14–1.
The 20-kg crate is subjected to a force having a constant
direction and a magnitude F = 100 N. When s = 15 m, the
crate is moving to the right with a speed of 8 m/s. Determine
its speed when s = 25 m. The coefficient of kinetic friction
between the crate and the ground is mk = 0.25.
F
30°
SOLUTION
Equation of Motion: Since the crate slides, the friction force developed between the
crate and its contact surface is Ff = mkN = 0.25N. Applying Eq. 13–7, we have
+ c a Fy = may ;
N + 100 sin 30° - 20(9.81) = 20(0)
N = 146.2 N
Principle of Work and Energy: The horizontal component of force F which acts
in the direction of displacement does positive work, whereas the friction force
Ff = 0.25(146.2) = 36.55 N does negative work since it acts in the opposite direction
to that of displacement. The normal reaction N, the vertical component of force F
and the weight of the crate do not displace hence do no work. Applying Eq.14–7,
we have
T1 + a U1 - 2 = T2
25 m
1
(20)(8 2) +
2
L15 m
100 cos 30° ds
25 m
-
L15 m
v = 10.7 m s
36.55 ds =
1
(20)v2
2
Ans.
14–2.
F (N)
For protection, the barrel barrier is placed in front of the
bridge pier. If the relation between the force and deflection
of the barrier is F = (800(103)x1>2) N, where x is in m,
determine the car’s maximum penetration in the barrier.
The car has a weight of 20 000 N and it is traveling with
a speed of 20 m>s just before it hits the barrier.
F ⫽ 800(10)3 x1/2
x (m)
SOLUTION
Principle of Work and Energy: The speed of the car just before it crashes into the
barrier is v1 = 20 m>s. The maximum penetration occurs when the car is brought to a
stop, i.e., v2 = 0. Referring to the free-body diagram of the car, Fig. a, W and N do no
work; however, Fb does negative work.
T1 + ©U1 - 2 = T2
1 ⎛ 20 000 ⎞ 2 ⎡ xmax
⎤
(20 ) + ⎢ −
800(103 ) x1/2 dx ⎥ = 0
L
0
2 ⎜⎝ 9.81 ⎟⎠
⎣
⎦
x max = 0.836 m
Ans.
W = 20 000 N
14–3.
The crate, which has a mass of 100 kg, is subjected to the
action of the two forces. If it is originally at rest, determine
the distance it slides in order to attain a speed of 6 m>s. The
coefficient of kinetic friction between the crate and the
surface is mk = 0.2.
5
30
SOLUTION
Equations of Motion: Since the crate slides, the friction force developed between
the crate and its contact surface is Ff = mk N = 0.2N. Applying Eq. 13–7, we have
+ c ©Fy = may;
3
N + 1000 a b - 800 sin 30° - 100(9.81) = 100(0)
5
N = 781 N
Principle of Work and Energy: The horizontal components of force 800 N and
1000 N which act in the direction of displacement do positive work, whereas the
friction force Ff = 0.2(781) = 156.2 N does negative work since it acts in the
opposite direction to that of displacement. The normal reaction N, the vertical
component of 800 N and 1000 N force and the weight of the crate do not displace,
hence they do no work. Since the crate is originally at rest, T1 = 0. Applying
Eq. 14–7, we have
T1 + a U1-2 = T2
4
1
0 + 800 cos 30°(s) + 1000 a bs - 156.2s = (100) A 62 B
5
2
s = 1.35m
1000 N
800 N
Ans.
3
4
14–4.
The 2-kg block is subjected to a force having a constant
direction and a magnitude F = 1300>11 + s22 N, where s is
in meters. When s = 4 m, the block is moving to the left
with a speed of 8 m>s. Determine its speed when s = 12 m.
The coefficient of kinetic friction between the block and the
ground is mk = 0.25.
s
SOLUTION
+ c ©Fy = 0;
NB = 2(9.81) +
150
1 + s
T1 + ©U1 - 2 = T2
12
12
300
1
1
150
(2)(8)2 - 0.25[2(9.81)(12 - 4)] - 0.25
ds +
b ds cos 30° = (2)(v22 )
a
2
1
+
s
1
+
s
2
L4
L4
v22 = 24.76 - 37.5 lna
v2 = 15.4 m>s
F
30
v
1 + 12
1 + 12
b + 259.81 lna
b
1 + 4
1 + 4
Ans.
14–5.
When a 7-kg projectile is fired from a cannon barrel that
has a length of 2 m, the explosive force exerted on the
projectile, while it is in the barrel, varies in the manner
shown. Determine the approximate muzzle velocity of the
projectile at the instant it leaves the barrel. Neglect the
effects of friction inside the barrel and assume the barrel is
horizontal.
F (MN)
15
10
5
SOLUTION
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
The work done is measured as the area under the force–displacement curve. This
area is approximately 31.5 squares. Since each square has an area of 2.5 A 106 B (0.2),
T1 + ©U1-2 = T2
0 + C (31.5)(2.5) A 106 B (0.2) D =
v2 = 2121 m>s = 2.12 km>s
1
(7)(v2)2
2
(approx.)
Ans.
s (m)
14–6.
The spring in the toy gun has an unstretched length of
100 mm. It is compressed and locked in the position shown.
When the trigger is pulled, the spring unstretches 12.5 mm,
and the 20-g ball moves along the barrel. Determine the
speed of the ball when it leaves the gun. Neglect friction.
50 mm
k
150 mm
D
A
B
SOLUTION
Principle of Work and Energy: Referring to the free-body diagram of the
ball bearing shown in Fig. a, notice that Fsp does positive work. The spring
has an initial and final compression of s1 = 0.1 - 0.05 = 0.05 m and
s2 = 0.1 - (0.05 + 0.0125) = 0.0375 m.
T1 + ©U1-2 = T2
1
1
1
0 + B ks1 2 - ks2 2 R = mvA 2
2
2
2
1
1
1
0 + B (2000)(0.05)2 - (2000)(0.03752) R = (0.02)vA 2
2
2
2
vA = 10.5 m>s
2 kN/m
Ans.
14–7.
As indicated by the derivation, the principle of work and
energy is valid for observers in any inertial reference frame.
Show that this is so, by considering the 10-kg block which
rests on the smooth surface and is subjected to a horizontal
force of 6 N. If observer A is in a fixed frame x, determine the
final speed of the block if it has an initial speed of 5 m>s and
travels 10 m, both directed to the right and measured from
the fixed frame. Compare the result with that obtained by an
observer B, attached to the x¿ axis and moving at a constant
velocity of 2 m>s relative to A. Hint: The distance the block
travels will first have to be computed for observer B before
applying the principle of work and energy.
A
B
5 m/s
6N
10 m
Observer A:
T1 + ©U1 - 2 = T2
1
1
(10)(5)2 + 6(10) = (10)v22
2
2
Ans.
Observer B:
F = ma
6 = 10a
+ B
A:
a = 0.6 m>s2
s = s0 + v0t +
1 2
at
2 c
10 = 0 + 5t +
1
(0.6)t2
2
t2 + 16.67t - 33.33 = 0
t = 1.805 s
At v = 2 m>s, s¿ = 2(1.805) = 3.609 m
Block moves 10 - 3.609 = 6.391 m
Thus
T1 + ©U1 - 2 = T2
1
1
(10)(3)2 + 6(6.391) = (10)v22
2
2
v2 = 4.08 m>s
Note that this result is 2 m>s less than that observed by A.
x¿
2 m/s
SOLUTION
v2 = 6.08 m>s
x
Ans.
14–8.
If the 50-kg crate is subjected to a force of P = 200 N,
determine its speed when it has traveled 15 m starting from
rest. The coefficient of kinetic friction between the crate
and the ground is mk = 0.3.
P
SOLUTION
Free-Body Diagram: Referring to the free-body diagram of the crate, Fig. a,
+ c Fy = may;
N - 50(9.81) = 50(0)
N = 490.5 N
Thus, the frictional force acting on the crate is Ff = mkN = 0.3(490.5) = 147.15 N.
Principle of Work and Energy: Referring to Fig. a, only P and Ff do work.The work of
P will be positive, whereas Ff does negative work.
T1 + g U1 - 2 = T2
0 + 200(15) - 147.15(15) =
v = 5.63 m>s
1
(50)v2
2
Ans.
14–9.
If the 50-kg crate starts from rest and attains a speed of
6 m>s when it has traveled a distance of 15 m, determine the
force P acting on the crate. The coefficient of kinetic friction
between the crate and the ground is mk = 0.3.
P
SOLUTION
Free-Body Diagram: Referring to the free-body diagram of the crate, Fig. a,
+ c Fy = may;
N - 50(9.81) = 50(0)
N = 490.5 N
Thus, the frictional force acting on the crate is Ff = mkN = 0.3(490.5) = 147.15 N.
Principle of Work and Energy: Referring to Fig. a, only P and Ff do work.The work of
P will be positive, whereas Ff does negative work.
T1 + gU1 - 2 = T2
0 + P(15) - 147.15(15) =
P = 207 N
1
(50)(62)
2
Ans.
14–10.
The 2-Mg car has a velocity of v1 = 100 km>h when the
driver sees an obstacle in front of the car. If it takes 0.75 s for
him to react and lock the brakes, causing the car to skid,
determine the distance the car travels before it stops. The
coefficient of kinetic friction between the tires and the road
is mk = 0.25.
v1
SOLUTION
Free-Body Diagram: The normal reaction N on the car can be determined by
writing the equation of motion along the y axis. By referring to the free-body
diagram of the car, Fig. a,
+ c ©Fy = may;
N - 2000(9.81) = 2000(0)
N = 19 620 N
Since the car skids, the frictional force acting on the car is
Ff = mkN = 0.25(19620) = 4905N.
Principle of Work and Energy: By referring to Fig. a, notice that only Ff does work,
m
1h
which is negative. The initial speed of the car is v1 = c 100(103) d a
b =
h 3600 s
27.78 m>s. Here, the skidding distance of the car is denoted as s¿ .
T1 + ©U1-2 = T2
1
(2000)(27.782) + (-4905s¿) = 0
2
s¿ = 157.31 m
The distance traveled by the car during the reaction time is
s– = v1t = 27.78(0.75) = 20.83 m. Thus, the total distance traveled by the car
before it stops is
s = s¿ + s– = 157.31 + 20.83 = 178.14 m = 178 m
Ans.
100 km/h
14–11.
The 2-Mg car has a velocity of v1 = 100 km>h when the
driver sees an obstacle in front of the car. It takes 0.75 s for
him to react and lock the brakes, causing the car to skid. If
the car stops when it has traveled a distance of 175 m,
determine the coefficient of kinetic friction between the
tires and the road.
v1
SOLUTION
Free-Body Diagram: The normal reaction N on the car can be determined by
writing the equation of motion along the y axis and referring to the free-body
diagram of the car, Fig. a,
+ c ©Fy = may;
N - 2000(9.81) = 2000(0)
N = 19 620 N
Since the car skids, the frictional force acting on the car can be computed from
Ff = mkN = mk(19 620).
Principle of Work and Energy: By referring to Fig. a, notice that only Ff does work,
1h
m
which is negative. The initial speed of the car is v1 = c 100(103) d a
b =
h 3600 s
27.78 m>s. Here, the skidding distance of the car is s¿ .
T1 + ©U1-2 = T2
1
(2000)(27.782) + C - mk(19 620)s¿ D = 0
2
s¿ =
39.327
mk
The distance traveled by the car during the reaction time is
s– = v1t = 27.78(0.75) = 20.83 m. Thus, the total distance traveled by the car
before it stops is
s = s¿ + s–
175 =
39.327
+ 20.83
mk
mk = 0.255
Ans.
100 km/h
14–12.
Design considerations for the bumper B on the 5-Mg train
car require use of a nonlinear spring having the loaddeflection characteristics shown in the graph. Select the
proper value of k so that the maximum deflection of the
spring is limited to 0.2 m when the car, traveling at 4 m>s,
strikes the rigid stop. Neglect the mass of the car wheels.
F (N)
F
ks2
s (m)
SOLUTION
B
0.2
1
ks2 ds = 0
(5000)(4)2—
2
L0
40 000 - k
(0.2)3
= 0
3
k = 15.0 MN>m2
Ans.
14–13.
A car is equipped with a bumper B designed to absorb collisions. The bumper is mounted to
the car using pieces of flexible tubing T. Upon collision with a rigid barrier at A, a constant
horizontal force F is developed which causes a car deceleration kg (the highest safe
deceleration for a passenger without a seatbelt). If the car and passenger have a total mass
M and the car is initially coasting with a speed v, determine the magnitude of F needed to
stop the car and the deformation x of the bumper tubing.
Units Used:
3
Mm
10 kg
3
kN
10 N
Given:
M
3
1.5 10 kg
v
1.5
k
3
m
s
Solution:
The average force needed to decelerate the car is
F avg
M kg
F avg
44.1 kN
Ans.
The deformation is
T1 U12
T2
1
2
M v F avg x
2
x
1
M
2
§ v2 ·
¨
¸
© Favg ¹
0
x
38.2 mm
Ans.
14–14.
If the cord is subjected to a constant force of F = 300 N
and the 15-kg smooth collar starts from rest at A, determine
the velocity of the collar when it reaches point B. Neglect
the size of the pulley.
200 mm
B
C
30⬚
200 mm
F ⫽ 300 N
200 mm
SOLUTION
300 mm
Free-Body Diagram: The free-body diagram of the collar and cord system at an
arbitrary position is shown in Fig. a.
Principle of Work and Energy: Referring to Fig. a, only N does no work since it
always acts perpendicular to the motion. When the collar moves from position A to
position B, W displaces vertically upward a distance h = (0.3 + 0.2) m = 0.5 m,
while
force
F
displaces
a
distance
of
s = AC - BC = 20.72 + 0.42 -
20.2 + 0.2 = 0.5234 m . Here, the work of F is positive, whereas W does
negative work.
2
2
TA + g UA - B = TB
0 + 300(0.5234) + [- 15(9.81)(0.5)] =
vB = 3.335 m>s = 3.34 m>s
1
(15)vB2
2
Ans.
A
14–15.
Determine the required height h of the roller coaster so that when it is essentially at rest at
the crest of the hill it will reach a speed v when it comes to the bottom. Also, what should
be the minimum radius of curvature U for the track at B so that the passengers do not
experience a normal force greater than kmg? Neglect the size of the car and passengers.
Given:
v
100
k
4
km
hr
Solution:
T1 U12
mgh
T2
1 2
mv
2
2
h
1 v
2 g
h
39.3 m
Ans.
2
kmg mg
mv
U
2
U
U
v
g( k 1)
26.2 m
Ans.
14–16.
The “flying car” is a ride at an amusement park which consists of a car having wheels that
roll along a track mounted inside a rotating drum. By design the car cannot fall off the
track, however motion of the car is developed by applying the car’s brake, thereby
gripping the car to the track and allowing it to move with a constant speed of the track, vt.
If the rider applies the brake when going from B to A and then releases it at the top of the
drum, A, so that the car coasts freely down along the track to B (T = S rad),determine the
speed of the car at B and the normal reaction which the drum exerts on the car at B.
Neglect friction during the motion from A to B. The rider and car have a total mass M and
the center of mass of the car and rider moves along a circular path having a radius r.
Units Used:
3
kN
10 N
Given:
M
250 kg
r
8m
vt
3
m
s
Solution:
1
2
M vt Mg2r
2
vB
vB
1
2
M vB
2
2
vt 4 g r
18.0
NB M g
m
s
Ans.
§ vB2 ·
¸
© r ¹
M¨
NB
§ vB2 ·
¸
M¨g r ¹
©
NB
12.5 kN
Ans.
14–17.
The collar has a mass M and is supported on the rod having a coefficient of kinetic friction
Pk. The attached spring has an unstretched length l and a stiffness k. Determine the speed
of the collar after the applied force F causes it to be displaced a distance s = s1 from point
A. When s = 0 the collar is held at rest.
Given:
M
30 kg
a
0.5 m
F
200 N
l
0.2 m
k
N
50
m
Pk
0.4
T
45 deg
s1
1.5 m
g
9.81
m
2
s
Solution:
Guesses
NC
1N
v
1
m
s
Given
NC M g F sin T
F cos T s1 P k NC s1 § NC ·
¨ ¸
© v ¹
Find NC v
0
1
2 1
2
k ( a l) k s1 a l
2
2
NC
152.9 N
v
1
2
Mv
2
1.666
m
s
Ans.
14–18.
The block has a mass M and moves within the smooth vertical slot. If it starts from rest
when the attached spring is in the unstretched position at A, determine the constant vertical
force F which must be applied to the cord so that the block attains a speed vB when it
reaches sB. Neglect the size and mass of the pulley. Hint: The work of F can be
determined by finding the difference 'l in cord lengths AC and BC and using UF = F 'l.
Given:
M
0.8 kg
l
0.4 m
vB
2.5
m
s
b
0.3 m
sB
0.15 m
k
100
N
m
Solution:
2
'l
Guess
2
l b F
l sB
2
b
2
1N
Given
F 'l M g sB F
Find ( F)
1
2
k sB
2
F
1
2
M vB
2
43.9 N
Ans.
14–19.
The collar has a mass M and is moving at speed v1 when x = 0 and a force of F is applied
to it. The direction T of this force varies such that T = ax, where T is clockwise, measured
in degrees. Determine the speed of the collar when x = x1. The coefficient of kinetic friction
between the collar and the rod is Pk.
Given:
M
5 kg
F
60 N
Pk
0.3
x1
3m
m
s
v1
8
g
9.81
m
2
s
a
10
deg
m
Solution:
N
F sin T M g
Guess
v
5
m
s
Given
x1
x1
´
1
2 ´
M v1 µ F cos ( a x) dx P k µ F sin ( a x) M g dx
2
¶0
¶0
v
Find ( v)
v
10.47
m
s
Ans.
1
2
Mv
2
14–20.
The steel ingot has a mass of 1800 kg. It travels along the
conveyor at a speed v = 0.5 m>s when it collides with the
“nested” spring assembly. Determine the maximum
deflection in each spring needed to stop the motion of the
ingot. Take kA = 5 kN>m, kB = 3 kN>m.
0.5 m
0.45 m
kB
A
SOLUTION
B
Assume both springs compress
T1 + ©U1 - 2 = T2
1
1
1
(1800)(0.5)2 - (5000)s2 - (3000)(s - 0.05)2 = 0
2
2
2
225 - 2500 s2 - 1500(s2 - 0.1 s + 0.0025) = 0
s2 - 0.0375 s - 0.05531 = 0
s = 0.2547 m 7 0.05 m
kA
(O.K!)
sA = 0.255 m
Ans.
sB = 0.205 m
Ans.
C
14–21.
The steel ingot has a mass of 1800 kg. It travels along the
conveyor at a speed v = 0.5 m>s when it collides with the
“nested” spring assembly. If the stiffness of the outer spring
is kA = 5 kN>m, determine the required stiffness kB of the
inner spring so that the motion of the ingot is stopped at
the moment the front, C, of the ingot is 0.3 m from the wall.
0.5 m
0.45 m
kB
A
SOLUTION
B
T1 + ©U1 - 2 = T2
1
1
1
(1800)(0.5)2 - (5000)(0.5 - 0.3)2 - (kB)(0.45 - 0.3)2 = 0
2
2
2
kB = 11.1 kN m
kA
Ans.
C
14–22.
When the driver applies the brakes of a light truck traveling at speed v1 it skids a distance d1
before stopping. How far will the truck skid if it is traveling at speed v2 when the brakes are
applied?
Given:
km
hr
v1
40
d1
3m
v2
80
km
hr
Solution:
1
2
M v1 P k M g d1
2
0
1
2
M v2 P k M g d2
2
0
Pk
v1
2
2g d1
Pk
2.10
d2
12.00 m
2
d2
v2
2P k g
Ans.
14–23.
The train car has a mass of 10 Mg and is traveling at 5 m> s
when it reaches A. If the rolling resistance is 1> 100 of the
weight of the car, determine the compression of each spring
when the car is momentarily brought to rest.
300 kN/m
500 kN/m
A
SOLUTION
Free-Body Diagram: The free-body diagram of the train in contact with the spring
1
[10 000(9.81)] = 981 N.
is shown in Fig. a. Here, the rolling resistance is Fr =
100
The compression of springs 1 and 2 at the instant the train is momentarily at rest
will be denoted as s1 and s2. Thus, the force developed in springs 1 and 2 are
A Fsp B 1 = k1s1 = 300(103)s1 and A Fsp B 2 = 500(103)s2. Since action is equal to
reaction,
A Fsp B 1 = A Fsp B 2
300(103)s1 = 500(103)s2
s1 = 1.6667s2
Principle of Work and Energy: Referring to Fig. a, W and N do no work, and Fsp and
Fr do negative work.
T1 + g U1 - 2 = T2
1
(10 000)(52) + [- 981(30 + s1 + s2)] +
2
e-
1
1
C 300(103) D s12 f + e - C 500(103) D s22 f = 0
2
2
150(103)s12 + 250(103)s22 + 981(s1 + s2) - 95570 = 0
Substituting Eq. (1) into Eq. (2),
666.67(103)s22 + 2616s2 - 95570 = 0
Solving for the positive root of the above equation,
s2 = 0.3767 m = 0.377 m
Substituting the result of s2 into Eq. (1),
s1 = 0.6278 m = 0.628 m
vA ⫽ 5 m/s
30 m
Ans.
14–24.
The 0.5-kg ball is fired up the smooth vertical circular track
using the spring plunger. The plunger keeps the spring
compressed 0.08 m when s = 0. Determine how far s it must
be pulled back and released so that the ball will begin to
leave the track when u = 135°.
B
135
SOLUTION
u
s
Equations of Motion:
A
©Fn = man;
y2B
0.5(9.81) cos 45° = 0.5 a
b
1.5
y2B = 10.41 m2>s2
k
Principle of Work and Energy: Here, the weight of the ball is being displaced
vertically by s = 1.5 + 1.5 sin 45° = 2.561 m and so it does negative work. The
spring force, given by Fsp = 500(s + 0.08), does positive work. Since the ball is at
rest initially, T1 = 0. Applying Eq. 14–7, we have
TA + a UA-B = TB
s
0 +
L0
500(s + 0.08) ds - 0.5(9.81)(2.561) =
s = 0.1789 m = 179 mm
1
(0.5)(10.41)
2
Ans.
500 N/m
1.5 m
14–25.
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, find 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.
A
50 m
B
4m
s
C
SOLUTION
30
TA + © UA - B = TB
0 + 70(9.81)(46) =
1
(70)(vB)2
2
vB = 30.04 m>s = 30.0 m>s
+ B
A:
Ans.
s = s0 + v0 t
s cos 30° = 0 + 30.04t
A+TB
s = s0 + v0 t +
1 2
a t
2 c
s sin 30° + 4 = 0 + 0 +
1
(9.81)t2
2
Eliminating t,
s2 - 122.67s - 981.33 = 0
Solving for the positive root
s = 130 m
Ans.
14–26.
The catapulting mechanism is used to propel the 10-kg
slider A to the right along the smooth track. The propelling
action is obtained by drawing the pulley attached to rod BC
rapidly to the left by means of a piston P. If the piston
applies a constant force F = 20 kN to rod BC such that it
moves it 0.2 m, determine the speed attained by the slider if
it was originally at rest. Neglect the mass of the pulleys,
cable, piston, and rod BC.
A
P
F
SOLUTION
2 sC + sA = l
2 ¢ sC + ¢ sA = 0
2(0.2) = - ¢ sA
-0.4 = ¢ sA
T1 + ©U1 - 2 = T2
0 + (10 000)(0.4) =
vA = 28.3 m>s
1
(10)(vA)2
2
Ans.
B
C
14–27.
The conveyor belt delivers crate each of mass M to the ramp at A such that the crate’s
velocity is vA, directed down along the ramp. If the coefficient of kinetic friction between
each crate and the ramp is Pk. determine the speed at which each crate slides off the ramp
at B. Assume that no tipping occurs.
Given:
M
12 kg
vA
2.5
Pk
0.3
g
9.81
m
s
m
2
s
T
30 deg
a
3m
Solution:
Nc
M g cos T
1
2
M vA ( M g a)sin T P k Nc a
2
vB
vB
2
1
2
M vB
2
vA ( 2g a)sin T 2P k g cos T a
4.52
m
s
Ans.
14–28.
The cyclist travels to point A, pedaling until he reaches a
speed vA = 4 m>s. He then coasts freely up the curved
surface. Determine how high he reaches up the surface
before he comes to a stop. Also, what are the resultant normal
force on the surface at this point and his acceleration? The
total mass of the bike and man is 75 kg. Neglect friction, the
mass of the wheels, and the size of the bicycle.
y
C
SOLUTION
1
2
x1/2
y1/2
B y
x
45
A
1
2
x + y = 2
4m
1 -1
1 1 dy
x 2 + y-2
= 0
2
2
dx
1
dy
-x - 2
=
1
dx
y-2
T1 + ©U1 - 2 = T2
1
(75)(4)2 - 75(9.81)(y) = 0
2
y = 0.81549 m = 0.815 m
Ans.
x1>2 + (0.81549)1>2 = 2
x = 1.2033 m
tan u =
-(1.2033) - 1>2
dy
=
= - 0.82323
dx
(0.81549) - 1>2
u = -39.46°
Q+ ©Fn = m an ;
Nb - 9.81(75) cos 39.46° = 0
Nb = 568 N
+ R©Ft = m at ;
Ans.
75(9.81) sin 39.46° = 75 at
a = at = 6.23 m>s2
2
4m
Ans.
x
14–29.
The collar has a mass of 20 kg and slides along the smooth
rod. Two springs are attached to it and the ends of the rod as
shown. If each spring has an uncompressed length of 1 m
and the collar has a speed of 2 m>s when s = 0, determine
the maximum compression of each spring due to the backand-forth (oscillating) motion of the collar.
s
kA
50 N/m
kB
1m
1m
0.25 m
SOLUTION
T1 + ©U1 - 2 = T2
1
1
1
(20)(2)2 - (50)(s)2 - (100)(s)2 = 0
2
2
2
s = 0.730 m
Ans.
100 N/m
14–30.
The block of mass M is subjected to a force having a constant direction and a magnitude F =
k/(a+bx). When x = x1, the block is moving to the left with a speed v1. Determine its speed
when x = x2. The coefficient of kinetic friction between the block and the ground is Pk .
Given:
M
b
2 kg
1m
1
k
600 N
x1
6m
a
2
v1
8
x2
m
s
g
18 m
9.81
m
2
s
T
30 deg
Pk
0.25
0
NB
Solution:
NB M g x2
U
´
µ
µ
¶x
1
§ k · sin T
¨
¸
© a b x¹
x2
´
k cos T
dx P k µ
a bx
µ
¶x
1
2
M v1 U
2
1
2
M v2
2
Mg Mg k sin T
dx
a bx
v2
v1 k sin T
a bx
U
348.54 N m
1
2
2U
M
v2
20.31
m
s
Ans.
14–31.
Marbles having a mass of 5 g are dropped from rest at A through
the smooth glass tube and accumulate in the can at C. Determine
the placement R of the can from the end of the tube and the
speed at which the marbles fall into the can. Neglect the size
of the can.
A
B
3m
2m
SOLUTION
C
TA + © UA-B = TB
R
1
0 + [0.005(9.81)(3 - 2)] = (0.005)v2B
2
vB = 4.429 m>s
A+TB
s = s0 + v0t +
2 = 0 + 0 =
1
a t2
2 c
1
(9.81)t2
2
t = 0.6386 s
+ b
a:
s = s0 + v0 t
R = 0 + 4.429(0.6386) = 2.83 m
Ans.
TA + © UA-C = T1
0 + [0.005(9.81)(3) =
vC = 7.67 m>s
1
(0.005)v2C
2
Ans.
14–32.
The cyclist travels to point A, pedaling until he reaches a
speed vA = 8 m>s. He then coasts freely up the curved
surface. Determine the normal force he exerts on the
surface when he reaches point B. The total mass of the bike
and man is 75 kg. Neglect friction, the mass of the wheels,
and the size of the bicycle.
y
C
x1/2
y1/2
B y
x
2
4m
SOLUTION
1
2
45
A
1
2
x + y = 2
4m
1 1 dy
1 -1
x 2 + y- 2
= 0
2
2
dx
1
dy
- x- 2
=
1
dx
y- 2
For y = x,
1
2 x2 = 2
x = 1, y = 1 (Point B)
Thus,
tan u =
dy
= -1
dx
u = - 45°
dy
1
1
= (- x - 2)(y2)
dx
d2y
dy
1
1
1
1 3
1
= y2 a x - 2 b - x - 2 a b ay - 2 b a b
2
2
dx
dx
2
d2y
dx2
=
1 1 -3
1 1
y2 x 2 + a b
2
2 x
For x = y = 1
dy
= -1,
dx
r =
d2y
dx2
= 1
[1 + ( -1)2]3>2
= 2.828 m
1
T1 + ©U1 - 2 = T2
1
1
(75)(8)2 - 75(9.81)(1) = (75)(v2B)
2
2
v2B = 44.38
Q+ ©Fn = man ;
NB - 9.81(75) cos 45° = 75 a
NB = 1.70 kN
44.38
b
2.828
Ans.
x
14–33.
The man at the window A wishes to throw the 30-kg sack on
the ground. To do this he allows it to swing from rest at B to
point C, when he releases the cord at u = 30°. Determine
the speed at which it strikes the ground and the distance R.
A
B
8m
8m
u
16 m
C
SOLUTION
TB + ©UB - C = TC
D
0 + 30(9.81)8 cos 30° =
R
1
(30)v2C
2
yC = 11.659 m>s
TB + ©UB - D = TD
0 + 30(9.81)(16) =
1
(30) v2D
2
vD = 17.7 m>s
Ans.
During free flight:
( + T) s = s0 + v0t +
1 2
act
2
16 = 8 cos 30° - 11.659 sin 30°t +
1
(9.81)t2
2
t2 - 1.18848 t - 1.8495 = 0
Solving for the positive root:
t = 2.0784 s
+ )s = s + v t
(:
0
0
s = 8 sin 30° + 11.659 cos 30°(2.0784)
s = 24.985 m
Thus,
R = 8 + 24.985 = 33.0 m
Ans.
Also,
(vD)x = 11.659 cos 30° = 10.097 m>s
(+ T) (vD)x = -11.659 sin 30° + 9.81(2.0784) = 14.559 m>s
nD = 2(10.097)2 + (14.559) 2 = 7.7 m>s
Ans.
14–34. The 1.5-kg block slides along a smooth plane and
strikes a nonlinear spring with a speed of W 4 ms. The
spring is termed “nonlinear” because it has a resistance of
'T Ls 2, where L 900 Nm2. Determine the speed of the
block after it has compressed the spring T 0.2 m.
v
k
Principle of Work and Energy: The spring force Fsp which acts in the opposite direction
to that of displacement does negative work. The normal reaction N and the weight of
the block do not displace hence do no work. Applying Eq. 14–7, we have
51
1
(1.5) 42 2
< 612 52
0.2 m
5
(0
900T2 ET 6 y 3.58 ms
1
(1.5) y2
2
Ans.
14–35.
The collar has a mass of 20 kg and is supported on the
smooth rod. The attached springs are undeformed when
d = 0.5 m. Determine the speed of the collar after the
applied force F = 100 N causes it to be displaced so that
d = 0.3 m. When d = 0.5 m the collar is at rest.
k' = 15 N/m
F = 100 N
60°
SOLUTION
d
T1 + a U1 - 2 = T2
0 + 100 sin 60°(0.5 - 0.3) + 196.2(0.5 - 0.3) -
k = 25 N/m
1
(15)(0.5 - 0.3)2
2
1
1
(25)(0.5 - 0.3)2 = (20)v2C
2
2
vC = 2.36 m>s
Ans.
14–36.
If the force exerted by the motor M on the cable is 250 N,
determine the speed of the 100-kg crate when it is hoisted
to s = 3 m. The crate is at rest when s = 0.
M
SOLUTION
Kinematics: Expressing the length of the cable in terms of position coordinates sC and
sP referring to Fig. a,
s
3sC + (sC - sP) = l
4sC - sP = l
(1)
Using Eq. (1), the change in position of the crate and point P on the cable can be
written as
(+ T)
4¢sC - ¢sP = 0
Here, ¢sC = - 3 m. Thus,
(+ T )
¢sp = - 12 m = 12 m c
4(- 3) - ¢sP = 0
Principle of Work and Energy: Referring to the free-body diagram of the pulley
system, Fig. b, F1 and F2 do no work since it acts at the support; however, T does
positive work and WC does negative work.
T1 + g U1 - 2 = T2
0 + T¢sP + [ - WC ¢sC] =
1
m v2
2 C
0 + 250(12) + [-100(9.81)(3)] =
v = 1.07 m>s
1
(100)v2
2
Ans.
14–37.
If the track is to be designed so that the passengers of the
roller coaster do not experience a normal force equal to
zero or more than 4 times their weight, determine the
limiting heights hA and hC so that this does not occur. The
roller coaster starts from rest at position A. Neglect friction.
A
C
rC
20 m
rB
hC
B
SOLUTION
Free-Body Diagram:The free-body diagram of the passenger at positions B and C
are shown in Figs. a and b, respectively.
Equations of Motion: Here, an =
NB = 4mg. By referring to Fig. a,
+ c ©Fn = man;
v2
. The requirement at position B is that
r
4mg - mg = m ¢
vB 2
≤
15
vB 2 = 45g
At position C, NC is required to be zero. By referring to Fig. b,
+ T ©Fn = man;
mg - 0 = m ¢
vC 2
≤
20
vC 2 = 20g
Principle of Work and Energy: The normal reaction N does no work since it always
acts perpendicular to the motion. When the rollercoaster moves from position A
to B, W displaces vertically downward h = hA and does positive work.
We have
TA + ©UA-B = TB
0 + mghA =
1
m(45g)
2
hA = 22.5 m
Ans.
When the rollercoaster moves from position A to C, W displaces vertically
downward h = hA - hC = (22.5 - hC) m.
TA + ©UA-B = TB
0 + mg(22.5 - hC) =
hC = 12.5 m
1
m(20g)
2
Ans.
15 m
hA
14–38.
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, find the distance d
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 M.
Given:
M
75 kg
h1
50 m
h2
4m
T
30 deg
Solution:
Guesses
Given
§¨ vB ·¸
¨t ¸
¨d¸
© ¹
vB
1
m
s
M g h1 h2
Find vB t d
t
d
1s
1
2
M vB
2
t
vB t
3.754 s
1m
d cos T
vB
h.2 d sin T =
30.0
m
s
d
1 2
gt
2
130.2 m
Ans.
14–39.
The 8-kg cylinder A and 3-kg cylinder B are released from
rest. Determine the speed of A after it has moved 2 m
starting from rest. Neglect the mass of the cord and pulleys.
SOLUTION
Kinematics: Express the length of cord in terms of position coordinates sA and sB by
referring to Fig. a
2sA + sB = l
B
Thus
2¢sA + ¢sB = 0
(2)
If we assume that cylinder A is moving downward through a distance of ¢sA = 2 m,
Eq. (2) gives
(+ T )
2(2) + ¢sB = 0
¢sB = - 4 m = 4 m c
Taking the time derivative of Eq. (1),
(+ T )
A
(1)
2vA + vB = 0
(3)
g T1 + gU1 - 2 = g T2
0 + 8(2)9.81 - 3(4)9.81 =
1
1
(8)vA2 + (3)vB2
2
2
Positive net work on left means assumption of A moving down is correct. Since
vB = - 2vA,
vA = 1.98 m>s T
vB = - 3.96 m>s = 3.96 m>s c
Ans.
14–40.
Cylinder A has a mass of 3 kg and cylinder B has a mass of
8 kg. Determine the speed of A after it moves upwards 2 m
starting from rest. Neglect the mass of the cord and pulleys.
SOLUTION
a T1 + a U1 - 2 = a T2
0 + 2[F1 - 3(9.81)] + 4[8(9.81) - F2] =
A
1
1
(3)v2A + (8)v2B
2
2
B
Also, vB = 2vA, and because the pulleys are massless, F1 = 2F2. The F1 and F2
terms drop out and the work-energy equation reduces to
255.06 = 17.5v2A
vA = 3.82 m>s
Ans.
14–41.
The catapulting mechanism is used to propel slider A of mass M to the right along the smooth
track. The propelling action is obtained by drawing the pulley attached to rod BC rapidly to the
left by means of a piston P. If the piston applies constant force F to rod BC such that it moves it
a distance d, determine the speed attained by the slider if it was originally at rest. Neglect the
mass of the pulleys, cable, piston, and rod BC.
Units Used:
kN
3
10 N
Given:
M
12 kg
F
30 kN
d
0.3 m
Solution:
0 Fd
1
2
Mv
2
v
2F d
M
v
38.73
m
s
Ans.
14–42. The 2-Mg car increases its speed uniformly from
rest to 25 ms in 30 s up the inclined road. Determine the
maximum power that must be supplied by the engine, which
operates with an efficiency of 1 0.75. Also, find the
average power supplied by the engine.
1
10
Kinematics: The constant acceleration of the car can be determined from
v v0
BD U
25 0
BD (30)
BD 0.8333 ms2
Equations of Motion: By referring to the free-body diagram of the car shown in
Fig. a,
j'Y NBY ;
' 2000(9.81) sin 5.711° 2000(0.8333)
' 3618.93N
Power: The maximum power output of the motor can be determined from
(1out)max F vmax 3618.93(25) 90 473.24 W
Thus, the maximum power input is given by
1in 1out
90473.24
120 631 W 120.6 kW
e
0.7 5
Ans.
The average power output can be determined from
(1out)avg F vavg 3618.93 3
25
4 45 236.62 W
2
Thus,
(1in)avg (1out)avg
e
45236.62
60 315.5 W 60.3 kW
0.75
Ans.
14–43.
A car has a mass M and accelerates along a horizontal straight road from rest such that
the power is always a constant amount P. Determine how far it must travel to reach a
speed of v.
Solution:
Power: Since the power output is constant, then the traction force F varies
with v. Applying Eq. 14-10, we have
P
Fv
P
v
F
Equation of Motion:
P
v
Ma
Kinematics: Applying equation ds
s
´
µ 1 ds
¶0
a
vdv
a
P
Mv
, we have
v
´
2
µ Mv
dv
µ
P
¶
0
3
s
Mv
3P
Ans.
14–44.
An automobile having a mass of 2 Mg travels up a 7° slope
at a constant speed of v = 100 km>h. If mechanical friction
and wind resistance are neglected, determine the power
developed by the engine if the automobile has an efficiency
P = 0.65.
7⬚
SOLUTION
Equation of Motion: The force F which is required to maintain the car’s constant
speed up the slope must be determined first.
+ ©Fx¿ = max¿;
F - 2(103)(9.81) sin 7° = 2(103)(0)
F = 2391.08 N
100(103) m
1h
b = 27.78 m>s.
R * a
h
3600 s
The power output can be obtained using Eq. 14–10.
Power: Here, the speed of the car is y = B
P = F # v = 2391.08(27.78) = 66.418(103) W = 66.418 kW
Using Eq. 14–11, the required power input from the engine to provide the above
power output is
power input =
=
power output
e
66.418
= 102 kW
0.65
Ans.
14–45.
A spring having a stiffness k is compressed a distance G. The stored energy in the spring is
used to drive a machine which requires power P. Determine how long the spring can
supply energy at the required rate.
Units Used:
kN
Given:
k
Solution:
U12
3
10 N
5
kN
m
1 2
kG
2
Pt
G
400 mm
P
t
1 §G
k¨
2 ©P
t
2·
¸
¹
90 W
4.44 s
Ans.
14–46.
To dramatize the loss of energy in an automobile, consider a
car having a weight of 2 5 000 N that is traveling at 5 6 km>h.
If the car is brought to a stop, determine how long a 100-W light
bulb must burn to expend the same amount of energy.
SOLUTION
Energy: Here, the speed of the car is y = a
1000 m
1h
5 6 km
b * a
b * a
b =
h
1 km
3600 s
15.5 6 m>s. Thus, the kinetic energy of the car is
U =
1 2
1 2 5 000
my = a
b A 15 .5 6 2 B = 308 504 J
2
2 9.81
The power of the bulb is
Pbulb = 100 W.
Thus,
t =
U
=
Pbulb
308 504
10 0
= 3085.0 4 s = 51.4 min
Ans.
14–47.
The escalator steps move with a constant speed of 0.6 m>s.
If the steps are 125 mm high and 250 mm in length,
determine the power of a motor needed to lift an average
mass of 150 kg per step. There are 32 steps.
250 mm
125 mm
v
SOLUTION
Step height: 0.125 m
The number of steps:
4
= 32
0.125
Total load: 32(150)(9.81) = 47 088 N
If load is placed at the center height, h =
4
= 2 m, then
2
4
U = 47 088 a b = 94.18 kJ
2
ny = n sin u = 0.6 ¢
4
2(32(0.25))2 + 42
t =
2
h
=
= 7.454 s
ny
0.2683
P =
94.18
U
=
= 12.6 kW
t
7.454
≤ = 0.2683 m>s
Ans.
Also,
P = F # v = 47 088(0.2683) = 12.6 kW
Ans.
0.6 m/s
4m
14–48.
If the escalator in Prob. 14–47 is not moving, determine the
constant speed at which a man having a mass of 80 kg must
walk up the steps to generate 100 W of power—the same
amount that is needed to power a standard light bulb.
250 mm
125 mm
v
SOLUTION
P =
(80)(9.81)(4)
U1 - 2
=
= 100
t
t
n =
2(32(0.25))2 + 42
s
=
= 0.285 m>s
t
31.4
t = 31.4 s
Ans.
0.6 m/s
4m
14–49.
The 2-Mg car increases its speed uniformly from rest to
25 m>s in 30 s up the inclined road. Determine the
maximum power that must be supplied by the engine, which
operates with an efficiency of P = 0.8. Also, find the
average power supplied by the engine.
1
10
SOLUTION
Kinematics:The constant acceleration of the car can be determined from
+ B
A:
v = v0 + ac t
25 = 0 + ac (30)
ac = 0.8333 m>s2
Equations of Motion: By referring to the free-body diagram of the car shown in
Fig. a,
©Fx¿ = max¿ ;
F - 2000(9.81) sin 5.711° = 2000(0.8333)
F = 3618.93N
Power: The maximum power output of the motor can be determined from
(Pout)max = F # vmax = 3618.93(25) = 90 473.24 W
Thus, the maximum power input is given by
Pin =
Pout
90473.24
= 113 091.55 W = 113 kW
=
e
0.8
Ans.
The average power output can be determined from
(Pout)avg = F # vavg = 3618.93 a
25
b = 45 236.62 W
2
Thus,
(Pin)avg =
(Pout)avg
e
=
45236.62
= 56 545.78 W = 56.5 kW
0.8
Ans.
14–50.
The crate has a mass of 150 kg and rests on a surface for
which the coefficients of static and kinetic friction are
ms = 0.3 and mk = 0.2, respectively. If the motor M supplies
a cable force of F = (8t2 + 20) N, where t is in seconds,
determine the power output developed by the motor when
t = 5 s.
M
SOLUTION
Equations of Equilibrium: If the crate is on the verge of slipping, Ff = ms N = 0.3N.
From FBD(a),
+ c ©Fy = 0;
N - 150(9.81) = 0
N = 1471.5 N
0.3(1471.5) - 3 A 8 t2 + 20 B = 0
+ ©F = 0;
:
x
t = 3.9867 s
Equations of Motion: Since the crate moves 3.9867 s later, Ff = mk N = 0.2N.
From FBD(b),
+ c ©Fy = may ;
+ ©F = ma ;
:
x
x
N - 150(9.81) = 150 (0)
N = 1471.5 N
0.2 (1471.5) - 3 A 8 t2 + 20 B = 150 ( -a)
a = A 0.160 t2 - 1.562 B m>s2
Kinematics:Applying dy = adt, we have
y
L0
5
dy =
L3.9867 s
A 0.160 t2 - 1.562 B dt
y = 1.7045 m>s
Power: At t = 5 s, F = 8 A 52 B + 20 = 220 N. The power can be obtained using
Eq. 14–10.
P = F # v = 3 (220) (1.7045) = 1124.97 W = 1.12 kW
Ans.
14–51.
The 50-kg crate is hoisted up the 30° incline by the pulley
system and motor M. If the crate starts from rest and, by
constant acceleration, attains a speed of 4 m>s after
traveling 8 m along the plane, determine the power that
must be supplied to the motor at the instant the crate has
moved 8 m. Neglect friction along the plane. The motor has
an efficiency of P = 0.74.
M
SOLUTION
30
Kinematics:Applying equation y2 = y20 + 2ac (s - s0), we have
42 = 02 + 2a(8 - 0)
a = 1.00 m>s2
Equations of Motion:
+ ©Fx¿ = max¿ ;
F - 50(9.81) sin 30° = 50(1.00)
F = 295.25 N
Power: The power output at the instant when y = 4 m>s can be obtained using
Eq. 14–10.
P = F # v = 295.25 (4) = 1181 W = 1.181 kW
Using Eq. 14–11, the required power input to the motor in order to provide the
above power output is
power input =
=
power output
e
1.181
= 1.60 kW
0.74
Ans.
14–52. A rocket having a total mass of 8 Mg is fired
vertically from rest. If the engines provide a constant thrust
of 4 300 kN, determine the power output of the engines
as a function of time. Neglect the effect of drag resistance
and the loss of fuel mass and weight.
C i&Y MAY ;
( C)
2 20
0
300(103)
8(103)(9.81) 8(103)A
A 27.69 ms2
AC T
T 300 kN
27.69T 27.69T
0 T v 300 (103) (27.69T) 8.31 T MW
Ans.
14–53. The material hoist and the load have a total mass
of 800 kg and the counterweight C has a mass of 150 kg. At
a given instant, the hoist has an upward velocity of 2 ms
and an acceleration of 1.5 ms2. Determine the power
generated by the motor M at this instant if it operates with
an efficiency of 1 0.8.
M
Equations of Motion: Here, B 1.5 ms2. By referring to the free-body diagram of
the hoist and counterweight shown in Fig. a,
D j'Z NBZ ;
25
5 j'Z NBZ ;
150 9.81 5 150 1.5 5 800(9.81) 800(1.5)
(1)
C
Solving,
5 1246.5 N
5 3900.75 N
Power:
1out 2T v 2(3900.75)(2) 15 603 W
Thus,
1in 1out
15603
19.5(103) W 19.5 kW
e
0.8
Ans.
14–54. The material hoist and the load have a total mass
of 800 kg and the counterweight C has a mass of 150 kg. If
the upward speed of the hoist increases uniformly from
0.5 ms to 1.5 ms in 1.5 s, determine the average power
generated by the motor M during this time. The motor
operates with an efficiency of 1 0.8.
M
Kinematics: The acceleration of the hoist can be determined from
D
v v0
BD U
1.5 0.5
B(1.5)
B 0.6667 ms2
C
Equations of Motion: Using the result of a and referring to the free-body diagram of
the hoist and block shown in Fig. a,
D j'Z NBZ ;
25
5 j'Z NBZ ;
150 9.81 5 150(0.6667)
5 800(9.81) 800(0.6667)
Solving,
5 1371.5 N
5 3504.92 N
Power:
(1out)avg 2T vavg 2(3504.92) 3
1.5
0.5
2
4 7009.8 W
Thus,
1in 1out
7009.8
8762.3
e
0.8
W 8.76 kW
Ans.
14–55.
The elevator E and its freight have a total mass of 400 kg.
Hoisting is provided by the motor M and the 60-kg block C.
If the motor has an efficiency of P = 0.6, determine the
power that must be supplied to the motor when the elevator
is hoisted upward at a constant speed of vE = 4 m>s.
M
C
SOLUTION
vE
Elevator:
E
Since a = 0,
+ c ©Fy = 0;
60(9.81) + 3T - 400(9.81) = 0
T = 1111.8 N
2sE + (sE - sP) = l
3vE = vP
Since vE = -4 m>s,
Pi =
vP = - 12 m>s
(1111.8)(12)
F # vP
=
= 22.2 kW
e
0.6
Ans.
14–56.
The sports car has a mass of 2.3 Mg, and while it is traveling
at 28 m/s the driver causes it to accelerate at 5 m>s2. If the
drag resistance on the car due to the wind is FD = 10.3v22 N,
where v is the velocity in m/s, determine the power supplied
to the engine at this instant. The engine has a running
efficiency of P = 0.68.
FD
SOLUTION
+ ©F = m a ;
:
x
x
F - 0.3v2 = 2.3(103)(5)
F = 0.3v2 + 11.5(103)
At v = 28 m>s
F = 11 735.2 N
PO = (11 735.2)(28) = 328.59 kW
Pi =
PO
328.59
= 438 kW
=
e
0.68
Ans.
14–57.
The sports car has a mass of 2.3 Mg and accelerates at
6 m>s2, starting from rest. If the drag resistance on the car
due to the wind is FD = 110v2 N, where v is the velocity in
m/s, determine the power supplied to the engine when
t = 5 s. The engine has a running efficiency of P = 0.68.
FD
SOLUTION
+ ©F = m a ;
:
x
x
F - 10v = 2.3(103)(6)
F = 13.8(103) + 10 v
+ )v = v + a t
(:
0
c
v = 0 + 6(5) = 30 m>s
PO = F # v = [13.8(103) + 10(30)](30) = 423.0 kW
Pi =
PO
423.0
= 622 kW
=
e
0.68
Ans.
14–58.
The block has a mass of 150 kg and rests on a surface for
which the coefficients of static and kinetic friction are
ms = 0.5 and mk = 0.4, respectively. If a force
F = 160t22 N, where t is in seconds, is applied to the cable,
determine the power developed by the force when t = 5 s.
Hint: First determine the time needed for the force to
cause motion.
F
SOLUTION
+ ©F = 0;
:
x
2F - 0.5(150)(9.81) = 0
F = 367.875 = 60t2
t = 2.476 s
+ ©F = m a ;
:
x
x
2(60t2)- 0.4(150)(9.81) = 150ap
ap = 0.8t2 - 3.924
dv = a dt
v
L0
5
dv =
v = a
L2.476
A 0.8t2 - 3.924 B dt
5
0.8 3
b t - 3.924t `
= 19.38 m>s
3
2.476
sP + (sP - sF) = l
2vP = vF
vF = 2(19.38) = 38.76 m>s
F = 60(5)2 = 1500 N
P = F # v = 1500(38.76) = 58.1 kW
Ans.
14–59.
v
The rocket sled has a mass of 4 Mg and travels from rest
along the horizontal track for which the coefficient of
kinetic friction is mk = 0.20. If the engine provides a
constant thrust T = 150 kN, determine the power output of
the engine as a function of time. Neglect the loss of fuel
mass and air resistance.
T
SOLUTION
+ ©F = ma ;
:
x
x
150(10)3 - 0.2(4)(10)3(9.81) = 4(10)3 a
a = 35.54 m>s2
+ Bv = v + a t
A:
0
c
= 0 + 35.54t = 35.54t
P = T # v = 150(10)3 (35.54t) = 5.33t MW
Ans.
14–60. If the engine of a 1.5-Mg car generates a constant
power of 15 kW, determine the speed of the car after it has
traveled a distance of 200 m on a level road starting from
rest. Neglect friction.
Equations of Motion: Here, A v
Dv
. By referring to the free-body diagram of the
ds
car shown in Fig. a,
i&X MAX ;
& 1500 2 v
Dv
3
DS
Power:
0out F v
15(103) 15002 v
200 m
'0
Dv
3v
DS
v
10DS 200 m
10S 0
v 18.7 ms
'0
v2 Dv
v3 v
3 0
Ans.
14–61.
If the jet on the dragster supplies a constant thrust of
T = 20 kN, determine the power generated by the jet as a
function of time. Neglect drag and rolling resistance, and
the loss of fuel. The dragster has a mass of 1 Mg and starts
from rest.
T
SOLUTION
Equations of Motion: By referring to the free-body diagram of the dragster shown
in Fig. a,
+ ©F = ma ;
:
x
x
20(103) = 1000(a)
a = 20 m>s2
Kinematics: The velocity of the dragster can be determined from
+ b
a:
v = v0 + ac t
v = 0 + 20 t = (20 t) m>s
Power:
P = F # v = 20(103)(20 t)
= C 400(103)t D W
Ans.
14–62.
An athlete pushes against an exercise machine with a force
that varies with time as shown in the first graph. Also, the
velocity of the athlete’s arm acting in the same direction as
the force varies with time as shown in the second graph.
Determine the power applied as a function of time and the
work done in t = 0.3 s.
F (N)
800
t (s)
0.2
0.3
SOLUTION
v (m/s)
For 0 … t … 0.2
F = 800 N
v =
20
t = 66.67t
0.3
P =
F#v
20
t (s)
= 53.3 t kW
Ans.
For 0.2 … t … 0.3
F = 2400 - 8000 t
v = 66.67t
P = F # v = 1160 t - 533t22 kW
Ans.
0.3
u =
P dt
L0
0.2
u =
=
L0
0.3
53.3t dt +
1160t - 533t 2 dt
2
L0.2
53.3
160
533
(0.2)2 +
[(0.3)2 - (0.2)2] [(0.3)3 - (0.2)3]
2
2
3
= 1.69 kJ
Ans.
0.3
14–63.
An athlete pushes against an exercise machine with a
force that varies with time as shown in the first graph.
Also, the velocity of the athlete’s arm acting in the same
direction as the force varies with time as shown in the
second graph. Determine the maximum power developed
during the 0.3-second time period.
F (N)
800
t (s)
0.2
0.3
SOLUTION
v (m/s)
See solution to Prob. 14–62.
P = 160 t - 533 t2
20
dP
= 160 - 1066.6 t = 0
dt
t (s)
0.3
t = 0.15 s 6 0.2 s
Thus maximum occurs at t = 0.2 s
Pmax = 53.3(0.2) = 10.7 kW
Ans.
14–64.
The 500-kg elevator starts from rest and travels upward
with a constant acceleration a c = 2 m>s2. Determine the
power output of the motor M when t = 3 s. Neglect the
mass of the pulleys and cable.
M
SOLUTION
+ c ©Fy = m ay ;
3T - 500(9.81) = 500(2)
E
T = 1968.33 N
3sE - sP = l
3 vE = vP
When t = 3 s,
(+ c ) v0 + ac t
vE = 0 + 2(3) = 6 m>s
vP = 3(6) = 18 m>s
PO = 1968.33(18)
PO = 35.4 kW
Ans.
14–65. A motor hoists a 60-kg crate at a constant velocity
to a height of H 5 m in 2 s. If the indicated power of the
motor is 3.2 kW, determine the motor’s efficiency.
h
Equations of Motion:
C i&Y MAY ;
&
60(9.81) 60(0)
Power: The crate travels at a constant speed of 2 & 588.6 N
5
2.50 ms. The power output
2
can be obtained using Eq. 14–10.
0 F v 588.6 (2.50) 1471.5 W
Thus, from Eq. 14–11, the efficiency of the motor is given by
power output
1471.5
0.460
power input
3200
Ans.
14–66.
The girl has a mass of 40 kg and center of mass at G. If she
is swinging to a maximum height defined by u = 60°,
determine the force developed along each of the four
supporting posts such as AB at the instant u = 0°. The
swing is centrally located between the posts.
A
2m
␪
G
SOLUTION
The maximum tension in the cable occurs when u = 0°.
B
T1 + V1 = T2 + V2
0 + 40(9.81)(- 2 cos 60°) =
1
(40)v2 + 40(9.81)(- 2)
2
v = 4.429 m>s
4.4292
b
2
+ c ©Fn = ma n;
T - 40(9.81) = (40)a
+ c ©Fy = 0;
2(2F) cos 30° - 784.8 = 0
T = 784.8 N
F = 227 N
Ans.
30⬚
30⬚
14–67.
Two equal-length springs are “nested” together in order to
form a shock absorber. If it is designed to arrest the motion
of a 2-kg mass that is dropped s = 0.5 m above the top of
the springs from an at-rest position, and the maximum
compression of the springs is to be 0.2 m, determine the
required stiffness of the inner spring, kB, if the outer spring
has a stiffness kA = 400 N>m.
s
A
SOLUTION
B
T1 + V1 = T2 + V2
0 + 0 = 0 - 2(9.81)(0.5 + 0.2) +
kB = 287 N>m
1
1
(400)(0.2)2 + (kB)(0.2)2
2
2
Ans.
14–68. Each of the two elastic rubber bands of the
slingshot has an unstretched length of 180 mm. If they are
pulled back to the position shown and released from rest,
determine the speed of the 30-g pellet just after the
rubber bands become unstretched. Neglect the mass of
the rubber bands. Each rubber band has a stiffness of
K 80 Nm.
41
0
61 42
240 mm
62
1
(2)2 3 (80)[ (0.05)2
2
v 4.76 ms
50 mm
50 mm
(0.240)2
0.18]2 1
(0.030)v2
2
Ans.
14–69. Each of the two elastic rubber bands of the
slingshot has an unstretched length of 180 mm. If they are
pulled back to the position shown and released from rest,
determine the maximum height the 30-g pellet will reach if
it is fired vertically upward. Neglect the mass of the rubber
bands and the change in elevation of the pellet while it is
constrained by the rubber bands. Each rubber band has a
stiffness K 80 Nm.
41
0
61 42
50 mm
50 mm
240 mm
62
1
2 2 3 (80)[ (0.05)2
2
H 1.15 4 m 115 4 mm
(0.240)2
0.18] 2 0
0.030(9.81)H
Ans.
14–70.
The 2-kg ball of negligible size is fired from point A with
an initial velocity of 10 m/s up the smooth inclined plane.
Determine the distance from point C to where it hits the
horizontal surface at D. Also, what is its velocity when it
strikes the surface?
B
10 m/s
1.5 m
A
SOLUTION
2m
Datum at A:
TA + VA = TB + VB
1
1
(2)(10)2 + 0 = (2)(nB)2 + 2(9.81)(1.5)
2
2
vB = 8.401 m>s
+ B s = s + v t
A:
0
0
4
d = 0 + 8.401 a b t
5
(+ c )
s = s0 + v0 t +
1 2
at
2 c
1
3
-1.5 = 0 + 8.401 a b t + ( - 9.81)t2
5
2
-4.905t2 + 5.040t + 1.5 = 0
Solving for the positive root,
t = 1.269 s
4
d = 8.401a b (1.269) = 8.53 m
5
Ans.
Datum at A:
TA + VA = TD + VD
1
1
(2)(10)2 + 0 = (2)(nD)2 + 0
2
2
vD = 10 m s
C
Ans.
D
d
14–71.
The girl has mass M and center of mass at G. If she is swinging to a maximum height defined by
T T1, determine the force developed along each of the four supporting posts such as AB at the
instant T = 0°. The swing is centrally located between the posts.
Given:
M
45 kg
T1
70 deg
I
30 deg
L
2m
g
9.81
m
2
s
Solution:
T1 V 1
v
2g L 1 cos T 1
T Mg
§ v2 ·
¸
©L¹
M¨
4FAB cos I T
1
2
Mv MgL
2
0 M g L cos T 1
T2 V 2
v
T
0
F AB
4
m
s
§ v2 ·
¸
©L¹
M g M¨
T
4 cos I
T
1022.4 N
F AB
295.1 N
Ans.
14–72.
The 2-kg collar is attached to a spring that has an
unstretched length of 3 m. If the collar is drawn to point B
and released from rest, determine its speed when it arrives
at point A.
k = 3 N/m
3m
B
A
SOLUTION
Potential Energy: The initial and final elastic potential energy are
1
1
(3) A 232 + 42 - 3 B 2 = 6.00 J and (3)(3 - 3)2 = 0, respectively.The gravitational
2
2
potential energy remains the same since the elevation of collar does not change when it
moves from B to A.
Conservation of Energy:
TB + VB = TA + VA
0 + 6.00 =
1
(2) v2A + 0
2
vA = 2.45 m s
Ans.
4m
14–73.
The 2-kg collar is attached to a spring that has an unstretched
length of 2 m. If the collar is drawn to point B and released
from rest, determine its speed when it arrives at point A.
k ⫽ 3 N/m
3m
B
A
SOLUTION
4m
Potential Energy: The stretches of the spring when the collar is at B and A are
sB = 232 + 4 2 - 2 = 3 m and sA = 3 - 2 = 1 m, respectively. Thus, the elastic
potential energy of the system at B and A are
(Ve)B =
(Ve)A =
1
1
ks 2 = (3)(32) = 13.5 J
2 B
2
1
1
ks 2 = (3)(12) = 1.5 J
2 A
2
There is no change in gravitational potential energy since the elevation of the collar
does not change during the motion.
Conservation of Energy:
TB + VB = TA + VA
1
1
mvB2 + (Ve)B = mvA2 + (Ve)A
2
2
0 + 13.5 =
1
(2)vA2 + 1.5
2
vA = 3.46 m>s
Ans.
14 –74.
The roller-coaster car has mass M, including its passenger, and starts from the top of the hill A
with a speed vA. Determine the minimum height h of the hill crest so that the car travels around
both inside loops without leaving the track. Neglect friction, the mass of the wheels, and the size
of the car. What is the normal reaction on the car when the car is at B and when it is at C?
Units Used:
3
kN
10 N
Given:
M
800 kg
rB
10 m
rC
7m
m
s
vA
3
g
9.81
m
2
s
Solution:
Check the loop at B first
§¨ vB2 ·¸
M¨
¸
© rB ¹
NB M g
TA V A
TB V B
2
h
We require that.
vB
g rB
1
2
M vA M g h
2
NB
0
vB
9.9
Ans.
m
s
1
2
M vB M g2rB
2
2
vB vA
2g
2rB
h
24.5 m
Ans.
Now check the loop at C
TA V A
vC
TC V C
2
vA 2g h 2rC
NC M g
§¨ vC2 ·¸
M¨
¸
© rC ¹
1
2
M vA M g h
2
vC
NC
14.7
1
2
M vC M g2rC
2
m
s
§¨ vC2 ·¸
M¨
¸ Mg
© rC ¹
NC
16.8 kN
Since NC > 0 then the coaster successfully
passes through loop C.
Ans.
14–75.
The roller-coaster car has mass M, including its passenger, and starts from the top of the hill A
with a speed vA. Determine the minimum height h of the hill crest so that the car travels around
both inside loops without leaving the track. Neglect friction, the mass of the wheels, and the size
of the car. What is the normal reaction on the car when the car is at B and when it is at C?
Units Used:
3
kN
10 N
Given:
M
800 kg
rB
10 m
rC
7m
m
s
vA
0
g
9.81
m
2
s
Solution:
Check the loop at B first
§¨ vB2 ·¸
M¨
¸
© rB ¹
NB M g
TA V A
TB V B
2
h
We require that.
vB
g rB
1
2
M vA M g h
2
NB
0
vB
9.9
Ans.
m
s
1
2
M vB M g2rB
2
2
vB vA
2g
2rB
h
Ans.
25.0 m
Now check the loop at C
TA V A
vC
TC V C
2
vA 2g h 2rC
NC M g
§¨ vC2 ·¸
M¨
¸
© rC ¹
1
2
M vA M g h
2
vC
NC
14.7
1
2
M vC M g2rC
2
m
s
§¨ vC 2 ¸·
M¨
¸ Mg
© rC ¹
NC
16.8 kN
Ans.
Since NC > 0 then the coaster successfully
passes through loop C.
14–76.
The roller coaster car having a mass m is released from rest
at point A. If the track is to be designed so that the car does
not leave it at B, determine the required height h. Also, find
the speed of the car when it reaches point C. Neglect
friction.
A
B
7.5 m
20 m
C
SOLUTION
Equation of Motion: Since it is required that the roller coaster car is about to leave
vB 2
vB 2
=
the track at B, NB = 0. Here, an =
. By referring to the free-body
rB
7.5
diagram of the roller coaster car shown in Fig. a,
©Fn = ma n;
m(9.81) = m ¢
vB 2
≤
7.5
vB 2 = 73.575 m2>s2
Potential Energy: With reference to the datum set in Fig. b, the gravitational
potential energy of the rollercoaster car at positions A, B, and C are
A Vg B A = mghA = m(9.81)h = 9.81mh, A Vg B B = mghB = m(9.81)(20) = 196.2 m,
and A Vg B C = mghC = m(9.81)(0) = 0.
Conservation of Energy: Using the result of vB 2 and considering the motion of the
car from position A to B,
TA + VA = TB + VB
1
1
mvA 2 + A Vg B A = mvB 2 + A Vg B B
2
2
0 + 9.81mh =
1
m(73.575) + 196.2m
2
h = 23.75 m
Ans.
Also, considering the motion of the car from position B to C,
TB + VB = TC + VC
1
1
mvB 2 + A Vg B B = mvC 2 + A Vg B C
2
2
1
1
m(73.575) + 196.2m = mvC 2 + 0
2
2
vC = 21.6 m>s
Ans.
h
14–77.
A 750-mm-long spring is compressed and confined by the
plate P, which can slide freely along the vertical 600-mm-long
rods. The 40-kg block is given a speed of v = 5 m>s when it is
h = 2 m above the plate. Determine how far the plate moves
downwards when the block momentarily stops after striking
it. Neglect the mass of the plate.
v ⫽ 5 m/s
A
h⫽2m
SOLUTION
Potential Energy: With reference to the datum set in Fig. a, the gravitational potential
energy of the block at positions (1) and (2) are A Vg B 1 = mgh1 = 40(9.81)(0) = 0
and A Vg B 2 = mgh2 = 40(9.81) C -(2 + y) D = C -392.4(2 + y) D , respectively. The
compression of the spring when the block is at positions (1) and (2) are
s1 = (0.75 - 0.6) = 0.15 m and s2 = s1 + y = (0.15 + y) m. Thus, the initial and
final elastic potential energy of the spring are
A Ve B 1 = ks21 = (25)(103)(0.152) = 281.25 J
1
2
1
2
A Ve B 2 = ks22 = (25)(103)(0.15 + y)2
1
2
1
2
Conservation of Energy:
T1 + V1 = T2 + V2
1
1
mv21 + c A Vg B 1 + A Ve B 1 d = mv22 + c A Vg B 2 + A Ve B 2 d
2
2
1
(40)(52) + (0 + 281.25) = 0 + [ - 392.4(2 + y)] +
2
1
(25)(103)(0.15 + y)2
2
12500y2 + 3357.6y - 1284.8 = 0
Solving for the positive root of the above equation,
y = 0.2133 m = 213 mm
Ans.
P
k ⫽ 25 kN/m
600 mm
14 –78.
Two equal-length springs are “nested” together in order to form a shock absorber. If it is
designed to arrest the motion of mass M that is dropped from a height s1 above the top of the
springs from an at-rest position, and the maximum compression of the springs is to be G,
determine the required stiffness of the inner spring, kB, if the outer spring has a stiffness kA.
Given:
M
4 kg
kA
500
s1
N
m
G
0.2 m
g
9.81
m
2
s
0.6 m
Solution:
T1 V 1
T2 V 2
0 M g s1 G
kB
2M g s1 G
G
2
0
1
2
kA kB G
2
kA
kB
1070
N
m
Ans.
14–79. The vertical guide is smooth and the 5-kg collar is
released from rest at A. Determine the speed of the collar
when it is at position C. The spring has an unstretched
length of 300 mm.
0.4 m
B
A
Potential Energy: With reference to the datum set in Fig. a, the gravitational potential
k ⫽ 250 N/m
0.3 m
energy of the collar at positions A and C are 7H " NHI" 5(9.81)(0) 0 and
7H $ NHI$ 5(9.81)( 0.3) 14.715 J. When the collar is at positions A and
C, the spring stretches T" 0.4 0.3 0.1 m and T$ 20.4
2
2
0.3 0.3 0.2 m. The elastic potential energy of the spring when the collar is at these two
1
1
positions are 7F " LT" 2 (250)(0.12) 1.25 J and
2
2
1
1
7F $ LT$ 2 (250)(0.22) 5 J.
2
2
Conservation of Energy:
5"
7" 5$
1
Nv" 2
2
5 7H "
0
1.25 0
v$ 2.09 ms
7$
(7F)" 6 1
(5)v$ 2
2
1
Nv 2
2 $
14.715
5 7H $
(7F)$ 6
5
Ans.
C
14–80.
Each of the two elastic rubber bands of the slingshot has
an unstretched length of 200 mm. If they are pulled back
to the position shown and released from rest, determine
the speed of the 25-g pellet just after the rubber bands
become unstretched. Neglect the mass of the rubber
bands. Each rubber band has a stiffness of k = 50 N>m.
50 mm
50 mm
240 mm
SOLUTION
T1 + V1 = T2 + V2
1
1
0 + (2) a b(50)[2(0.05)2 + (0.240)2 - 0.2]2 = (0.025)v2
2
2
v = 2.86 m>s
Ans.
14–81.
Each of the two elastic rubber bands of the slingshot has an
unstretched length of 200 mm. If they are pulled back to the
position shown and released from rest, determine the
maximum height the 25-g pellet will reach if it is fired
vertically upward. Neglect the mass of the rubber bands and
the change in elevation of the pellet while it is constrained
by the rubber bands. Each rubber band has a stiffness
k = 50 N>m.
50 mm
50 mm
240 mm
SOLUTION
T1 + V1 = T2 + V2
1
0 + 2a b (50)[2(0.05)2 + (0.240)2 - 0.2]2 = 0 + 0.025(9.81)h
2
h = 0.416 m = 416 mm
Ans.
14–82.
If the mass of the earth is Me, show that the gravitational
potential energy of a body of mass m located a distance r
from the center of the earth is Vg = −GMem>r. Recall that
the gravitational force acting between the earth and the
body is F = G(Mem>r 2), Eq. 13–1. For the calculation, locate
the datum at r : q. Also, prove that F is a conservative
force.
r2
r
SOLUTION
r1
The work is computed by moving F from position r1 to a farther position r2.
Vg = - U = -
L
F dr
r2
= -G Me m
Lr1
= - G Me m a
dr
r2
1
1
- b
r2
r1
As r1 : q , let r2 = r1, F2 = F1, then
Vg :
-G Me m
r
To be conservative, require
F = - § Vg = =
G Me m
0
b
ar
0r
-G Me m
r2
Q.E.D.
14–83.
A rocket of mass m is fired vertically from the surface of the
earth, i.e., at r = r1 . Assuming no mass is lost as it travels
upward, determine the work it must do against gravity to
reach a distance r2 . The force of gravity is F = GMem>r2
(Eq. 13–1), where Me is the mass of the earth and r the
distance between the rocket and the center of the earth.
r2
r
SOLUTION
F = G
r1
Mem
r2
r2
F1-2 =
L
F dr = GMem
= GMema
1
1
- b
r1
r2
dr
2
Lr1 r
Ans.
14–84.
The firing mechanism of a pinball machine consists of a
plunger P having a mass of 0.25 kg and a spring of stiffness
k = 300 N>m. When s = 0, the spring is compressed
50 mm. If the arm is pulled back such that s = 100 mm and
released, determine the speed of the 0.3-kg pinball B just
before the plunger strikes the stop, i.e., s = 0. Assume all
surfaces of contact to be smooth. The ball moves in the
horizontal plane. Neglect friction, the mass of the spring,
and the rolling motion of the ball.
s
B
P
k
SOLUTION
T1 + V1 = T2 + V2
0 +
1
1
1
1
(300)(0.1 + 0.05)2 = (0.25)(n2)2 + (0.3)(n2)2 + (300)(0.05)2
2
2
2
2
n2 = 3.30 m s
Ans.
300 N/m
14–85.
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).
B
vB
rB
80 Mm
rA
SOLUTION
yA = 40 Mm>h = 11 111.1 m>s
Since V = -
GMe m
r
T1 + V1 = T2 + V2
66.73(10) - 12(5.976)(10)23(60)
66.73(10) - 12(5.976)(10)24(60)
1
1
2
(60)(11 111.1)2 (60)v
=
B
2
2
20(10)6
80(10)6
vB = 9672 m>s = 34.8 Mm>h
Ans.
20 Mm
vA
A
14–86.
The bob of mass M of a pendulum is fired from rest at position A by a spring which has a
stiffness k and is compressed a distance G. Determine the speed of the bob and the tension
in the cord when the bob is at positions B and C. Point B is located on the path where the
radius of curvature is still r, i.e., just before the cord becomes horizontal.
Units Used:
kN
3
10 N
Given:
M
0.75 kg
kN
m
k
6
G
125 mm
r
0.6 m
Solution:
At B:
0
vB
TB
1 2
kG
2
1
2
M vB M g r
2
§ k · G 2 2g r
¨ ¸
© M¹
vB
10.6
§ vB2 ·
¸
© r ¹
TB
142 N
vC
9.47
m
s
Ans.
TC
48.7 N
Ans.
M¨
m
s
Ans.
Ans.
At C:
0
vC
1 2
kG
2
§ k · G 2 6g r
¨ ¸
© M¹
TC M g
TC
1
2
M vC M g3r
2
§ vC2 ·
¸
M¨
© 2r ¹
§ vC2 ·
M¨
g¸
© 2r
¹
14–87.
A block having a mass M is attached to
four springs. If each spring has a stiffness
k and an unstretched length G, determine
the maximum downward vertical
displacement smax of the block if it is
released from rest at s = 0.
Units Used:
kN
3
10 N
Given:
M
20 kg
kN
m
k
2
l
100 mm
G
150 mm
Solution:
Guess
smax
100 mm
Given
1
2
4 k lG
2
smax
Find smax
2º
2º
ª1
ª1
M g smax 2« k l G smax » 2« k l G smax »
¬2
¼ ¬2
¼
smax
49.0 mm
Ans.
14–88.
Two equal-length springs having a stiffness kA = 300 N>m
and kB = 200 N>m are “nested” together in order to form
a shock absorber. If a 2-kg block is dropped from an at-rest
position 0.6 m above the top of the springs, determine
their deformation when the block momentarily stops.
0.6 m
B
SOLUTION
Datum at initial position:
T1 + V1 = T2 + V2
0 + 0 = 0 - 2(9.81)(0.6 + x) +
1
(300 + 200)(x)2
2
250x2 - 19.62x - 11.772 = 0
Solving for the positive root,
x = 0.260 m
Ans.
A
14–89.
When the 6-kg box reaches point A it has a speed of
vA = 2 m>s. Determine the angle u at which it leaves the
smooth circular ramp and the distance s to where it falls
into the cart. Neglect friction.
vA = 2 m/s
20°
A
θ
B
1.2 m
SOLUTION
s
At point B:
+b©Fn = man;
6(9.81) cos f = 6 a
n2B
b
1.2
(1)
Datum at bottom of curve:
TA + VA = TB + VB
1
1
(6)(2)2 + 6(9.81)(1.2 cos 20°) = (6)(vB)2 + 6(9.81)(1.2 cos f)
2
2
13.062 = 0.5v2B + 11.772 cos f
(2)
Substitute Eq. (1) into Eq. (2), and solving for vB,
vB = 2.951 m>s
Thus,
f = cos - 1 a
(2.951)2
b = 42.29°
1.2(9.81)
u = f - 20° = 22.3°
A+cB
Ans.
s = s0 + v0 t + 12 ac t2
-1.2 cos 42.29° = 0 - 2.951(sin 42.29°)t +
1
( -9.81)t2
2
4.905t2 + 1.9857t - 0.8877 = 0
Solving for the positive root: t = 0.2687 s
+ b
a:
s = s0 + v0 t
s = 0 + (2.951 cos 42.29°)(0.2687)
s = 0.587 m
Ans.
14–90.
The Raptor is an outside loop roller coaster in which riders
are belted into seats resembling ski-lift chairs. Determine
the minimum speed v0 at which the cars should coast down
from the top of the hill, so that passengers can just make the
loop without leaving contact with their seats. Neglect
friction, the size of the car and passenger, and assume each
passenger and car has a mass m.
v0
h
SOLUTION
Datum at ground:
T1 + V1 = T2 + V2
1 2
1
mv + mgh = mv21 + mg2r
2 0
2
v1 = 2v20 + 2g(h-2r)
+ T ©Fn = man;
mg = ma
v21
b
r
v1 = 2gr
Thus,
gr = v20 + 2gh - 4gr
v0 =
g(5r - 2h)
Ans.
r
14–91.
The Raptor is an outside loop roller coaster in which riders
are belted into seats resembling ski-lift chairs. If the cars
travel at v0 = 4 m>s when they are at the top of the hill,
determine their speed when they are at the top of the loop
and the reaction of the 70-kg passenger on his seat at this
instant. The car has a mass of 50 kg. Take h = 12 m, r = 5 m.
Neglect friction and the size of the car and passenger.
v0
h
SOLUTION
Datum at ground:
T1 + V1 = T2 + V2
1
1
(120)(4)2 + 120(9.81)(12) = (120)(v1)2 + 120(9.81)(10)
2
2
v1 = 7.432 m>s
+ T ©Fn = man;
Ans.
70(9.81) + N = 70 a
N = 86.7 N
(7.432)2
b
5
Ans.
r
14–92.
The 75-kg man bungee jumps off the bridge at A with an
initial downward speed of 1.5 m>s. Determine the required
unstretched length of the elastic cord to which he is
attached in order that he stops momentarily just above the
surface of the water. The stiffness of the elastic cord is
k = 80 N>m. Neglect the size of the man.
A
150 m
SOLUTION
B
Potential Energy: With reference to the datum set at the surface of the water, the
gravitational potential energy of the man at positions A and B are A Vg B A = mghA =
75(9.81)(150) = 110362.5 J and A Vg B B = mghB = 75(9.81)(0) = 0. When the man
is at position A, the elastic cord is unstretched (sA = 0), whereas the elastic cord
stretches sB = A 150 - l0 B m, where l0 is the unstretched length of the cord.Thus, the
elastic potential energy of the elastic cord when the man is at these two positions are
1
1
1
A Ve B A = ksA 2 = 0 and A Ve B B = ksB 2 = (80)(150 - l0)2 = 40(150 - l0)2.
2
2
2
Conservation of Energy:
TA + VA = TB + VB
1
1
mvA 2 + B a Vg b + A Ve B A R = mvB 2 + B a Vg b + A Ve B B R
2
2
A
B
1
(75)(1.52) + A 110362.5 + 0 B = 0 + C 0 + 40(150 - l0)2 D
2
l0 = 97.5 m
Ans.
14–93.
The 10-kg sphere C is released from rest when u = 0° and
the tension in the spring is 100 N. Determine the speed of
the sphere at the instant u = 90°. Neglect the mass of rod
AB and the size of the sphere.
E
0.4 m
SOLUTION
A
Potential Energy: With reference to the datum set in Fig. a, the gravitational potential
energy of the sphere at positions (1) and (2) are A Vg B 1 = mgh1 = 10(9.81)(0.45) =
44.145 J and A Vg B 2 = mgh2 = 10(9.81)(0) = 0. When the sphere is at position (1),
100
the spring stretches s1 =
= 0.2 m. Thus, the unstretched length of the spring is
500
l0 = 30.32 + 0.42 - 0.2 = 0.3 m, and the elastic potential energy of the spring is
1
1
A Ve B 1 = ks12 = (500)(0.22) = 10 J. When the sphere is at position (2), the spring
2
2
stretches s2 = 0.7 - 0.3 = 0.4 m, and the elastic potential energy of the spring is
1
1
A Ve B 2 = ks22 = (500)(0.42) = 40 J.
2
2
Conservation of Energy:
T1 + V1 = T2 + V2
1
1
m A v B 2 + c A Vg B 1 + A Ve B 1 d = ms A vs B 2 2 + c A Vg B 2 + A Ve B 2 d
2 s s 1
2
0 + (44.145 + 10) =
(vs)2 = 1.68 m>s
k ⫽ 500 N/m
1
(10)(vs)22 + (0 + 40)
2
Ans.
u
D
B
0.3 m
C
0.15 m
14–94.
A quarter-circular tube AB of mean radius r
contains a smooth chain that has a mass per unit length of N0.
If the chain is released from rest from the position shown,
determine its speed when it emerges completely from the tube.
r
B
Potential Energy: The location of the center of gravity G of the chain at
positions (1) and (2) are shown in Fig. a. The mass of the chain is
p
2S
p 2
p
N N 0 3 S4 N 0S . Thus, the center of mass is at I1 S 3
4 S.
p
p
2
2
With reference to the datum set in Fig. a the gravitational potential energy of the
chain at positions (1) and (2) are
p
2
7H 1 NHI1 3 N 0SH 4 3
p 2
p 2
4S 3
4N 0S2H
p
2
and
7H 2 NHI 2 0
Conservation of Energy:
51
71 52
1
Nv1 2
2
0
3
v2 7H 1 72
1
Nv2 2
2
7H 2
p 2
1 p
4 N 0S2H 3 N 0S 4v 2 2
2
2 2
2
(p 2)HS
Ap
A
O
0
Ans.
14–95. The ball of mass m is given a speed of
W" 23HS at position A. When it reaches B, the cord hits
the small peg P, after which the ball describes a smaller
circular path. Determine the position x of P so that the
ball will just be able to reach point C.
C
x
B
O
P
r
A
vA
Equation of Motion: If the ball is just about to complete the small circular path, the
v$ 2
v2
cord will become slack at position C, i.e., 5 0. Here, BO . By
r
S Y
referring to the free-body diagram of the ball shown in Fig. a,
j'O NB O ;
NH N 3
v$ 2
4
S Y
v$ 2 H(S Y)
(1)
Potential Energy: With reference to the datum set in Fig. b, the gravitational
potential energy of the ball at positions A and C are 7H " NHI" NH(0) 0
and 7H $ NHI$ NH(2S Y).
Conservation of Energy:
5"
7" 5$
1
Nv " 2
2
1
N(3HS)
2
7$
7H " 0 1
Nv$ 2
2
1
Nv$ 2
2
v$ 2 H(2Y S)
7H $
NH(2S Y)
(2)
Solving Eqs. (1) and (2) yields
Y v$ 2
S
3
1
HS
A2
Ans.
14–96. The ball of mass m is given a speed of
W" 25HS at position A. When it reaches B, the cord hits
the peg P, after which the ball describes a smaller circular
path. If Y 23S, determine the speed of the ball and the
tension in the cord when it is at the highest point C.
C
x
B
O
P
r
A
vA
Potential Energy: With reference to the datum set in Fig. a, the gravitational
potential energy of the ball at positions A and C are 7H " NHI" NH(0) 0
4
4
and 7H $ NHI$ NH3 S 4 NHS.
3
3
Conservation of Energy:
5"
7" 5$
1
Nv " 2
2
1
N(5HS)
2
v$ 7$
7H " 0 1
Nv$ 2
2
1
Nv $ 2
2
7H $
4
NHS
3
7
HS
A3
Ans.
7
HS
v$ 2
3
Equations of Motion: Here, BO . By referring to the free-body diagram
r
S3
of the ball shown in Fig. b,
j'O NBO;
5
NH N(7H)
5 6NH
Ans.
14–97.
A pan of negligible mass is attached to two identical springs of
stiffness k = 250 N>m. 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.
1m
k
250 N/m
1m
k
250 N/m
0.5 m
d
SOLUTION
Potential Energy: With reference to the datum set in Fig. a, the gravitational potential
energy of the box at positions (1) and (2) are A Vg B 1 = mgh1 = 10(9.81)(0) = 0 and
A Vg B 2 = mgh2 = 10(9.81) C - A 0.5 + d B D = - 98.1 A 0.5 + d B . Initially, the spring
50
= 0.2 m. Thus, the unstretched length of the spring
250
is l0 = 1 - 0.2 = 0.8 m and the initial elastic potential of each spring is
1
A Ve B 1 = (2) ks1 2 = 2(250 > 2)(0.22) = 10 J . When the box is at position (2), the
2
stretches s1 =
spring stretches s2 = a 2d2 + 12 - 0.8 b m. The elastic potential energy of the
springs when the box is at this position is
A Ve B 2 = (2) ks2 2 = 2(250 > 2) c 2d2 + 1 - 0.8 d = 250a d2 - 1.62d2 + 1 + 1.64 b .
2
1
2
Conservation of Energy:
T1 + V 1 + T2 + V2
1
1
mv1 2 + B a Vg b + A Ve B 1 R = mv2 2 + B aVg b + A Ve B 2 R
2
2
1
2
0 + A 0 + 10 B = 0 + B - 98.1 A 0.5 + d B + 250 ¢ d2 - 1.6 2d2 + 1 + 1.64 ≤ R
250d2 - 98.1d - 400 2d2 + 1 + 350.95 = 0
Solving the above equation by trial and error,
d = 1.34 m
Ans.