2=14.715 sin (
u=0°
u=0°
(Ve)1=
1
2
kx1
2=
1
2
(1500)
(
0.1
2
)
=7.50 J
(Ve)2=0
Conservation of Energy. Since the ball starts from rest,
T1=0
.
T1+V1=T2+V2
0+(–4.4145) +7.50 =
1
2
(0.3)v2+4.4145 sin (u–90°)+0
v
2=
20.57
–
29.43 sin (u
–
90°) (2)
Equating Eqs. (1) and (2),
14.715 sin (
u
–90°)=20.57 –29.43 sin (
u
–90°)
sin (
u
–90°)=0.4660
u–90°=27.77°
u=117.77°=118°
Ans.
14–83.
A rocket of mass mis fired vertically from the surface of the
earth, i.e., at Assuming no mass is lost as it travels
upward, determine the work it must do against gravity to
reach a distance The force of gravity is
(Eq. 13–1), where is the mass of the earth and rthe
distance between the rocket and the center of the earth.
Me
F=GMem>r2
r2.
r=r1.
SOLUTION
Ans.=GMema1
r1
–1
r2b
F1–2=LFdr=GMemLr2
r1
dr
r2
F=GMem
r2
r2
r
14–85.
SOLUTION
Since
Ans.vB=9672 m>s=34.8 Mm>h
1
2(60)(11 111.1)2–66.73(10)–12(5.976)(10)23(60)
20(10)6=1
2(60)vB
2–66.73(10)–12(5.976)(10)24(60)
80(10)6
T1+V1=T2+V2
V=-
GMem
r
yA=40 Mm>h=11 111.1 m>s
A
B
v
A
vBrB80 Mm
rA20 Mm
A 60-kg satellite travels in free flight along an elliptical orbit
such that at A, where rA = 20 Mm, it has a speed vA = 40 Mm>h.
What is the speed of the satellite when it reaches point B, where
rB = 80 Mm? Hint: See Prob. 14–82, where Me = 5.976(1024) kg
and G = 66.73(10–12) m3>(kg # s2).
14–86.
The skier starts from rest at A and travels down the ramp. If
friction and air resistance can be neglected, determine his
speed
vB
when he reaches B. Also, compute the distance s to
where he strikes the ground at C, if he makes the jump
traveling horizontally at B. Neglect the skier’s size. He has a
mass of 70 kg.
SOLUTION
4 m
vB
B
A
50 m
14–87.
SOLUTION
(1)
Also, (2)
Solving Eqs. (1) and (2) yields:
Ans.
Ans.sA=1.02 m
sB=0.638 m
Fs=500sA=800sBsA=1.6sB
0+0=0+1
2(500)sA
2+1
2(800)sB
2+20(9.81)
C
–(sA+sB)–0.5
D
The block has a mass of 20 kg and is released from rest
when s0.5m. If the mass of the bumpers Aand Bcan
be neglected, determine the maximum deformation of each
spring due to the collision.
s= 0.5 m
A
B
k
A
= 500 N/m
14–89.
SOLUTION
Datum at bottom of curve:
(2)
Substitute Eq. (1) into Eq. (2), and solving for ,
Ans.
Solving for the positive root:
Ans.s=0.587 m
s=0+(2.951 cos 42.29°)(0.2687)
a:
+bs=s0+v0t
t=0.2687 s
4.905t2+1.9857t–0.8877 =0
–1.2 cos 42.29° =0–2.951(sin 42.29°)t+1
2(–9.81)t2
A
+c
B
s=s0+v0t+1
2act2
u=f–20° =22.3°
Thus,
f=cos–1a(2.951)2
1.2(9.81) b=42.29°
vB=2.951 m>s
vB
13.062 =0.5v2
B+11.772 cos f
1
2(6)(2)2+6(9.81)(1.2 cos 20°) =1
2(6)(vB)2+6(9.81)(1.2 cos f)
TA+VA=TB+VB
When the 6-kg box reaches point Ait has a speed of
Determine the angle at which it leaves the
smooth circular ramp and the distance sto where it falls
into the cart. Neglect friction.
uvA=2m>s. vA=2m/s
1.2 m
B
A
θ
20°
466
14–90.
When the 5-kg box reaches point A it has a speed
vA
=10 m>s.
Determine the normal force the box exerts
on the surface when it reaches point B. Neglect friction and
the size of the box.
9 m
B
y
y x
x1/2 y1/2 3
Ans:
N=78.6 N
467
SOLUTION
14–91.
When the 5-kg box reaches point A it has a speed
vA
=10 m>s.
Determine how high the box reaches up the
surface before it comes to a stop. Also, what is the resultant
normal force on the surface at this point and the
acceleration? Neglect friction and the size of the box.
9 m
B
y
x
A
y x
x1/2 y1/2 3
14–93.
k 500 N/m
E
0.4 m
The 10-kg sphere Cis released from rest when and
the tension in the spring is .Determine the speed of
the sphere at the instant .Neglect the mass of rod
and the size of the sphere.AB
u=90°
100 N
u=0°
SOLUTION
470
14–94.
SOLUTION
With reference to the datum set in Fig. athe gravitational potential energy of the
chain at positions (1) and (2) are
and
Conservation of Energy:
Ans.v2=
A
2
p
(p–2)gr
0+ap–2
2bm0r2g=1
2ap
2m0rbv22+0
1
2mv12+
A
Vg
B
1=1
2mv22+
A
Vg
B
2
T1+V1=T2+V2
A
Vg
B
2=mgh 2=0
A
Vg
B
1=mgh1=ap
2m0rgbap–2
pbr=ap–2
2bm0r2g
Aquarter-circular tube AB of mean radius rcontains a smooth
chain that has a mass per unit length of . If the chain is
released from rest from the position shown, determine its
speed when it emerges completely from the tube.
m0
A
B
O
r
SOLUTION
T1+V1=T2+V2
0+0=0+2
c1
2 (40)
(
2
32+22–22
)
d
–20(9.81)(3) +
1
2(20)v2
v
=6.97 m>s
Ans.
14–95.
The cylinder has a mass of 20 kg and is released from rest
when
h=0.
Determine its speed when
h=3 m.
Each
spring has a stiffness
k=40 N>m
and an unstretched
length of 2 m.
kk
h
2 m 2 m
473
14–97.
Apan of negligible mass is attached to two identical springs of
stiffness .If a 10-kg box is dropped from a height
of 0.5 m above the pan, determine the maximum vertical
displacement d.Initially each spring has a tension of 50 N.
k=250 N>m
SOLUTION
.Initially,the spring
stretches . Thus, the unstretched length of the spring
is and the initial elastic potential of each spring
.When the box is at position (2), the
A
Ve
B
1=(2)1
2ks12=2(250 >2)(0.22)=10 J
l0=1–0.2 =0.8 m
s1=50
250 =0.2 m
A
Vg
B
2=mgh2=10(9.81)
C
–
A
0.5 +d
BD
=-98.1
A
0.5 +d
B
1m 1m
0.5 m
k250 N/m k250 N/m
d
spring stretches .The elastic potential energy of the
springs when the box is at this position is
.
Conservation of Energy:
Solving the above equation by trial and error,
Ans.d=1.34 m
250d2–98.1d–4002d2+1+350.95 =0
0+
A
0+10
B
=0+
B
–98.1
A
0.5 +d
B
+250
¢
d2–1.62d2+1+1.64
≤R
1
2mv12+
B
aVgb1
+
A
Ve
B
1
R
=1
2mv22+
B
aVgb2
+
A
Ve
B
2
R
T1+V1+T2+V2
A
Ve
B
2=(2) 1
2ks22=2(250 >2)c2d2+1–0.8 d2
=250ad2–1.62d2+1+1.64 b
s2=a2d2+12–0.8bm
is
Ans:
d=1.34
m