P6.2
215 N, horizontally inward
P6.4
(a) 1.65 × 103 m/s; (b) 6.84 × 103 s
P6.6
(a) 0.233 m/s 2 ˆi  0.163 m/sˆj ; (b) 6.53 m/s,  0.181 m/s2 ˆi  0.181m/s2 ˆj
P6.8
(a)  68.6 N ˆi   784 N ˆj ; (b) a  0.857 m/s2
P6.10
The situation is impossible because the speed of the coin is too small,
requiring that the lower string act like a rod and push rather than like a string and pull.
P6.12
(a) the gravitational force and the contact force exerted on the water by the
pail; (b) contact force exerted by the pail; (c) 3.13 m/s; (d) the water would follow the
parabolic path of a projectile
P6.14
(a) v = 4.81 m/s; (b) 700 N
P6.16
(a) 2.49 × 104 N; (b) 12.1 m/s
P6.18
(a) 20.6 N; (b) 32.0 m/s2 inward, 3.35 m/s2 downward tangent to the circle;
(c) 32.2 m/s2 inward and below the cord at 5.98˚; (d) no change; (e) acceleration is
regardless of the direction of swing
P6.20
(a) 3.60 m/s2; (b) T = 0; (c) noninertial observer in car claims that the forces
on the mass along x are T and a fictitious force (−Ma); (d) inertial observer outside the
car claims that T is the only force on M in the x-direction
2  vt  L 
 k
P6.22
 g  at2
P6.24
93.8 N
P6.26
(a) 6.27 m/s2 downward; (b) 784 N directed up; (c) 283 N upward
P6.28
(a) 53.8 m/s; (b) 148 m
P6.30
(a) B 
P6.32
(a) 2.03 N down; (b) 3.18 m/s2 down; (c) 0.205 m/s down
P6.34
36.5 m/s
P6.36
101 N
P6.38
1.2 × 103 N
P6.40
(a) 1.15 × 104 N up; (b) 14.1 m/s
9.80 m s 2
 32.7 s  ; (b) 9.80 m/s2 down; (c) 4.90 m/s2 down
0.300 m s
P6.42
1
n sin  mv 2 r
v2

 tan  
 v 2  gr tan   v 2  g  L cos   tan   v   gL sin   2
n cos 
gr
gr
P6.44
(a) 217 N; (b) 283 N; (c) From above T2 > T1 always, so string 2 will break
first
P6.46
The situation is impossible because the speed of the child given in the
problem is too large: static friction could not keep the child in place on the incline
P6.48
0.835 rev/s
P6.50
(a) v  Rg tan 35.0 
6.86 m/s  R ; (b) the mass is unnecessary; (c)
2
increasing the radius will make the required speed increase; (d) when the radius
increases, the period increases; (e) the time interval required is proportional to
R/ R  R
P6.52
mv
´
(a) 967 lb; (b) −647 lb; (c) When Fg  0, then mg 
R
P6.54
(a) m2g; (b) m2g; (c)
2
 m2 

 gR ; (d) The puck will spiral inward, gaining
 m1 
speed as it does so;
(e) The puck will spiral outward, slowing down as it does so
P6.56
θ = 84.7°
P6.58
(a) v   Rg ; (b) m g
P6.60
(a) See table in P6.60 (a); (b) See graph in P6.60 (b); (c) 53.0 m/s
P6.62
(a) a 
dv
dx
 0k
  kv ; (b)
dt
dt

 F  kmv ; (c) some feedback mechanism
could be used to impose such a force on an object; (d) think of a duck landing on a
lake, where the water exerts a resistive force on the duck proportional to its speed
P6.64
0.0928°
P6.66
(a) x 
P6.68
(a)   70.4 and   0 ; (b)   0 ; (c) the period is too large; (d) Zero is
1
ln 1  vi kt  ; (b) vi e kx  v
k
always a solution for the angle; (e) there are never more than two solutions