.62) 1.62m h mh= − = −
Substituting into (1) 180,000 1.62 0m mh− =
3
111.11 10 mh = ×
111.1 kmh = !
(b) 12
so 180,000 (same as above)
GMm
F T m
R
= =
2
2
GMm
W mg GM gR
R
= = ⇒ =
At some elevation r 2
GMm
F
r
= −
so,
1 2 2
R h
R
GMm
U dr
r
+
→ = −∫
2
2
2
1 1 1
r
R
GMm gR m
r r R
  
= = −  
   
6 2
6
2
1 1
(1.62)(1.740 10 )
r 1.740 10
m
 
= × − 
× 

12
6
2
1 1
180,000 4.9047 10 m 0
1.74 10
m
r
 
+ × − = 
× 
Solve for r2
6
2 1.8587 10 mr = ×
1858.7 km=
2so, 1858.7 1740h r R= − = −
118.7 km=
118.7 kmh = !

Assume the blocks move as one: 2 1 1 2T T U −− =
2 2
1 2friction
1 1
( )
2 2
A Bm m v kx U −+ = −
2 21 1
(3 kg) (180 N/m)(0.1 m) (3)(9.81)(0.1)(0.1 m)
2 2
v = −
2
0.4038 0.63545 m/sv v= =
Check assumption at release, s18 N > (3)(9.81) 4.41 Nµ =
∴ Slips at the horizontal surface
At release
18 NsF = x xF maΣ =
18 2.94 3a− = 2
5.02 m/sa =
For A alone:
x xF maΣ =
18 (1.5)(5.02)fF− =
10.47 NfF =
s18 7.53 10.47 N < (1.5)(9.81) (0.95)(14.175) 13.98 NfF µ= − = = =
∴ A and B move as one, thus (a) 0.635 m/sv = !
(b) vmax is max at a = 0,
0 180 2.943, 0.01635 ms fF F x x= = = − =
2 2 2
max 0 0
1 1
( ) ( ) ( )
2 2
A B fm m v k x x F x x+ = − − −
2 2 2
max
1 1
(3) (180)[(0.1) (0.01635) ] 2.943(0.1 0.01635)
2 2
v = − − −
max 0.648 m/sv = !
0

From problem 13.39 assuming the blocks move together,
2
5.02 m/s at release.a =
2
(1.5 kg)(9.81 m/s ) 14.715 NA BW W= = =
s18 7.53 10.47 N < (1.5)(9.81) 0.35(14.715) 5.15 NfF µ= − = = =
∴ Block A slides on Block B
A alone:
(a) 2
2 1 1 2 1 2 friction
1 1
, ( )
2 2
AT T U m v kx U− −− = = −
2 21 1
(1.5)( (180 N/m)(0.1 m ) 14.175(0.3)(0.1 m)
2 2
v = −
2
0.6114v = 0.782 m/sv = !
(b) max at acceleration 0,v v= =
0s f k AF F kx Wµ− = = −
180 (0.30)(14.715) 4.4145, 0.0245 mx x= = =
2 2
2 1 0 0
1
( ) ( )
2
A kT T k x x W x xµ− = − − −
2
max
1
(1.5) 0.84598 0.33329 0.51268
2
v = − =
max 0.827 m/sv = !
0
0

2
1 0
1
2
T mv=
2
2
1
2
T mv=
1 2U mgl− = −
2 2
1 1 2 2 0
1 1
2 2
T U T mv mgl mv−+ = − =
2 2
0 2v v gl= +
Newtons’ law at 2
(a) For minimum v, tension in the cord must be zero.
Thus 2
v gl=
2 2
0 2 3v v gl gl= + =
0 3v gl= !
(b) Force in the rod can support the weight so that v can be zero.
Thus 2
0 0 2v gl= +
0 2v gl= !

( )22
1 0
1 1
16 128 m
2 2
T mv m= = =
2
2
1
2
T mv=
( )1 2 6sinU mg θ− =
2
1 1 2 2
1
: 128 6 sin
2
T U T m mg mvθ−+ = + =
2
256 12 sing vθ+ = (a)
Newton’s law
2
: 2 sin
6
n
mv
F ma mg mg θΣ = − =
Using (a) ( )12 6sin 256 12 sing gθ θ− = +
18 sin 12 256g gθ = −
( )
( )
12 32.2 256
sin 0.22498
18 32.2
θ
−
= = 13.00θ = ° !

Use work - energy : position 1 is bottom and position 2 is the top
1 1 2 2T U T→+ = (1)
where,
2
1 0
1
2
T mv=
2
2 2
1
2
T mv=
1 2 (0.5)U mgh mg→ = − = −
2 2
0 2
1 1
(0.5)
2 2
mv mg mv− =
so
2 2
0 2v v g= + (2)
(a) Slender rod 2 00v v g= ⇒ =
0 3.13 m/sv = !
(b) Cord, so the critical condition is tension = 0 at the top
2
22
2n
mv
F ma mg v gρ
ρ
Σ = ⇒ = ⇒ =
2
0 9.81(0.25 1)v g gρ= + = +
12.2625=
0 3.502 m/sv =
0 3.50 m/sv = !

Use work - energy : position 1 is at A, position 2 is at B.
1 1 2 2T U T→+ = (1)
Where 2
1 1 2 2
1
0; sin ;
2
BT U mg l T mvθ→= = =
Substitute
21
0 sin
2
Bmg l mvθ+ =
2
2 sinBv g l θ= (2)
For T = 2 W use Newtons 2nd law.
2
2 sin B
n n
mv
F ma W W
l
θΣ = ⇒ − = (3)
Substitute (2) into (3)
sin
2 sin 2
l
mg mg mg
l
θ
θ− =
2 3sinθ=
2
or sin 41.81
3
θ θ= ⇒ = °
41.8θ = °!

( )2 2 21 1
0 0 250 kg 125
2 2
A A B B B Bv T T mv v v= = = = =
( ) ( )2
1250 kg 9.81 m/sW = ×
( )( )27 1 cos40A BU W− = − °
( )( )( )2
250 kg 9.81m/s 27 m 0.234A BU − = ×
15495 JA BU − =
2
0 15495 125A A B A BT U T v−+ = + =
( )
( )
2 15495 J
125 kg
Bv =
2 2 2
124.0 m /sBv =
Newtons Law at B
2
2 2 2
cos40 ; 124.0 m /sB
B
mv
N W v
R
−
− ° = =
( )( )
( )( )2 2
2
250 kg 124.0 m /s
250 kg 9.81m/s cos40
27 m
N = × ° −
1879 1148 731 NN = − = 731 NN = !

Normal force at B
See solution to Problem 13.45, 731.0 NBN =
Newtons Law
From B to C (car moves in a straight line)
cos40 0BN W′ − ° =
( )2
250 kg 9.81m/s cos40 1878.7 NBN′ = × ° =
At C and D (car in the curve at C)
At C
2
cos C
C
W v
N W
g R
θ− =
( )
2
2
250 kg 9.81m/s cos C
C
v
N
gR
θ
 
= × +  
 
At D
2
D
D
W v
N W
g R
− = +
continued

( )
2
250 9.81 1 D
D
v
N
gR
 
= × +  
 
Since and cos 1,D C D Cv v N Nθ> < >
Work and energy from A to D
2 21
0, 0 125
2
A A D D Dv T T mv v= = = =
( ) ( )( )( )2
27 18 250 kg 9.81m/s 45 m 110362.5A DU W− = + = =
2
0 110362.5 125A A D D DT U T v−+ = + =
2
882.90Dv =
( )
( )
2
882.90
250 1 250 9.81 1 5518.1 N
72 72 9.81
D
D
v
N g
g
  
= + = + =     
   
min max731 N; 5520 NB DN N N N= = = = !

Kinematics: 21
constant, ,
2
a v at x at= = =
( )2 21
110 5.4 , 7.5446 ft/s
2
x a a= = =
( )2
2
150 lb
constant 7.5446 ft/s 35.1456 lb
32.2 ft/s
F ma
 
= = = = 
 
7.5446v t=
( )2150
Power 7.5446
32.2
Fv mav t
 
= = =   
( )( ) ( )2 2
5.4
0
150/32.2 7.5446 5.41
Average power 0
5.4 5.4 2
Fv dt
 
 = = −
 
 
∫
Average power 715.93 ft lb/s= ⋅
Average power 1.302 hp= !

3
tan
100
θ =
1.718θ = °
( ) 2
7 60 kg 9.81m/sB WW W W= + = + ×
657.3 NW =
( )( )sinWP W v W vθ= ⋅ =
( )( )( )657.3 sin1.718 2WP = °
39.41 WWP =
39.4 WWP = !
( ) 2
9 90 kg 9.81m/sB mW W W= + = + ×
971.2 NW =
Brake must dissipate the power generated by the bike and the man going
down the slope at 6 m/s.
( )( )sinBP W v W vθ= ⋅ =
( )( )( )971.2 sin1.718 6 174.701BP = ° =
174.7 WBP = !

(a) ( ) ( )( ) ( )( ) ( )2800 650p A A L A AA
P F v W W v v= = + = +
6.5 ft
s/t 0.40625 ft/s
16 s
Av = = =
( ) ( )( )3450 lb 0.40625 ft/s 1401.56 lb ft/sp A
P = = ⋅
( )1hp 550 ft lb/s, 2.548 hpp A
P= ⋅ =
( ) 2.55 hpp A
P = !
(b) ( )
( ) 2.55
0.82
p A
E A
P
P
η
= =
( ) 3.11hpE A
P = !

(a) Material is lifted to a height b at a rate, ( )( ) ( )2
kg/h m/s N/hm g mg =  
Thus,
( ) ( )
( )
N/h
N m/s
3600 s/h 3600
mg b mU mgb
t
   ∆     = = ⋅ 
∆  
1000 N m/s 1kw⋅ =
Thus, including motor efficiency, η
( )
( )
( ) ( )
N m/s
kw
1000 N m/s
3600
kw
mgb
P
η
⋅
=
⋅ 
 
 
( ) 6
kw 0.278 10
mgb
P
η
−
= × !
(b)
( )( ) ( )tons/h 2000 lb/ton ft
3600 s/h
W bU
t
   ∆    =
∆
ft lb/s; 1hp 550 ft lb/s
1.8
Wb
= ⋅ = ⋅
With ,η ( )
1hp 1
ft lb/s
1.8 550 ft lb/s
Wb
hp
η
    
= ⋅      ⋅    
3
1.010 10 Wb
hp
η
−
×
= !

For constant power, P:
dv
P Fv mav m v
dt
= = =
Separate variables
1
2 2
1 0
0 0
2 2
t v m dv m v v
dt t
P v P
 
= ⇒ = −  
 
∫ ∫ (1)
Distance
2dv dx dv
P mv mv
dx dt dx
= =
Separate variables
1
0
3 3
2 1 0
0
3 3
x v
v
m m v v
dx v dv x
P P
 
= ⇒ = −  
 
∫ ∫ (2)
with numbers
(a) 0 136 km/h 10 m/s; 54 km/h 15 m/s, so,v v= = = =
( )
( ) ( )
3
2 2
3
15 10 kg
15 m/s 10 m/s
2 50 ×10 W
t
×  = −  
18.75 st⇒ = !
( )( )
3
3 3
3
15 10
15 10 237.5 m
3 50 10
x
×  = − = ×
238 mx⇒ = !
(b) 0 154 km/h 15 m/s; 72 km/h 20 m/sv v= = = =
( )( )
3
2 2
3
15 10
20 15 26.25 s
2 50 10
t
×  = − = ×
26.2 st⇒ = !
( )( )
3
3 3
3
15 10
20 15 462.5 m
3 50 10
x
×  = − = ×
462 mx⇒ = !

(a) constant
dv
P F v m v
dt
 
= ⋅ = = 
 
4.3 5
2.0 0
m v dv P dt∴ =∫ ∫
( ) ( )2 2
4.3 m/s 20 m/s
60 kg 5 , 86.94 W
2
P P
 −
  = =
 
 
86.9 WP = !
(b) constant
dv
P F v mv v
dx
 
= ⋅ = = 
 
4.3 2
2.0 0
x
m v dv P dx∴ =∫ ∫
( ) ( )3 3
4.3 2.0
60 86.94
3
x
 −
  =
 
 
16.45 mx = !

Motion is determined as a function of time as,
( )12000 ln cosh 0.03x t=
Velocity ( )( )
1
12000 sinh 0.03 0.03
cosh 0.03
dx
v t
dt t
 
= =  
 
360 sinh 0.03
cosh 0.03
t
v
t
=
Power dissipated ( )2 3
0.01 0.01P Dv v v v= = =
( )
33 0.03 0.03
3 3
0.03 0.03
sinh 0.03
0.01 360 466.56 10
cosh 0.03
t t
t t
t e e
P
t e e
−
−
   −
= = ×   
+   
(a) 10 s,t = 11534 ft lb/sP = ⋅ 21.0 hpP = !
(b) 15 s,t = 35037 ft lb/sP = ⋅ 63.7 hpP = !

Motion is defined by the following function:
0.0005
11 x dv
a e v
dx
−
= =
0.00050.0005
0 0 0
11
11
0.0005
v x xx u
vdv e dx e du
−− −
= =∫ ∫ ∫
( )
2
0.0005
22000 1
2
xv
e−
= −
( )2 0.0005
44000 1 x
v e−
= −
( )
1
0.0005 2209.76 1 x
v e−
= −
3
Power dissipated 0.01P Dv v= =
3
0.0005 292295 1 x
P e− = − 
(a) 600 ftx = , 12178 ft lb/sP = ⋅ 22.1 hpP = !
(b) 1200 ft,x = 27971 ft lb/sP = ⋅ 50.9 hpP = !

System is in equilibrium in deflected 0x position.
Case (a) Force in both springs is the same P=
0 1 2x x x= +
0
e
P
x
k
=
1 2
1 2
P P
x x
k k
= =
Thus
1 2e
P P P
k k k
= +
1 2
1 1 1
ek k k
= +
1 2
1 2
e
k k
k
k k
=
+
!
Case (b) Deflection in both springs is the same 0x=
1 0 2 0P k x k x= +
( )1 2 0P k k x= +
0eP k x=
Equating the two expressions for
( )1 2 0 0eP k k x k x= + =
1 2ek k k= + !

Use conservation of energy
Let position 1 be where A is compressed 0.1 m; position 2 when B is compressed a maximum distance
So
1 1 2 2T V T V+ = + (1)
Where 1 0;T = ( )( )22
1 1
1 1
1600 N/m 0.1 m 8 J
2 2
AV k x= = =
2 0;T = ( )2 2 2
2 2 2 2
1 1
2800 N/m 1400
2 2
BV k x x x= = =
2
2 20 8 0 1400 0.07559 mx x+ = + ⇒ =
This answer is independent of mass
Distance traveled 0.5 m 0.05 m 0.07559 m 0.526 m= − + =
The maximum velocity will occur when the mass is between the two springs
where 1 1 2 2T V T V+ = +
1 0;T = ( )1 8 J same as beforeV =
2
2 max
1
;
2
T mv= 2 0V =
2
max
1
0 8 0;
2
mv+ = + 2
max
16
v
m
=
For 1 kgm = 2
max 16v = (a) max 4 m/sv = !
For 2.5 kgm = 2
max
16
6.4
2.5
v = = (b) max 2.53 m/sv = !

2 2
1 6 9 10.817 in.= + =l
( ) ( )2 2
0 6 8 10 in. 0.8333 ft= + = =l
Stretch 10.817 10 0.817 in.= − =
1 0.06805 ftS =
( ) ( )2 2
2 7 6 9.215 in.= + =l
Stretch 9.2195 10 0.7805 in.= − = −
2 0.06504 ftS =
1 20, 0T V= =
2 2
2 2 2
1 1 4
2 2 32.2
T mv v
 
= =  
 
( )( )2 2
1 1 2
1
33,600 lb/ft
2
V S S= +
( )( )1 16,800 0.008861 148.86 ft lbV = = ⋅
1 1 2 2T V T V+ = +
2
2 2396.7v = 2 49.0 ft/sv = !
0 0

2800 lb/in. 33,600 lb/ftk = =
( ) ( )
2
2
1 1 1
1 1 1
0 33,600
2 2 12
T V k l
 
= = ∆ =  
 
116.667 ft lb= ⋅
2 2
1 6 9 10.817 in. 0.9014 ft= + = =l
1 Stretch 10.817 10 0.817 in. 0.06808 ftS = = − = =
2 2
2 6 7 9.2195 in.= + =l
2 Stretch 9.2195 9 0.2195 in. 0.018295 ftS = = − = =
2 2 2
2 2 2 2
1 1 4
0.0621
2 2 32.2
T mv v v
 
= = = 
 
( )( )2 2
2 1 2
1
33,600
2
V S S= +
( ) ( )2 2
16800 0.06808 0.018295 = +  
83.489 ft lb= ⋅
1 1 2 2T V T V+ = +
2
2116.667 0.06211 83.489v= +
2 23.1 ft/sv = !
0

Use conservation of energy 1 1 2 2T V T V+ = + (1)
(a) Position 1 is at A and position 2 is at B
1 0;T = 2
1 1
1
2
V kx= where 1 0x = −l l
1
2 2 2 2500 400 350 729.726 mm = + + = l
So 0 429.726 mm− =l l
( )( )2
1
1
1500 N/m 0.429726 m 13.8498 J
2
V = =
At B
1
2 2 2
0350 400 531.507 mm 231.507 mm = + = ⇒ − = l l l
( )2 2 2
2 2 2 2
1 1
0.75 0.375
2 2
T mv v v= = =
( )( )2
2
1
150 0.231507 4.01966 J
2
V = =
Substituting into (1) 2
20 13.8498 0.375 4.01966v+ = +
2 5.12 m/sBv v= = !
(b) At E ( )1 10; 13.8498 same as beforeT V= =
At E
1
2 2 2
0350 500 610.328 mm 310.328 mm = + = ⇒ − = l l l
2 2
3
1
0.375
2
E ET mv v= =
( )( )221 1
150 0.310328 7.223 J
2 2
Ev kx= = =
2
0 13.8498 0.375 7.223 4.20 m/sE Ev v+ = + ⇒ = !

Conservation of energy 1 1 2 2T V T V+ = +
At A
1
2 2 2 2
0500 400 350 729.726 mm 729.726 450 279.726 mm = + + = ⇒ − = − = l l l
So ( )( )221 1
0; 150 0.279726 5.8685 J
2 2
A AT V kx= = = =
At B
1
2 2 2
0350 400 531.507 81.507 mm = + = ⇒ − = l l l
( )( )22 2 21 1 1
0.375 ; 150 0.081507 0.49825 J
2 2 2
B B B BT mv v V kx= = = = =
0 5.8685 0.375 0.49825Bv+ = +
3.78 m/sBv = !
At E ( ) ( )
1
2 2 2
0350 500 610.328 mm 160.328 mm = + = ⇒ − =  
l l l
So ( )( )22 2 21 1 1
0.375 ; 150 0.160328 1.9279 J
2 2 2
E E E ET mv v V kx= = = = =
0 5.8685 0.375 1.9279EV+ = +
3.24 m/sEv = !
The fact the cord becomes slack doesn’t matter.

(a) Maximum height when 2 0v =
1 2 0T T∴ = =
g eV V V= +
Position 1 ( )1
0gV =
1
6 lb
6 in. 0.4 6 6.4 in.
15 lb/in.
x = + = + =
( ) ( )( )22
11
1 1
15 lb/in. 6.4 in.
2 2
eV kx= =
307.2 lb in. 25.6 lb ft= ⋅ = ⋅
Position 2 ( ) ( )2
6
6 0.5
12
gV mg h h
 
= + = + 
 
( )2
0eV =
( ) ( ) ( ) ( )1 1 2 2 1 21 2
: g e g eT V T V V V V V+ = + + = +
( )25.6 6 0.5 h= +
3.767 fth = 45.2 in.h = !
(b) Maximum velocity occurs when acceleration is 0, equilibrium
position
2 2 2
3 3 3 3
1 1 6
0.093167
2 2 32.2
T mv v v
 
= = = 
 
( ) ( ) ( ) ( ) ( )2 2
3 133
1
6 6 6 36 7.5 6.4 6
2
g eV V V k x= + = + − = + −
37.2 lb in. 3.1lb ft= ⋅ = ⋅
2
1 1 3 3 3: 25.6 0.093167 3.1T V T V v+ = + = +
max 15.54 ft/sv = !

(a) Collar is in equilibrium.
( )15 lb/in. 6 lbF δΣ = −
( )
( )
6 lb
0.4 in.
15 lb/in.
δ = =
max 0.4 in.δ = !
(b) Maximum compression occurs when velocity at 2 is zero.
1 10 0T V= =
2
2 2 max max
1
0
2
T V W kδ δ= = − +
max
1
2
W kδ=
( )
( )max
2 6 lb
0.8 in.
15 lb/in.
δ = =
max 0.8 in.δ = !

30 rad
6
π
θ = ° =
0.3 mR =
(a) Maximum height
Above B is reached when the velocity at E is zero
0CT =
0ET =
e gV V V= +
Point C
( )0.3 m rad
6
BCL
π 
∆ =  
 
m
20
BCL
π
∆ =
( ) ( ) ( )
2
21 1
40 N/m m 0.4935 J
2 2 20
C BCe
V k L
π 
= ∆ = = 
 
( ) ( ) ( )( )( )2
1 cos 0.2 kg 9.81m/s 0.3 m 1 cos30C g
V WR θ= − = × − °
( ) 0.07886 JC g
V =
( ) 0 (spring is unattached)E e
V =
( ) ( )( ) ( )0.2 9.81 1.962 JE g
V WH H H= = × =
C C E ET V T V+ = +
0 0.4935 0.07886 0 0 1.962H+ + = + +
0.292 mH = !
continued

(b) The maximum velocity is at B where the potential energy is
zero, maxBv v=
0 0.4935 0.07886 0.5724 JC CT V= = + =
( )2 2
max
1 1
0.2 kg
2 2
B BT mv v= =
2
max0.1BT v=
0BV =
( ) 2
max0 0.5724 0.1C C B BT V T V v+ = + + =
2 2 2
max 5.72 m /sv =
max 2.39 m/sv = !

0.3 mR =
( ) ( )2
0.2 kg 9.81m/sW = ×
1.962 N=
(a) Smallest angleθ occurs when the velocity at D is close to zero
0 0C Dv v= =
0 0C DT T= =
e gV V V= +
Point C
( )0.3 m 0.3 mBCL θ θ∆ = =
( ) ( )21
2
C BCe
V k L= ∆
( ) 2
1.8C e
V θ=
( ) ( )1 cosC g
V WR θ= −
( ) ( )( )( )1.962 N 0.3 m 1 cosC g
V θ= −
( ) ( ) ( )2
1.8 0.5886 1 cosC C Ce g
V V V θ θ= + = + −
Point D
( ) 0 (spring is unattached)D e
V =
( ) ( ) ( )( )( )2 2 1.962 N 0.3 m 1.1772 JD g
V W R= = =
( )2
; 0 1.8 0.5586 1 cos 1.1772 JC C D DT V T V θ θ+ = + + + − =
( ) ( )2
1.8 0.5886 cos 0.5886θ θ− =
By trial 0.7522 radθ =
43.1θ = ° !
continued

(b) Velocity at A
Point D
( )0 0 1.1772 J see Part ( )D D DV T V a= = =
Point A
( )2 21 1
0.2 kg
2 2
A A AT mv v= =
2
0.1A AT v=
( ) ( ) ( )( )1.962 N 0.3 m 0.5886 JA A g
V V W R= = = =
A A D DT V T V+ = +
2
0.1 0.5886 0 1.1772Av + = +
2 2 2
5.886 m /sAv =
2.43 m/sAv = !

Conservation of energy
Position (1) is at the top of the incline; position (2) is when the spring has maximum deformation
1500 lb/ftk =
Where 1 1 2 2T V T V+ = +
At (1) ( )22
1 1
1 1 200
8 198.76 ft lb
2 2 32.2
T mv
 
+ = = ⋅ 
 
( )2
1 1 1 1 1
1
datum at point 2
2
g eV V V mgz k x= + = =
( ) ( )( )21
200 25 sin 20 1500 0.5
2
x= − ° +
Deformation of the springx =
1 1710.1 68.404 187.5V x= + +
At (2) ( )( )22
2 2 2 2 2
1 1
0; 1500 0.5
2 2
g eT V V V k x x= = + = = +
Substituting into (1) ( )2
198.78 1710.1 68.404 187.5 750 0.5x x+ + + = +
Solve 2
750 681.596 1908.9 0x x+ − =
2.11 or +1.2044 ftx = −
1.204 ftx = !
14.45 in.= !
0

Spring length 2 2
14 28= +
31.305 in.=
Stretch
31.305 in. 14 in.
12 in./ft
−
=
1.44208 ft=
( )( )2
0 0
1
0 0 48 lb/ft 1.44208 ft 49.910 lb ft
2
T V+ = + + = ⋅
(a) At A: ( )
2
2
2
1 10 lb 1 14 2 14
49.910 48 lb/ft
2 2 1232.2 ft/s
Av
   −
= +     
   
16.89 ft/sAv = !
(b) At B: ( )
2
21 10 1 14 14
49.910 48 10
2 32.2 2 12 12
Bv
     
= + −          
13.64 ft/sBv = !

( ) ( )1 1 10, 0 Constraint: 2e g B AT V V y x= = = ↓ = →
2 2
2
1 1
2 2
A A B BT M v M v= +
( ) ( )
2
2 21 1
4 kg 1.5 kg 1.25
2 2 2
B
B B
v
v v
 
= + =  
(a) ( )( )2
2
1
0.15 m, 0.075 m, 300 N/m 0.075 m 0.84375 N m
2
B A ey x V= = = = ⋅
( )( )( )2 1.5 9.81 0.15 m 2.2073 JgV = − = −
2
1 1 2 2 ; 0 1.25 0.84375 2.20725BT V T V v+ = + = + −
1.044 m/sBv = !
(b) Maximum velocity when acceleration 0=
; 2 0; Cord tensile forcex x AF ma k x T T= − = =∑
( )2 14.7152
; 0.0981 m; 2 0.1962 m
300
A A B A
T
x x x x
k
= = = = =
( )( )2 14.715 N 0.1962 m 2.8871gV = − = −
( )( )2
2
1
300 N/m 0.0981 m 1.4435
2
eV = =
2
1 1 2 2 ; 0 1.25 1.44354BT V T V v+ = + = − 1.075 m/sBv = !
(c) ( )
2
2 2
1
0; 0 300 14.715
2 2
B
B
y
T V y
 
= = = −  
0.392 m 392 mmBy = = !

(a) Calculate spring lengths after deflection.
Original spring length 0.75 m,= collar moved 100 mm 0.1 m=
( ) 2 2
1 1 2
1
0, 4 2
2
T V T v v= = = =
( )( )( )2
2 4 kg 9.81 m/s 0.1 m 3.924 JgV = − = −
( ) ( ) ( )2 2
2
1
300 0.8322 0.75 0.6727 0.75
2
eV  = − + −
 
1.9098 J=
2
1 1 2 2 : 0 2 3.924 1.9098T V T V v+ = + = − +
1.0035 m/sv = 1.004 m/sv = !
(b) Calculate spring lengths after deflection, collar moved190 mm 0.19 m=
( )( )( )2
2 4 kg 9.81 m/s 0.19 m 7.4556 JgV = − = −
( ) ( ) ( )2 2
2
1
300 0.90918 0.75 0.60877 0.75
2
eV  = − + −
 
6.7927 J=
2
1 1 2 2 : 0 2 7.4556 6.7927T V T V v+ = + = − +
0.576 m/sv = 0.576 m/sv = !

(a) 0, 0C Cv T= =
21
2
B BT mv=
( ) 21
0.2 kg
2
B BT v=
2
0.1B BT v= ( ) ( )C C Ce g
V V V= +
arc BCBC L Rθ= ∆ =
( )( )
( )0.3 m 30
180
BCL
π
∆ = °
°
0.15708 mBCL∆ =
( ) ( ) ( )( )2 21 1
40 N/m 0.15708 m 0.49348 J
2 2
C BCe
V k L= ∆ = =
( ) ( ) ( )( )( )( )2
1 cos 0.2 kg 9.81 m/s 0.3 m 1 cos30C g
V WR θ= − = − °
( ) 0.078857 JC g
V =
( ) ( ) 0.49348 J 0.078857 J 0.57234 JC C Ce g
V V V= + = + =
( ) ( ) 0 0 0B B Be g
V V V= + = + =
2
; 0 0.57234 0.1C C B B BT V T V v+ = + + =
2 2 2
5.7234 m /sBv = 2.39 m/sBv = !
(b)
2
B
R
mv
F F W
R
Σ = − =
( )
( )
( )
2 2
5.7234 m /s
1.962 N 0.2 kg
0.3 m
RF = +
1.962 N 3.8156 N 5.7776 NRF = + = 5.78 NRF = !

(a) Speed at C
( ) ( ) ( )2 2 2
300 150 75 343.69318 mmABL = + + =
320 N/mk =
At B 0 0B Bv T= =
( ) ( )B B Be g
V V V= +
343.69318 mm 200 mmABL∆ = −
143.69318 mm 0.14369318 mABL∆ = =
( ) ( ) ( )( )2 21 1
320 N/m 0.1436932 m
2 2
B ABe
V k L= ∆ =
( ) 3.303637 JB e
V =
( ) ( )( )( )2
0.5 kg 9.81m/s 0.15 m 0.73575 JB g
V Wr= = =
( ) ( ) 3.303637 J 0.73575 J 4.03939 JB B Be g
V V V= + = + =
At C ( )( )2 21 1
0.5 kg
2 2
C C CT mv v= =
2
0.25C CT v=
( ) ( )21
2
C ACe
V k L= ∆
309.23 mm 200 mm 109.23 mm 0.10923 mACL∆ = − = =
( ) ( )( )21
320 N/m 0.10923 m 1.90909 J
2
C e
V = =

B B C CT V T V+ = +
2
0 4.0394 0.25 1.90909Cv+ = +
2 2 24.0394 1.90909
8.5212 m /s
0.25
Cv
−
= = 2.92 m/sCv = !
(b) Force of rod on collar AC
0zF = (no friction)
x yF F= +F i j
1 75
tan 14.04
300
θ −
= = °
( )( )cos sinACk L θ θ= ∆ +eF i k
( )( )( )320 0.10923 cos14.04 sin14.04= ° + °eF i k
33.909 8.4797 (N)= +eF i k
( ) ( )
2
33.909 4.905 8.4797x y
mv
F F mg
r
Σ = + + − + = +F i j k j k
( )
( )2 2
8.5212 m /s
33.909 N 0 4.905 N 0.5
0.15 m
x yF F+ = = +
33.909 NxF = −
33.309 NyF =
33.9 N 33.3 N= − +F i j !

Datum at point C.
2
2
2.5 lb 1b s
0.07764
ft32.2 ft/s
m
  ⋅
= = 
 
( ) ( ) 21 2.5
0, 0, 0,
2 32.2
C C C A Ag e
T V V T v
 
= = = =  
 
( ) ( )( )2.5 lb 7/12 ft 1.4583A g
V = − = −
(a) ( ) ( )( )21
20 lb/ft 0.63465 ft 0.3333 ft 0.90812
2
A C
V = − =
From conservation of energy:
( ) 21
0 0.07764 1.4583 0.90812
2
Av= − +
3.7646 ft/sAV = 3.76 ft/sAv = !
At point A,
( )( )20 lb/ft 0.63465 ft 0.3333 ft 6.0263 lbS CAF k L= ∆ = − =
( )( )
( )
222 2.5 lb/32.2 ft/s 3.76551 ft/s
1.8872 lb
7/12 ft
Amv
r
= =

( )
2
7
; 6.0263 1.8872
7.61577
A
A
mv
F N
r
= − =∑
3.65 lbAN = !
(b) Datum at C.
( ) ( )( )2.5 lb 14/12 ft 2.9167 ft lbB g
V = − = − ⋅
( ) ( ) ( )( )2 21 1
20 lb/ft 0.5 ft 2.5 ft lb
2 2
B CBe
V k L= ∆ = = ⋅
From Conservation of energy:
21 2.5
0 2.9167 2.5
2 32.2
Bv
 
+ − + 
 
3.2762 ft/sBv = 3.28 ft/sBv = !
( ) ( )( )20 0.5 10 lbS CBF k L= ∆ = =
( )
2
2.5
1.4286 lb
32.2 7/12
Bv 
=  
10 2.5 1.4286y BF N= − + − =∑
6.07 lbBN = !

(a) For maximum velocity,
7.61577 4 3.61577 in. 0.30131 ftL∆ = − = =
0ta s= =&& ( )sin 3/7.61577θ =
2.5 lbW =
( )0 0.30131 3/7.61577 2.5 0yF k= = − =∑
21.063 lb/ftk = 21.1 lb/ftk = !
(b) Put datum at C
( ) ( ) 0,C C Cg e
T V V= = = 21 2.5
2 32.2
AAT v
 
=   
( ) ( )2.5 7/12 1.4583A g
V = − = −
( ) ( )( )21
21.063 0.30131 0.9561
2
A e
V = =
Conservation of energy: 21 2.5
0 1.4583 0.9561
2 32.2
Av
 
= − +  
3.597Av = 3.60 ft/sAv = !

Loop 1
(a) The smallest velocity at B will occur when the force exerted by the
tube on the package is zero.
2
0 Bmv
F mg
r
Σ = + =
( )2 2
1.5 ft 32.2 ft/sBv rg= =
2
48.30Bv =
At A
2
0
1
2
AT m v=
A
0.5
0 8 oz 0.5 lb 0.01553
32.2
V
 
= = ⇒ = = 
 
At B ( )21 1
48.30 24.15 m
2 2
B BT mv m= = =
( ) ( )7.5 1.5 9 9 0.5 4.5 lb ftBV mg mg= + = = = ⋅
( ) ( )2
0
1
: 0.01553 24.15 0.01553 4.5
2
A A B BT V T V v+ = + = +
2
0 0627.82 25.056v v= = 0 25.1 ft/sv = "
At C
( )2 21
0.007765 7.5 7.5 0.5 3.75
2
C C C CT mv v v mg= = = = =
2 2
0: 0.007765 0.007765 3.75A A C C CT V T V v v+ = + = +
( )2 2
0.007765 25.056 3.75 0.007765 Cv− =
2
144.87Cv =
continued

Loop 2
(b)
( )144.87
: 0.01553
1.5
nF ma NΣ = =
1.49989N =
{Package in tube} 1.500 lbCN = "
(a) At B, tube supports the package so,
0Bv ≈
( )0, 0 7.5 1.5B B Bv T V mg= = = +
4.5 lb ft= ⋅
A A B BT V T V+ = +
( ) 21
0.01553 4.5 24.073
2
A Av v= ⇒ =
24.1 ft/sAv = "
(b) At C 2
0.007765 , 7.5 3.75C C CT v V mg= = =
( )2 2
: 0.007765 24.073 0.007765 3.75A A C C CT V T V v+ = + = +
2
96.573Cv =
96.573
0.01553 0.99985
1.5
CN
 
= = 
 
{Package on tube} 1.000 lbCN = "

(a) Loop 1
From 13.75, at B
2 2 2
48.3 ft /s 6.9498 ft/sB Bv gr v= = ⇒ =
( )( )21 1
0.01553 48.3 0.37505
2 2
B BT mv= = =
( ) ( )( )7.5 1.5 0.5 9 4.5 lb ftBV mg= + = = ⋅
( )2 2 21 1
0.01553 0.007765
2 2
C C C CT mv v v= = =
( )7.5 0.5 3.75 lb ftCV = = ⋅
2
: 0.37505 4.5 0.007765 3.75B B C C CT V T V v+ = + + = +
2
144.887 12.039 ft/sC Cv v= ⇒ =
12.04 ft/s 10 ft/s> ⇒ Loop (1) does not work !
(b) Loop 2 at A 2 2
0 0
1
0.007765
2
AT mv v= =
0AV =
At C assume 10 ft/sCv =
( )221
0.007765 10 0.7765
2
C CT mv= = =
( )7.5 0.5 3.75Cv = =
2
0: 0.007765 0.7765 3.75A A C CT V T V v+ = + = +
0 24.144v = 0 24.1 ft/sv = !

Use conservation of energy from the point of release (A) and the top of the circle.
1 1 2 2T V T V+ = + (1) (datum at lowest point)
where
1 10;T V mg= = l
At 2 ( )( )2
2 2
1
; 2
2
T mv V mgz mg a= = = −l
0 2
2
mg mv mg a+ = + −l l (2)
We need another equation – use Newton’s 2nd
law at the top. ( )0Tension, 0 at topT =
n nF ma m= ⇒∑
m
g =
2
v
ρ
( )2
v g g aρ= = −l
( ) ( )
1
2
2
mg mg a mg a= − + −l l l
2 4 4
5 3
a a
a
= − + −
=
l l l
l
3
5
a = l !

( ) ( )( )( )2 21
0, 70 , 70 kg 9.81 m/s 40 m 1 0.7071
2
A A B B BT V T v V= = = = − −
( ) 21
70 ,
2
C CT v= 2C BV V= −
Conservation of energy: 0, 15.161 m/sB B BT V v+ = =
0, 21.441 m/sC C CT V v+ = =
(a)
(b)
(c)
( )2
70 /40 + 70
402.26 686.7
BN v g
N
=
= +
1 1089 NBN = !
( )2
70 /40 70
402.26 686.7
BN v g
N
= − +
= − +
2 = 284 NBN !
( ) 2
70 kg 0.7071 70 /40CN v− = −
1 804.5 485.6 0CN = − + <
Skier airborne? Yes!
,
,
,

For a conservative force, Equation (13.22) must be satisfied
x y z
V V V
F F F
x y z
∂ ∂ ∂
= − = − = −
∂ ∂ ∂
We now write
2 2
yx
FF V V
y x y x y x
∂∂ ∂ ∂
= − = −
∂ ∂ ∂ ∂ ∂ ∂
Since
2 2
:
yx
FV V F
x y y x y x
∂∂ ∂ ∂
= =
∂ ∂ ∂ ∂ ∂ ∂
!
We obtain in a similar way
y z z x
F F F F
z y x z
∂ ∂ ∂ ∂
= =
∂ ∂ ∂ ∂
!

(a) x y
yz zx
F F
xyz xyz
= =
( ) ( )11
0 0
yyxx
FF
y y x x
∂∂ ∂∂
= = = =
∂ ∂ ∂ ∂
Thus
yx
FF
y x
∂∂
=
∂ ∂
The other two equations derived in Problem 13.80 are checked in a similar way.
(b) Recall that , ,x y z
v v v
F F F
x y z
∂ ∂ ∂
= − = − = −
∂ ∂ ∂
( )
1
ln ,x
v
F V x f y z
x x
∂
= = − = − +
∂
(1)
( )
1
ln ,y
v
F V y g z x
y y
∂
= = − = − +
∂
(2)
( )
1
ln ,z
v
F V z h x y
z z
∂
= = − = − +
∂
(3)
Equating (1) and (2)
( ) ( )ln , ln ,x f y z y g z x− + = − +
Thus ( ) ( ), lnf y z y k z= − + (4)
( ) ( ), lng z x x k z= − + (5)
Equating (2) and (3)
( ) ( )ln , ln ,z h x y y g z x− + = − +
( ) ( ), lng z x z l x= − +
From (5)
( ) ( ), lng z x x k z= − +
Thus
( ) lnk z z= −
( ) lnl x x= −
continued

From (4)
( ), ln lnf y z y z= − −
Substitute for ( ),f y z in (1)
ln ln lnV x y z= − − −
lnV xyz= − "

(a)
( ) ( )
3 1
2 2 2 2 2 22 2
x y
x y
F F
x y z x y z
= =
+ + + +
( )( )
( )
( )
( )
3 3
2 2
5 5
2 2 2 2 2 22 2
2 2yx
x y y yFF
y x
x y z x y z
− −∂∂
= =
∂ ∂
+ + + +
Thus yx
FF
y x
∂∂
=
∂ ∂
The other two equations derived in Problem 13.79 are checked in a similar fashion
(b) Recalling that , ,x y z
V V V
F F F
x y z
∂ ∂ ∂
= − = − = −
∂ ∂ ∂
( )
3
2 2 2 2
x
V x
F V dx
x
x y z
∂
= − = −
∂
+ +
∫
( ) ( )
1
2 2 2 2 ,V x y z f y z
−
= + + +
Similarly integrating
V
y
∂
∂
and
V
z
∂
∂
shows that the unknown function ( ),f x y is a constant.
( )
1
2 2 2 2
1
V
x y z
=
+ +
!

(a)
2
0 2
a
AB
a
U kxdx k= =∫
,x yF F F= ∴ is normal to BC, 0BCU =
( )
2
0 2
a
CA
a
U a u du
−
= − − =∫
( )
2
1 ,
2
ABCA
a
U k= − not conservative !
(b) From Problem 13.77, 1
yx
FF
y x
∂∂
= =
∂ ∂
Conservative, 0ABCAU = !

(a) ( )2 2
1 1
3
1 2 1 2
x x
x x
U Fdx k x k x dx→ = − = − +∫ ∫
( ) ( )2 2 4 41 2
2 1 2 1
2 4
k k
x x x x= − − −
1 2 1 2e eU V V→ = −
2 4
1 2
1 1
2 4
eV k x k x= + !
(b) Conservation of energy: 2
1 2
1
0,
2
T T mv= =
2 4
1 1 0 2 0 2
1 1
, 0
2 4
e eV k x k x V= + =
2 2 4
1 0 2 0
1 1 1
2 2 4
mv k x k x= +
2 41 2
0 0
2
k k
v x x
m m
   
= +   
   
!

(a) ( )2 2
1 1
3
1 2 1 2
x x
x x
U Fdx k x k x dx→ = − = − +∫ ∫
( ) ( )2 2 4 41 2
2 1 2 1
2 4
k k
x x x x= − − + −
1 2 1 2:e eU V V→ = −
2 4
1 2
1 1
2 4
eV k x k x= − !
(b) Conservation of energy: 2
1 2
1
0,
2
T T mv= =
2 4
1 1 0 2 0 2
1 1
, 0
2 4
e eV k x k x V= − =
2 2 4
1 0 2 0
1 1 1
2 2 4
mv k x k x= −
2 41 2
0 0
2
k k
v x x
m m
   
= +   
   
!
1
0
2
2
Requires
k
x
k
 
<  
 

Circular orbit velocity
6
3960 mi = 20.9088 10 ftR = ×
6
930 mi = 4.9104 10 ftAor = ×
2
2
2
,Cv GM
GM gR
r r
= =
( )( )
( )
2
62
2
6 6
32.2 ft/s 20.9088 10 ft
20.9088 10 ft + 4.9104 10 ft
C
GM gR
v
r r
×
= = =
× ×
2 6 2 2
545.22 10 ft /sCv = ×
23350 ft/sCv =
Velocity reduced to 60% of 14010 ft/sCv =
Conservation of energy:
A A B BT V T V+ = +
2 21 1
2 2
A B
A B
GMm GMm
mv mv
r r
− = −
( )
( )
( )
( )
( )
2 2
6 62
2
6 6
32.2 20.9088 10 32.2 20.9088 101
14010
2 225.819 10 20.9088 10
Bv× ×
− = −
× ×
21269 ft/sBv = 4.03 mi/sBv = !

Distance 6
OA 3960 mi + 376 mi = 4336 mi = 22.894 10 ft= ×
Distance 6
EOO 10,840 mi 4336 mi = 6504 mi = 34.341 10 ft= − ×
( ) ( )2 22
8670 6504Cr = +
6
10,838.4 mi = 57.2268 10 ftCr = ×
( )
2
21
= 15681.6
2 C
gR m
T V m
r
+ −
( ) 6 2 21
constant = 123.032 10 ft /sT V
m
+ = − ×
6
3960 mi = 20.9088 10 ft; 2.97 mi/s = 15681.6 ft/sCR v= × =
(a) At point A,
2
2
6
1
=
2 22.894 10 ft
A
T V gR
v
m
+
−
×
( )
( )( )
( )
2
2 6
6 2
6
32.2 ft/s 20.9088 10 ft1
123.032 10
2 22.894 10 ft
Av
×
− × = −
×
31364 ft/sAv = 5.94 mi/sAv = !
(b) At point B, ( ) ( )2 10,840 mi 4336 miBr = −
6
17344 mi = 91.5763 10 ft= ×
( )
( )( )
( )
2
6
6 2
6
32.2 20.9088 101
123.032 10
2 91.5763 10
Bv
×
− × = −
×
7834.3 ft/sBv = 1.484 mi/sBv = !

-
At A, 32.5 /h 9028 m/sAv Mm= =
( )2 61
9028 m/s 40.752 10
2
AT m m= = ×
2
A
A A
GMm gR m
V
r r
−
= − =
6
10.67 M 10.67 10 mAr m= = ×
6
6370 km 6.37 10 mR = = ×
( )( )
( )
22 6
6
6
9.81m/s 6.37 10 m
m 37.306 10 m
10.67 10 m
AV
×
= − = − ×
×
At B
2
21
;
2
B B B
B B
GMm gR m
T mv V
r r
−
= = − =
6
19.07 M 19.07 10 mBr m= = ×
( )( )
( )
22 6
6
6
9.81m/s 6.37 10 m
20.874 10 m
19.07 10 m
B
m
V
×
= − = − ×
×
6 6 2 61
; 40.752 10 m 37.306 10 m 20.874 10 m
2
A A B B BT V T V mv+ = + × − × = − ×
2 6 6 6
2 40.752 10 37.306 10 20.874 10Bv  = × − × + × 
2 6 2 2
48.64 10 m /sBv = ×
3
6.9742 10 m/s 25.107 Mm/hBv = × = 25.1Mm/hBv = !
4.3 6.37A Ar h R Mm Mm= + = +
10 67Ar Mm=
72.7 6.37B Br h R Mm Mm= + = +
19.07Br Mm=

Note: moon earth0.0123GM GM=
By Equation 12.30 2
moon 0.0123 EGM gR=
At ∞ distance from moon: 2 2, assume 0r v= ∞ =
lem
2 2 2 0 0 0 0mGM m
E T V= + = − = − =
∞
(a) On surface of moon: 6
1740 km 1.74 10 mMR = = ×
6
1 10, 0 6370 km 6.37 10 mEv T R= = = = ×
2
lem lem
1 1 1 1
0.0123
0m E
M m
GM m gR m
V E T V
R R
−
= = + = −
( )( )( )
( )
2
2 6
lem
1 6
0.0123 9.81m/s 6.37 10 m
1.740 10 m
m
E
×
= −
×
Where lem mass of the lemm =
( )6 2 2
1 lem2.814 10 m /sE m= − ×
( )6 2 2
2 1 lem0 2.814 10 m /sE E E m∆ = − = + ×
Energy per kilogram:
lem
2810 kj/kg
E
m
∆
= !
(b) 1 80 kmmr R= +
( ) 6
1 1740 km 80 km 1820 km 1.82 10 mr = + = = ×
Newton’s second law:
2
lem lem 1
lem lem 2
11
: mGM m m v
F m a
rr
= =
2 2 lem
1 1 lem 1
1 1
1 1
2 2
m mGM m GM
v T m v
r r
= = =

lem lem lem
1 1 1 1
1 1 1
1
2
m m mGM m GM m GM m
V E T V
r r r
−
= = + = −
( ) 2
lemlem
1
1 1
0.01231 1
2 2
Em gR mGM m
E
r r
= − = −
( )( )( )2 3
lem
1 6
0.0123 9.81m/s 6.37 10 m1
2 1.82 10 m
m
E
×
= −
×
( )6 2 2
1 lem1.345 10 m /sE m= − ×
( )6 2 2
2 1 lem0 1.345 10 m /sE E E m∆ = − = + ×
Energy per kilogram:
lem
1345 kJ/kg
E
m
∆
= !

Total energy per unit weight
2 2
0 0
1 1
;
2 2
A A p p A p
A p
GMm GMm
E T V T V E mv mv
r r
= + = + = − = −
Unit weightW mg=
2 2
0
2 2
A p
A p
W W GM W W GM
E v v
g g r g g r
= − = −
22
0
2 2
pA
A p
vE v GM GM
W g gr g gr
= − = − (1)
22
2
1 1
1
2
pA
A pA
vv GM
g g r rv
   
 − = − 
  
  
2
2
2
1 2
p p A
A
A pA
v r r
v GM
r rv
   −
 − =  
  
  
( )given
pA
p A
rv
v r
=
2
2
2
1 2
p AA
A
A pp
r rr
v GM
r rr
   −
 − =  
  
  
2 2
2
2
2
p A p A
A
A pp
r r r r
v GM
r rr
   − −
  =  
  
  
2 1
2
p
A
A p A
r
v GM
r r r
 
=  
 + 
(2)

Substitute Av from (2) into (1)
0 1 1 1 1
2
2
p p
A p A A A p A A
r rE GM GM
GM
W g r r r gr g r r r r
      
   = − = −   
   + +         
( )1
p p Ap
A p A A p A
r r rrGM GM
r g r r r g r r
   − +
 = − = 
 + +    
( )A B
GM
g r r
−
=
+
2
2 0 E
E
A p
E R
GM gR
W r r
= ⇒ = −
+
( )
( )
2
60 3960 mi 5280 ft/mi
1.65598 10 ft lb/lb
50,000 mi 5280 ft/mi
E
W
×
= − = − × ⋅
×
On earth: , 0, 0,E E E E E E
E
WGM
E T V V T V
gR
= + = = = −
2
6
20.9088 10 ft lb/lbE E
E
E E
E GM gR
R
W gR gR
= − = − = − = − × ⋅
For propulsion: 0p E
E E E
W W W
= −
( )6 6
1.65598 10 20.9088 10= − × − − ×
6
19.25 10 ft lb/lb
pE
W
= × ⋅ "

Geosynchronous orbit
6
1 3960 200 4160 mi 21.965 10 ftr = + = = ×
6
2 3960 22,000 25,960 mi 137.07 10 ftr = + = = ×
Total energy 21
2
GMm
E T V mv
r
= + = −
mass of earthM =
mass of satellitem =
Newton’s second law
2
2
2
;n
GMm mv GM
F ma v
r rr
= = ⇒ =
21
2 2
GM GMm
T mv m V
r r
= = = −
1 1
2 2
GMm GMm GMm
E T V
r r r
= + = − = −
( )
2 2
2 1 1
where
2 2
E E
E
gR m R W
GM gR E W mg
r r
= = − = − =
( )( )
2
6 186000 20.9088 10 ft1 1.3115 10
lb ft
2
E
r r
× ×
= − = − ⋅
Geosynchronous orbit at 6
2 137.07 10 ftr = ×
18
9
6
1.3115 10
9.5681 10 lb ft
137.07 10
GsE
− ×
= = − × ⋅
×
(a) At 200 mi, 6
1 21.965 10 ftr = ×
18
10
200 6
1.3115 10
5.9709 10
21.965 10
E
×
= − = − ×
×
10
300 200 5.0141 10GsE E E∆ = − = ×
9
300 50.1 10 ft lbE∆ = × ⋅ !

(b) Launch from earth
At launch pad
2
E
E E
E E
GMm gR m
E WR
R R
−
= − = = −
( ) 11
6000 3960 5280 1.25453 10EE = − × = − ×
9 9
9.5681 10 125.453 10E Gs EE E E∆ = − = − × + ×
9
115.9 10 ft lbE
E∆ = × ⋅ !

We know
Geosynchronous orbit: 2 35780 6370 42,150 kmr = + =
Orbit of shuttle: 1 6370 km + 296 km = 6666 kmr =
Radius of Earth: 2
6370km also =R GM = gR=
For any circular orbit
2
2
2n
GMm mv GM
F ma v
r rr
= ⇒ = ⇒ =
Energy
2
2
1 1
;
2 2
1 1 1
2 2 2
GMm GMm
T mv v
r r
GMm GMm GMm gR m
E T V
r r r r
= = = −
 
= + = − = − = −   
 
For Geosynchronous orbit
( )( ) ( )
2
2 6
9
2 6
9.81 m/s 6.370 10 m 3600 kg1
16.999 10 J
2 42.150 10 m
E
×
= − = − ×
×
For orbit of shuttle
( )( ) ( )
2
6
6
1 6
9.81 6.370 10 36001
107.487 10 J
2 6.666 10
E
×
= − = − ×
×
On the launching pad vo = 0
( )( )( )6 9
3600 9.81 6.370 10 224.963 10 Jo
GMm
E mgR
R
= − = − = − × = − ×
(a) From shuttle to orbit
( )9 9
2 1 16.999 10 J 107.487 10 JE E E∆ = − = − × − − ×
9
90.5 10 JE∆ = × "
(b) From surface to orbit
( )9 9
2 0 16.999 10 J 224.963 10 JE E E∆ = − = − × − − ×
9
208 10 JE∆ = × "

(a) Potential energy
2
constant
GMm gR m
V
r r
= − = − +
(cf. Equation 13.17)
Choosing the constant so that 0 for :V r R= =
1
R
V mgR
r
 
= − 
 
!
(b) Kinetic energy
Newton’s second law
2
2
:n
GMm v
F ma m
rr
= =
2
2 GM gR
v
r r
= =
21
2
T mv=
2
1
2
mgR
T
r
= !
Energy
(c) Total energy
2
1
1
2
gR R
E T V m mg
r r
 
= + = + − 
 
1
2
R
E mgR
r
 
= − 
 
!

In a circular orbit,
2
2N
mv GMm
F
r r
= =
mass of Venus, mass of Sunm M= =
2
,
GM
v
r
∴ = 21
2 2
GMm
T mv
r
= =
,
2
GMm GMm
V T V
r r
= − + = −
(a)
2
3 6
2
19
88
(78.3 10 ) (67.2 10 )(5280)
60
34.4 10
v r
M
G −
  
× ×  
  = =
×
27 2
136.0 10 lb s /ftM = × ⋅ !
(b) 21
2 2
GMm
T V mv
r
+ = − = −
227
31 136.029 10 88
78.3 10
2 60407 10
T V 3
   ×  
+ = − ×      ×   
33
2.20 10 lb ftT V+ = − × ⋅ !

2 2
setting :
1
g g y
R
WR WR WR
V r R y V
r R y
= − = + = − = −
+ +
( ) ( )( )1 2
1 1 2
1 1
1 1 2
g
y y y
V WR WR
R R R
−  − − −   
= − + = − + + +    
⋅     
!
We add the constant WR, which is equivalent to changing the datum from to :r r R= ∞ =
2
g
y y
V WR
R R
  
= − +  
   
!
(a) First order approximation:
g
y
V WR Wy
R
 
= = 
 
!
[ ]Equation 13.16
(b) Second order approximation:
2
g
y y
V WR
R R
  
= −  
   
2
g
Wy
V Wy
R
= − !

Use Conservation of energy and Conservation of angular momentum.
Conservation of angular momentum.
( )( )
2
1 1
1 1 2 2
2
0.2 6
0.5
= ⇒ = =
rv
rmv r mv v
r
θ
θ θ θ
2
2.4 m/s=vθ !
1 1 2 2T V T V+ = + (1)
At 1 ( )( )22
1 1
1 1
4 kg 6 m/s 72 J
2 2
T mv= = =
( )( )22
1 1
1 1
1500 N/m 0.2 m 0.4 m 72 J
2 2
V kx= = − =
At 2 ( )( ) ( )22 2 2
2 2 2 2
1 1 1 1
4 2.4 4
2 2 2 2
r rT mv mv v= + = +θ
2
211.52 2 rv= +
( )( )22
2 2
1 1
1500 N/m 0.5 m 0.4 m 7.5 J
2 2
V kx= = − =
272 30 11.52 2 7.5rv+ = + +
2
22 82.98rv =
2 6.44 m/srv = !

Conservation of angular momentum.
1r m 1 maxv r mθ = 2v θ
so
( )( )1 1
2
max max max
0.2 6 1.2rv
v
r r r
= = =θ
θ
Energy
1 1 2 2T V T V+ = + (1)
where,
At 1 ( )( )22
1 1
1 1
4 kg 6 m/s 72 J
2 2
T mv= = =
( )( )22
1 1
1 1
1500 0.2 0.4 30 J
2 2
V kx= = − =
At 2 ( )
2
2 2
2 2 2 2
max max
1 1 1 1.2 288
4
2 2 2
rT mv mv
r r
 
= + = = 
 
θ
( )( )22
2 2 max
1 1
1500 0.4
2 2
V kx r= = −
max2
max
2.88
72 30 750 0.4r
r
+ = + −
Solve by trial for maxr ⇒ max 0.760 mr = !
Solve for 2v θ
2
max
1.2 1.2
0.760
v
r
= = ⇒ 2 1.580 m/sv = !
0

Initial state
F maΣ =
2
2
0 0, A
A
mr θ
kx mr θ x
k
= =
!
!
( )( )( )2
0
4/32.2 3 ft 5 rad/s
1.331 ft
7 lb/ft
= =x
Unstretched length = 4.5 ft + 1.331 ft = 5.831 ft
Conservation of angular momentum
(A) ( ) ( )22
= 3 5 45= =!
A Ah r θ (1)
(B) ( ) ( ) ( )2 22
= 9 = 7.5 5 , = 3.4722= ! ! !
B Bh r θ θ θ (2)
(a) Substitute into (1) gives 3.6 ftAr = !
(b) = 3.47 rad/sθ! !
Conservation of energy ( )( )2
0 0 0
1
, 7 lb/ft 1.331 ft 6.200 ft lb
2
T V T V V+ = + = = ⋅
( ) ( )
2 2
0
1 4 1 4
5 3 5 7.5 101.320 ft lb
2 32.2 2 32.2
T
   
   = + = ⋅      
   
( ) ( )
21
7 9 3.6 5.831 0.6501 ft lb
2
V  = − − = ⋅ 
0 0= + −T T V V
101.32 6.200 0.6501= + −
(c) T = 106.9 ft ⋅ lb!

Initial state
F maΣ =
2
0 Akx mr θ= !
( )( )( )22
0
4/32.2 3 5
1.331 ft
7
Amr θ
x
k
= = =
!
Unstretched length = 4.5 + 1.331 = 5.831 ft
( ) ( )
2 2
0
1 4 1 4
5 3 5 7.5
2 32.2 2 32.2
T
   
   = +         
101.320 ft lb= ⋅
( )( )2
0
1
7 1.331 6.200 ft lb
2
V = = ⋅
( ) 2
1
1
100 7 101.32 6.200
2
T V x+ = + = +
1 1.4658 ftx = ±
For compression:
( ) 5.831 1.4658B Ar r− − = −
Conservation of angular momentum
(A) ( ) ( )22
= 3 5 45A Ah r θ= =!
(B) ( ) ( )22
= 7.5 5 = 281.25B Bh r θ= !
continued

( )22 2
2.5 , 2.5B A B Ar r r r= =
5.831 2.5 5.831 1.5 5.831B A A A Ar r r r r− − = − − = −
(a) 1.5 5.831 1.4658Ar∴ − = − 2.91 ftAr = !
(b) 2.5 7.2753 ftB Ar r= = 7.28 ftBr = !
(c) 2
= 45, = 5.31 rad/sAr θ θ! ! = 5.31 rad/sθ! !

solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 13

solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 13

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solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 13