14–41.
SOLUTION
Ans.l0=2.77 ft
0+1
2(2)
C
p(1.5) –l0
D
2–1
2(2)c3p
4(1.5) –l0d2
–2(1.5 sin 45°) =1
2a2
32.2 b(5.844)2
T1+© U1–2=T2
v=5.844 ft>s
+b©Fn=man;2sin 45° =2
32.2 av2
1.5 b
A 2-lb block rests on the smooth semicylindrical surface.An
elastic cord having a stiffness is attached to the
block at Band to the base of the semicylinder at point C.If
the block is released from rest at A (), determine the
unstretched length of the cord so that the block begins to
leave the semicylinder at the instant . Neglect the
size of the block.
u=45°
u=0°
k=2lb>ft
CA
u
B
k2lb/ft
1.5 ft
14–42.
The jeep has a weight of 2500 lb and an engine which
transmits a power of 100 hp to all the wheels.Assuming the
wheels do not slip on the ground, determine the angle of
the largest incline the jeep can climb at a constant speed
v=30 ft>s.
u
SOLUTION
Ans.u=47.2°
100(550) =2500 sin u(30)
P=FJv
θ
14–43.
SOLUTION
Using Eq. 14–11, the required power input for the motor to provide the above
power output is
Ans.=1500
0.65 =2307.7 ft #lb>s=4.20 hp
power input =power output
P
Determine the power input for a motor necessary to lift
300
lb
at a constant rate of The efficiency of the motor is
.P=0.65
5ft
>
s.
*14–44.
SOLUTION
Power: Here, the speed of the car is .
The power output can be obtained using Eq. 14–10.
Using Eq. 14–11, the required power input from the engine to provide the above
power output is
Ans.=66.418
0.65 =102 kW
power input =power output
e
P=F#v=2391.08(27.78) =66.418(103)W =66.418 kW
y=
B
100(103)m
h
R
*a1h
3600 s b=27.78 m>s
F=2391.08 N
An automobile having a mass of 2 Mg travels up a 7° slope
ataconstant speed of If mechanical friction
and wind resistance are neglected, determine the power
developed by the engine if the automobile has an efficiency
P=0.65.
v=100 km>h.
7
421
14–45.
SOLUTION
Ans.P=5200(600)a88 ft>s
60 m>hb1
550 =8.32 (103)hp
h
ilki
i
b
j
i
h
is placed in a plane having a weight of 13000 lb. If the engine
develops a constant thrust of 5200 lb, determine the power
output of the plane when it is just ready to take off with a
speed of 600 mi>h.
14–46.
SOLUTION
.Thus, the kinetic energy of the car is
The power of the bulb is
.Thus,
Ans.t=U
Pbulb
=204.59(103)
73.73 =2774.98 s =46.2 min
73.73 ft #lb>s
Pbulb =100 W *a1hp
746 W b*a550 ft #lb>s
1hpb=
U=1
2my2=1
2a5000
32.2 b
A
51.332
B
=204.59
A
103
B
ft #lb
51.33 ft>s
car having aweight of 5000 lb that is traveling at .If
the car is brought to astop,determine how long a100-W light
bulb must burn to expend the same amount of energy.
11mi=5280 ft.2
35 mi>h
14–47.
SOLUTION
15 ft
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.
*14–48.
SOLUTION
The man having the weight of 150 lb is able to run up
a 15-ft-high fl ight of stairs in 4 s. Determine the power
generated. How long would a 100-W light bulb have to burn
to expend the same amount of energy? Conclusion: Please
turn off the lights when they are not in use! 15 ft
14–49.
SOLUTION
Equations of Motion: By referring to the free-body diagram of the car shown in
Fig. a,
Power: The maximum power output of the motor can be determined from
Thus, the maximum power input is given by
Ans.
The average power output can be determined from
Thus,
Ans.(Pin)avg =(Pout)avg
e=45236.62
0.8 =56 545.78 W =56.5 kW
(Pout)avg =F#vavg =3618.93a25
2b=45 236.62 W
Pin =Pout
e=90473.24
0.8 =113 091.55 W =113 kW
(Pout)max=F#vmax=3618.93(25) =90 473.24 W
F=3618.93N
©Fx¿=max¿;F–2000(9.81) sin 5.711° =2000(0.8333)
ac=0.8333 m>s2
25 =0+ac(30)
The 2-Mg car increases its speed uniformly from rest to
in 30 s up the inclined road. Determine the
maximum power that must be supplied by the engine, which
operates with an efficiency of . Also, find the
average power supplied by the engine.
P=0.8
25 m>s
10
1
14–50.
SOLUTION
nM=8ft>s
2(–4) =-nM
2nP=-nM
2sP+sM=l
Determine the power output of the draw-works motor M
necessary to lift the 600-lb drill pipe upward with a constant
speed of The cable is tied to the top of the oil rig,
wraps around the lower pulley, then around the top pulley,
and then to the motor.
4ft>s.
M
4ft/s
14–51.
The 1000-lb elevator is hoisted by the pulley system and
motor M. If the motor exerts a constant force of 500 lb on
the cable, determine the power that must be supplied to the
motor at the instant the load has been hoisted
s=15 ft
starting from rest. The motor has an efficiency of
e=0.65.
SOLUTION
M
*14–52.
The 50-lb crate is given a speed of 10 ft
>
s in
t=4 s
starting
from rest. If the acceleration is constant, determine the
power that must be supplied to the motor when
t=2 s.
The motor has an efficiency
e=0.65.
Neglect the mass of
the pulley and cable.
SOLUTION
+
c
ΣF
y=
ma
y; 2T–50 =
50
32.2
a
(
+
c
)
v
=
v
0+
a
c
t
10 =0+a(4)
a=2.5 ft>s2
T=26.94 lb
In t=2 s
(
+
c
)
v
=
v
0+
a
c
l
v=0+2.5(2) =5 ft>s
sC+(sC–sP)=l
2 vC=vP
2(5) =vP=10 ft>s
P0=26.94(10) =269.4
P1=
269.4
0.65
=414.5 ft
#
lb
>
s
P1=0.754 hp
Ans.
s
M
14–53.
The sports car has a mass of 2.3 Mg,and while it is traveling
at 28 m/sthe driver causes it to accelerate at If the
drag resistance on the car due to the wind is
where vis 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=10.3v22N,
5m>s2.
SOLUTION
At
Ans.Pi=PO
e=328.59
0.68 =438 kW
PO=(11 735.2)(28) =328.59 kW
F=11 735.2 N
v=28 m>s
F=0.3v2+11.5(103)
:
+©Fx=ma
x;F–0.3v2=2.3(103)(5)
F
D
14–54.
SOLUTION
Ans.Pi=PO
e=423.0
0.68 =622 kW
PO=F#v=[13.8(103)+10(30)](30) =423.0 kW
v=0+6(5) =30 m>s
(:
+)v=v0+act
F=13.8(103)+10 v
:
+©Fx=ma
x;F–10v=2.3(103)(6)
starting from rest. If the drag resistance on the car
due to the wind is where vis the velocity in
m/s, determine the power supplied to the engine when
The engine has a running efficiency of P=0.68.t=5s.
FD=110v2N,
6m>s2,
D
14–55.
SOLUTION
Hoisting is provided by the motor Mand the 60-kg block C.
If the motor has an efficiencyofdetermine the
power that must be supplied to the motor when the elevator
is hoisted upward at aconstant speed of vE=4m>s.
P=0.6,
C
M
432
*14–56.
SOLUTION
s
Ans.P=0.229 hp
P=F#v=25 cos 60°(10.06) =125.76 ft #lb>s
v=10.06 ft>s
0+38.40 –10(4 –1.732) =1
2(10
32.2)v2
T1+©U1–2=T2
applying a constant vertical force of to the cord.
If the rod is smooth, determine the power developed by the
force at the instant u=60°.
F=25 lb
4ft
A
3ft
F
u
Ans:
P=0.229 hp
14–57.
SOLUTION
Ans.=0.0364 hp
P=F#v=12.5a4
5b(2) =20 lb #ft>s
F=12.5 lb
+c©Fy=ma
y;Fa4
5b–10 =0
T
h
e 10-
lb
co
ll
ar starts from rest at Aan
d
i
s
li
fte
d
w
i
t
h
a
constant speed of along the smooth rod. Determine the
power developed by the force Fat the instant shown.
2ft>s
4ft
A
3ft
F
u
14–58.
SOLUTION
The 50-lb block rests on the rough surface for which
the coeffi cient of kinetic friction is mk=0.2. A force
F=
(
40 +s2
)
lb, where s is in ft, acts on the block in the
direction shown. If the spring is originally unstretched
(s=0) and the block is at rest, determine the power
developed by the force the instant the block has moved
s=1.5 ft.
F
k = 20 lb ft
30°
14–59.
SOLUTION
Total load:
If load is placed at the center height, , then
Ans.
Also,
Ans.P=F#v=47 088(0.2683) =12.6 kW
P=U
t=94.18
7.454 =12.6 kW
t=h
ny
=2
0.2683 =7.454 s
ny=nsin u=0.6
¢
4
2(32(0.25))2+42
≤
=0.2683 m>s
U=47 088a4
2b=94.18 kJ
h=4
2=2m
32(150)(9.81) =47 088 N
The escalator steps move with a constant speed of
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.
0.6 m
>
s.
4 m
125 mm
250 mm
v0.6 m/s
*14–60.
SOLUTION
Ans.n=s
t=2(32(0.25))2+42
31.4 =0.285 m>s
P=U1–2
t=(80)(9.81)(4)
t=100 t=31.4 s
125 mm
250 mm
v0.6 m/s
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.