14–1.
SOLUTION
Principle of Work and Energy: The horizontal component of force Fwhich acts
in the direction of displacement does positive work, whereas the friction force
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
Ans.v=10.7 m s
–L25 m
15 m
36.55 ds =1
2(20)v2
1
2(20)(8 2)+L25 m
15 m
100 cos 30° ds
T1+aU1–2=T2
Ff=0.25(146.2) =36.55 N
N=146.2 N
+caFy=may;N+100 sin 30° –20(9.81) =20(0)
30°
F
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.
14–2.
F (lb)
F 90(10)3 x1/2
x (ft)
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 lb, where is in ft,
determine the car’s maximum penetration in the barrier.
The car has a weight of 4000 lb and it is traveling with a
speed of just before it hits the barrier.75 ft>s
xF =(90(103)x1>2)
SOLUTION
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 The
coefficient of kinetic friction between the crate and the
surface is .mk=0.2
6m>s.
SOLUTION
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 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, . Applying
Eq. 14–7, we have
Ans.s=1.35m
0+800 cos 30°(s)+1000a4
5bs–156.2s=1
2(100)
A
62
B
T1+aU1–2=T2
T1=0
Ff=0.2(781) =156.2 N
N=781 N
+c©Fy=may;N+1000a3
5b–800 sin 30° –100(9.81) =100(0)
3
4
5
1000 N
30
800 N
*14–4.
The 100-kg crate is subjected to the forces shown. If it is
originally at rest, determine the distance it slides in order to
attain a speed of v
=8 m>s.
The coefficient of kinetic
friction between the crate and the surface is m
k=0.2.
SOLUTION
400 N
30
45
500 N
14–5.
Determine the required height h of the roller coaster so that
when it is essentially at rest at the crest of the hill A it will
reach a speed of 100 km
>
h when it comes to the bottom B.
Also, what should be the minimum radius of curvature
r
for
the track at B so that the passengers do not experience a
normal force greater than
4mg =(39.24m) N?
Neglect the
size of the car and passenger.
SOLUTION
100
(
10
3
)
A
h
B
r
14–6.
When the driver applies the brakes of a light truck traveling
40 km
>
h, it skids 3 m before stopping. How far will the truck
skid if it is traveling 80 km
>
h when the brakes are applied?
SOLUTION
40
(
10
3
)
14–7.
Ans.
Observer B:
At ,
Block moves
Thus
Ans.
Note that this result is less than that observed by A.2 m>s
v2=4.08 m>s
1
2(10)(3)2+6(6.391) =1
2(10)v2
2
T1+©U1–2=T2
10 –3.609 =6.391 m
s¿=2(1.805) =3.609 mv=2m>s
t=1.805 s
t2+16.67t–33.33 =0
10 =0+5t+1
2(0.6)t2
A
:
+
B
s=s0+v0t+1
2act2
6=10aa=0.6 m>s2
F=ma
v2=6.08 m>s
1
2(10)(5)2+6(10) =1
2(10)v2
2
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 Ais in a fixed frame x,determine the
final speed of the block if it has an initial speed of 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 axis and moving at a constant
velocity of relative to A.Hint: The distance the block
travels will first have to be computed for observer Bbefore
applying the principle of work and energy.
2m>s
x¿
5m>s
6N
5m/s
2m/s
10 m
B
x
x¿
A
*14–8.
A force of
F=250 N
is applied to the end at B. Determine
the speed of the 10-kg block when it has moved 1.5 m,
starting from rest.
SOLUTION
14–9.
The “air spring” A is used to protect the support B and
prevent damage to the conveyor-belt tensioning weight C
in the event of a belt failure D. The force developed by
the air spring as a function of its deection is shown by the
graph. If the block has a mass of 20 kg and is suspended
a height
d=0.4 m
above the top of the spring, determine
the maximum deformation of the spring in the event the
conveyor belt fails. Neglect the mass of the pulley and belt.
SOLUTION
d
B
A
DF (N)
s (m)
C
1500
0.2
14–10.
The force F, acting in a constant direction on the 20-kg
block, has a magnitude which varies with the position s of
the block. Determine how far the block must slide before its
velocity becomes 15 m
>
s. When
s=0
the block is moving
to the right at v
=6 m>s.
The coefficient of kinetic friction
between the block and surface is m
k=0.3.
SOLUTION
s=20.52
m=20.5
m
Ans.
F (N)
F 50s1/2
F
v
387
14–11.
The force of
F=50 N
is applied to the cord when
s=2 m.
If the 6-kg collar is orginally at rest, determine its velocity at
s=0.
Neglect friction.
SOLUTION
A
s
1.5 m
F
Ans:
v=4.08 m>s
*14–12.
SOLUTION
Design considerations for the bumper Bon the 5-Mg train
car require use of a nonlinear spring having the load-
deflection characteristics shown in the graph. Select the
proper value of kso that the maximum deflection of the
spring is limited to 0.2 m when the car, traveling at
strikes the rigid stop. Neglect the mass of the car wheels.
4m>s,
F(N) Fks
2
s(m)
389
14–13.
SOLUTION
Solving for the positive root,
Ans.
Ans
.vC=54.1 ft>s
1
2a2
32.2 b(5)2+2(45) =1
2a2
32.2 bv2
C
TA+©UA–C=TC
d=31.48a4
5b(0.89916) =22.6 ft
t=0.89916 s
16.1t2+18.888t–30 =0
30 =0+31.48a3
5bt+1
2(32.2)t2
A
+T
B
s=s0+v0t–1
2act2
d=0+31.48a4
5bt
a:
+bs=s0+v0t
TA+©UA–B=TB
The 2-lb brick slides down a smooth roof, such that when it
is at Ait has a velocity of Determine the speed of the
brick just before it leaves the surface at B, the distance d
from the wall to where it strikes the ground, and the speed
at which it hits the ground.
5ft>s.
A
B
15 ft
5ft/s
5
y
3
4
Ans:
vB=31.5
ft>s
d=22.6
ft
vC=54.1
ft>s
14–14.
SOLUTION
B
A
5
4
3
Block A has a weight of 60 lb and block B has a weight of
10 lb. Determine the speed of block A after it moves 5 ft
down the plane, starting from rest. Neglect friction and the
mass of the cord and pulleys.
14–15.
The two blocks Aand Bhave weights and
If the kinetic coefficient of friction between the
incline and block Ais determine the speed of A
after it moves 3 ft down the plane starting from rest. Neglect
the mass of the cord and pulleys.
mk=0.2,
WB=10 lb.
WA
=60 lb
SOLUTION
B
A
392
*14–16.
SOLUTION
(1)
Equations of Motion: Here,.By
referring to Fig. a,
It is required that the block leave the track. Thus,.
Since ,
Ans.u=41.41° =41.4°
3 cos u–9
4=0
mg Z0
0=mga3 cos u–9
4b
N=0
N=mga3 cos u–9
4b
©Fn=man;mg cos u–N=mcga9
4–2 cos ubd
an=v2
r
=
gra9
4–2 cos ub
r=ga9
4–2 cos ub
v2=gra9
4–2 cos ub
1
2ma1
4grb+mg(r–rcos u)=1
2mv2
T1+©U1–2=T2
A small box of mass mis given a speed of at the
top of the smooth half cylinder. Determine the angle at
which the box leaves the cylinder.
u
v=2
1
4gr
r
O
A
u
Ans:
u=41.4°
14–17.
F 30 lb
A
C
B
y
y x2
x
4.5 ft
1
2
If the cord is subjected to a constant force of lb and
the smooth 10-lb collar starts from rest at A, determine its
speed when it passes point B.Neglect the size of pulley C.
F=30
SOLUTION
14–18.
When the 12-lb block A is released from rest it lifts the two
15-lb weights B and C. Determine the maximum distance
A will fall before its motion is momentarily stopped.
Neglect the weight of the cord and the size of the pulleys.
SOLUTION
Consider the entire system:
t
=2
y2
+
42
T1+ΣU1–2=T2
(0
+
0
+
0)
+
12y
–
2(15)
(
2
y2
+
42
–
4
)
=
(0
+
0
+
0)
0.4y
=2
y2
+
16
–
4
(0.4y+4)2=y2+16
–0.84y2+3.20y+16 =16
–0.84y+3.20 =0
y=3.81
ft
Ans.
BC
A
4 ft 4 ft
y=3.81
ft
395
14–19.
B
C
200 mm
200 mm
200 mm
F 300 N
30
If the cord is subjected to a constant force of
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.
F=300 N
SOLUTION
while force F displaces a distance of
.Here,the work of F is positive, whereas Wdoes
negative work.
Ans.vB=3.335 m>s=3.34 m>s
0+300(0.5234) +[–15(9.81)(0.5)] =1
2 (15)vB2
TA+gUA–B=TB
20.22+0.22=0.5234 m
s=AC –BC =20.72+0.42 –
Ans:
vB=3.34
m>s
*14–20.
SOLUTION
(O.K!)
Thus
Ans.s=5ft+4.2
9
ft =
9
.2
9
ft
x=4.29 ft 6(15 –5) ft
2(9) +(5 –2)(18) +x(27) =187.89
Area =187.89 kip #ft
1
2a4000
32.2 b(55)2–Area =0
T1+©U1–2=T2
The crash cushion for a highway barrier consists of a nest of
barrels filled with an impact-absorbing material.The barrier
stopping force is measured versus the vehicle penetration
into the barrier. Determine the distance a car having a
weight of 4000 lb will penetrate the barrier if it is originally
traveling at when it strikes the first barrel.55 ft>s
521015
Vehicle penetration (ft)
20 25
Barrier stopping force (kip)
36
27
18
0
9