P5.2
4.47 1015 m/s2 ;
(a)
(b)
2.09 1010 N
P5.4
8.71 N
P5.6
(a)
1
2
vt ;
(b)
magnitude:
1  gt 
m (v / t )2  g 2 , direction: tan  
 v 
P5.8
16.3ˆi 14.6ˆj N
P5.10
(a–c) See free-body diagrams
and corresponding forces in P5.10
P5.12
(a)
P5.14
(a)
1
; (b) 0.750 m/s2
3
~10−22
m/s2;
(b)
x 1023m
P5.16
(a) â is at 181; (b) 11.2 kg;
(c) 37.5 m/s; (d) (37.5ˆi  0.893ˆj)m/s
P5.18
(a) See ANS FIG P5.18; (b)
1.03 N; (c) 0.805 N to the right
P5.20
(a) 49.0 N; (b) 49.0 N; (c)
98.0 N; (d) 24.5 N
P5.22
112 N
P5.24
325 N, 253 N, 165 N
P5.26
(a) 78.4 N; (b) 105 N
P5.28
a = 6.30 m/s2 and T = 31.5 N
P5.30
(a) See ANS FIG P5.30; (b)
3.57 m/s2; (c) 26.7 N; (d) 7.14 m/s
P5.32
the upper string breaks first; (c) 0, 0
(a) 2m(g + a); (b) T1 = 2T2, so
P5.34
(a) a2  2a1 ;
(b)
m1m2
m1 g
m1m2
T2 
g
m1 g
T2 
g
1
1
1
m2  m1 ; (c)
and
and
2m2  m1
2m2  m1
4m2  m1
4
2
2
ANS FIG. P5.76
P5.36
(a) 0.529 m below its initial
level; (b) 7.40 m/s upward
P5.38
(a) 256 m; (b) 42.7 m
P5.40
The situation is impossible
because maximum static friction cannot provide the acceleration necessary to keep the
book stationary on the seat.
P5.42
(a) 4.18; (b) Time would
increase, as the wheels would skid and only kinetic friction would act; or perhaps the
car would flip over.
P5.44
(a) See ANS FIG P5.44; (b)
  55.2 ; (c) n = 167 N
P5.46
(a) See ANS FIG P5.46; (b)
See P5.46 for complete list of accelerations and directions; (c) 2.31 m/s2, down for m1,
left for m2, and up for m3; (d) T12 = 30.0 N and T23 = 24.2 N; (e) T12 decreases and T23
increases
P5.48
(a) 48.6 N, 31. 7 N; (b) If P >
48.6 N, the block slides up the wall. If P < 31.7 N, the block slides down the wall; (c)
62.7 N, P > 62.7 N, the block cannot slide up the wall. If P < 62.7 N, the block slides
down the wall
P5.50
834 N
P5.52
(a) 3.43 m/s2 toward the scrap
iron; (b) 3.43 m/s2 toward the scrap iron; (c) −6.86 m/s2 toward the magnet
P5.54
The situation is impossible
because these forces on the book cannot produce the acceleration described.
P5.56
a
(a) See ANS FIG P5.56(a); (b)
F
 mb 
; (c) T  
 F ; (d) the tension in a cord of negligible mass is
mb  mr
 mb  mr 
constant along its length
P5.58
(a) Nick and the seat, with
total weight 480 N, will accelerate down and the child, with smaller weight 440 N,
will accelerate up; (b) In P5.57, a rope tension of 250 N does not make the rope break.
In part (a), the rope is strong enough to support tension 459 N. But now the tension
everywhere in the rope is 480 N, so it can exceed the breaking strength of the rope.
P5.60
(a) At any instant they have
the same velocity and at all instants they have the same acceleration; (b) 1.61 × 104 N;
(c) 2.95 × 104 N
P5.62
(a) the system will not start to
move when released; (b and c) no answer; (d) f  m2 g sin   29.4 N
P5.64
See table in P5.64 and ANS
FIG P5.64; (b) 0.143 m/s2; (c) The acceleration values agree
P5.66
(a) a1  2a2 ; (b) a2 = 12.7 N
(1.30 kg + 4m1)-1 down; (c) 9.80 m/s2 down; (d) a2 approaches zero; (e) T = 6.37 N;
(f) yes
P5.68
(a) n = (8.23 N)cosθ; (b) a =
(9.80 m/s2)sinθ; (c) See ANS FIG P5.68; (d) At 0˚, the normal force is the full weight,
and the acceleration is zero. At 90˚ the mass is in free fall next to the vertical incline.
P5.70
(a) 3.00 s; (b) 20.1 m; (c)
(18.0m) î - (9.00m) ĵ
P5.72
The situation is impossible
because at the angle of minimum tension, the tension exceeds 4.00 N
(a) T1 
P5.74
2mg 2mg
 T3 ; (b)
,
sin 1 tan 1
mg
 tan 1  T  
2
,
; (c) See P5.74


sin  tan 1  12 tan 1 
 2 
 2  tan 1 
for complete explanation
P5.76

R = mg cos sin  to the right + (M + m cos2  ) g upward