52 4. WORK AND ENERGY OF PARTICLES
Remarks:
e main difference between the principle of work and energy, Equation (4.6), and the princi-
ple of conservation of energy, Equation (4.8), is that Equation (4.6) applies to cases in which
the forces can be non-conservative whereas Equation (4.8) applies to a particle acting on by a
conservative force or forces.
4.6 POWER AND MECHANICAL EFFICIENCY
Power is defined as the time rate of work. If U is the incremental work done during the
incremental interval t the average power during the interval is
D
U
t
so that power is given as
D lim
t !0
U
t
D
d U
dt
:
Substituting for Equation (4.1a), one has
D
d U
dt
D
*
F d
*
r
dt
D
*
F
*
v (4.9)
which states the power is the scalar or dot product of force and velocity.
e mechanical efficiency of a machine is defined as the ratio of output power
0
to the
input power
I
D
0
I
: (4.10)
Note that since power is a scalar quantity the mechanical efficiency of a machine is also a di-
mensionless scalar quantity.
Example 4.1
e cable system shown in Figure 3.3a is at rest when a constant force P D 300 N is applied to
block A. Disregarding the masses of the pulleys, effect of friction in the pulleys, and assuming
that the static and kinetic coefficients of friction between block A and the horizontal surface are
respectively,
s
D 0:30 and
k
D 0:25, determine
(a) the velocity of block B after block A has moved 3 m, and
(b) the tension in the cable.
4.6. POWER AND MECHANICAL EFFICIENCY 53
Solution:
e analysis in the solution to Example 3.3 is applied in the present problem. With reference
to Figure 3.3b, the constraint condition of the cable is
x
A
C 3y
B
D constant:
Taking the time derivative of this equation gives
v
A
C 3v
B
D 0 ) v
A
D 3v
B
: (4.11a)
Recall, m
A
D 30 kg, P D 300 N. e principle of work and energy defined by Equation (4.6) is
applicable.
For block A, .T
1
/
A
C .U
1!2
/
A
D .T
2
/
A
in which .T
1
/
A
D 0 since the cable system starts from
rest. e work done during the movement of 3 m is
.U
1!2
/
A
D .P F
A
F /.3 m/:
erefore, .P F
A
F /.3 m/ D
1
2
m
A
.v
2
A
/.
By Equation (4.11a) and the results in Example 3.3,
.300 N 168:25 N F /.3 m/ D
1
2
.30 kg/.9v
2
B
/
) 395:25 3F D 135.v
2
B
/; (4.11b)
For block B, m
B
D 25 kg, m
B
g D 245:25 N. Applying Equation (4.6), .T
1
/
B
C .U
1!2
/
B
D
.T
2
/
B
; .T
1
/
B
D 0 since the cable system starts from rest.
e work done during the movement of 3 m is
.U
1!2
/
B
D .3F m
B
g/.y
B
/
) .U
1!2
/
B
D .3F 245:25 N/
x
A
3
;
using the constraint condition. erefore,
.3F 245:25 N/
3 m
3
D
1
2
.25 kg/.v
2
B
/ D 12:5.v
2
B
/: (4.11c)
Adding Equations (4.11c) and (4.11b), one has
150 D 147:5.v
2
B
/ ) v
B
D 1:0084 m/s:
(a) e velocity of block B is
*
v
B
D 1:0084 m/s " :
54 4. WORK AND ENERGY OF PARTICLES
(b) Substituting v
B
into Equation (4.11b), one obtains the tension in the cable
F D 85:9873 N:
at is,
*
F D 85:9873 N ".
Example 4.2
As shown in Figure 4.6a, a spring device consisting of one rigid panel connected by cables on
both sides to hold on the initial 50 mm compression of the spring. is spring device is used as
a crash-resistant wall to prevent the important structure (not shown) further down the incline
plane of 20
ı
to the horizontal line. If a 30 kg block having a velocity v D 2 m/s traveling 8 m
away from and toward the panel of the spring device, determine the additional deflection of the
spring device in bringing the block to rest. Assume that the kinetic coefficient of friction between
the block and the incline is 0.2 and the stiffness constant of the spring device is k D 20 kN/m.
Solution:
e given data are:
k D 20,000 N/m, m D 30 kg,
v D 2 m/s, D 20
ı
, initial compression of spring device,
x
0
D 50 mm,
k
D 0:2, and let the additional deflection of the spring device be x.
Since friction is included in the present problem and therefore, work energy principle,
Equation (4.6) is applied
T
1
C U
1!2
D T
2
(4.12)
in which U
1!2
represents the work, strain energy, and energy attenuated by friction at the two
stages of motion.
Kinetic energy at stage 1
T
1
D
1
2
mv
2
D
1
2
.30/.2
2
/ N.m D 60 J.
Kinetic energy at stage 2
T
2
D
1
2
mv
2
2
D
1
2
.30/.0
2
/ N.m D 0 since the block is brought to rest.
Work in two stages
U
1!2
D U
1
U
2
.U
1!2
/
f
;
where
U
1
D U
1g
C U
1e
D mgh
1
C
1
2
k
x
2
0
;
4.6. POWER AND MECHANICAL EFFICIENCY 55
30 kg block
⃑ = 2 m/s
⃑ = 2 m/s
Cable
(a)
(b)
8 m
x
20˚
20˚
20˚
8 m
1
2
Figure 4.6: (a) Spring device and traveling block, and (b) stages of motion of traveling block.
in which
U
1g
D mgh
1
D 30.9:81/
8 sin 20
ı
J; U
1e
D
1
2
.20;000/.0:05
2
/ J:
Substituting, U
1
D .805:254 C 25/ J.
U
2
D U
2g
C U
2e
D mgh
2
C
1
2
k.x C 0:05/
2
) U
2
D 30.9:81/.x sin 20
ı
/ C 10;000x
2
C 1;000x C 25 J.
e energy attenuated by friction is
.U
1!2
/
f
D
k
mg.x C 8/ cos 20
ı
56 4. WORK AND ENERGY OF PARTICLES
) .U
1!2
/
f
D 0:2.30/.9:81/.x C 8/ cos 20
ı
D 55:31.8 C x/ J.
erefore,
U
1!2
D
.
805:254 C 25
/
30
.
9:81
/
x sin 20
ı
C 10;000x
2
C 1;000x C 25
55:31
.
8 C x
/
J:
Substituting all terms into Equation (4.12) and simplifying, one obtains
10;000x
2
C 954:65x 422:72 D 0:
is quadratic equation gives
x D
954:65
p
954:65
2
C 4
.
10;000
/
.422:72/
20;000
) x D
954:65
p
17;820;176
20;000
D
954:65 4;221:395
20;000
D 0:1633
with the negative value being disregarded. erefore, the additional deflection is x D 0.1633 m.
Example 4.3
A collar of mass m D 5 kg is allowed to slide along the upper horizontal uniform rod, as shown
in Figure 4.7a. e collar is attached to spring whose other end is anchored at a fixed point R.
e spring has an undeformed length of 0.25 m and spring coefficient k D 75 N/m. If the collar
is released from rest at P determine
(a) the speed of the collar when it reaches Q and
(b) the speed of the collar when it reaches S.
Assume that friction between the collar and rod can be disregarded.
Solution:
e given data are, k D 75 N/m, m D 5 kg, undeformed length of spring `
o
D 0:25 m. Consider
the length of the deformed spring at different configurations. e deformed length PR is given
by
`
PR
D
q
.
0:5
/
2
C
.
0:4
/
2
C
.
0:3
/
2
m D 0:7071 m:
Similarly,
`
QR
D
q
.
0:4
/
2
C
.
0:3
/
2
m D 0:50 m;
`
RS
D
q
.
0:5
/
2
C
.
0:3
/
2
m D 0:58309 m:
..................Content has been hidden....................
You can't read the all page of ebook, please click here login for view all page.