phys 218 sec. 517-520 review chap. 6 work and kinetic energy
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PHYS 218sec. 517-520
ReviewChap. 6
Work and Kinetic Energy
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Work
sr
Displacement vector
Fr
Constant forceStraight-line displacement
Force and displacement are in the same direction.
, : magnitude of , : magnitude of W Fs F s= F sr r
Unit of work
1 joule (1 Newton) (1 meter) or 1J 1N m= ´ = ×
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Work
Constant force in direction of straight-line displacement
W Fs=
Constant force, straight-line displacement
cosW Fs f= × =F sr r : force vector, : displacement vector,
: angle between the two vectorsf
F sr r
Varying x-component of force, straight-line displacement
2
1
x
xx
W F dx=ò
Work done on a curved path (general definition of work)
2 2 2
1 1 1
cosP P P
P P PW F dl Fdl df= = = ×ò ò ò FP
rrl
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Evaluation of Work
Ex 6.1
ˆ ˆ ˆ ˆConstant force (160 N) (40 N) , displacement (14m) (11m) .
Then (160 N) (14m) ( 40 N) (11m) 1800 J
F i j s i j
W F s
= - = +
= × = ´ + - ´ =
r r
r r
Ex 6.2
( )( )
displacement 20 m, total weight 14,700 N,
constant force exerted by the tractor 5000 N, (angle 36.9 )
friction force 3500 N
TF
f
f
= =
= =
=
o
TF
n
w
f f ( )
work done by the tractor
cos (5000 N)(20 m)cos36.9 80,000J 80 kJ
work done by the friction force
cos180 (3500 N)(0 m) 1 70 kJ
work done by the normal force work done by the weight 0
angle
T T
f
W F s
W fs
f= = = =
= = - =-
= =
=
o
o
Q( )90
total work 10 kJT f n wW W W W W= + + + =
o
rs
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Work done by a varying force, straight-line motion
m m
1x 2x
2xF
sr
1xF
1x 2x
( )xF x
1xF
2xF
1x 2x
1xF
2xF
axD
axF
bxD
bxF
2
1
To calculate the work,
divide the displacement into small segments , etc
is nearly constant for this small segment, and thena
x
ax a bx b xx
x
F
W F x F x F dx
D
= D + D + =òL
Work is the area under the curve between the initial and final positions in the graph of force as a function of position.
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Evaluation of Work: Stretched spring
xxF kx=
xF kx=-
Hooke's law: , : spring constantF kx k=-
Restoring force
2 2
0
Work done by the force which stretches the spring ( )
when the elongation goes from 0 to a maximum value is
1 1 (work done by the spring is )
2 2
Generally, from
X
kx
X
W kx dx kX kX
=
= = -ò
1 2
2 22 1
to ,
1 1
2 2
x x x x
W kx kx
= =
= -
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Varying x-component of force, straight-line displacement
2
1
x
xx
W F dx=ò
Evaluation of Work
( )2 2
1 12 1
x x
x x x xx x
W F dx F dx F x x F s= = = - =ò ò
When Fx is constant
2
1
In general, ,
so the work is obtained by calculating the above line integral,
which is defined on a given path.
P
PW d= ×ò F
rrl
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Kinetic energy and work-energy theorem
Kinetic energy (K)
21
2K mv=
Work-Energy Theorem
When v = 0, K = 0
The work done by the net force on a particle equals the change in the particle’s kinetic energy
2 1totW K K K= - =DThis is a very general theorem. This theorem is true
regardless of the nature of the force.
Kinetic energy at the initial point
Kinetic energy at the final point
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Proof of the work-energy theorem
Constant force in direction of straight-line displacement
m m
1x 2x
Fr
sr
1v 2v
2 22 2 2 12 1
2 22 22 12 1 2 1
For a constant force (acceleration ),
2 2
1 1
2 2 2
a
v vv v as a
s
v vW Fs mas m s mv mv K K
s
-= + Þ =
-Þ = = = = - = -
When a particle undergoes a displacement, it speeds up if W > 0, slows down if W < 0, and maintains the same speed if W = 0.
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Proof of the work-energy theorem
Varying force, straight-line motion
2 2 2 2
1 1 1 1
2 22 1
Start from the definition of acceleration,
Then,
1 1
2 2 The work-energy theorem is valid.
x x x xx x x
x x x vx
tot x x x x xx x x v
dv dv dv dvdxa v v
dt dx dt dx dx
dvW F dx ma dx mv dx mv dv
dx
mv mv
= = = =
= = = =
= -
\
ò ò ò ò
Eliminate time dependence since W is an integral over x.
1 1 2 2At , and at ,x x v v x x v v= = = =
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Ex 6.3 Same as Ex 6.2: suppose that the initial speed v1 is 2 m/s. What is the final speed?
Examples
2 1
We already know that 10kJ. To know the final speed, we use the work-energy theorem.
To compute kinetic energy, we need to know the mass. Since the total weight is 14,700 N,
14,700 N
tot
tot
W
W K K
wm
g
=
= -
= =
( )( )
2
221 1
2 1
2
22
1500 kg9.8 m/s
1 1Initial kinetic energy 1500 kg 2 m/s 3000 J
2 2Then ,
3000 J 10,000 J 13,000 J
Therefore,
2 2 13,000 J4.2 m/s
1500 kg
tot
K mv
K K W
K
Kv
m
=
= = =
= +
= + =
´= = =
Use Work-Energy Theorem to calculate speed.
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Forces on a hammerheadEx 6.4
hammerhead
I-beam Point 3
Point 2
Point 1
3 m7.4 cm
Falling hammerhead
y
v
60 Nf =
w mg=
1 212 12
1 2 22 1 2 2 1
2
Total work done on the hammerhead: ( ) ( ) 5700J.
1Work-Energy Theorem gives ( 0)
2Therefore,
2 2 (5700 J)7.55 m/s
200 kg
tot
tot
tot
W w f s mg f s
W K K K mv v
Wv
m
®
®
= - = - =
= - = = =
´= = =
Q
Speed of the hammerhead at point 2
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Forces on a hammerheadEx 6.4 (cont’d)
Hammerhead pushing I-beami.e., Point 2 Point 1y
60 Nf =
w mg=
n( )
( ) ( )
2 3 1 223 3 2
1 2
23
Total work done on the hammerhead:
5700 J1960 N 60 N 79,000 N
0.074 m
tot tot
tot
W w f n s K K W
Wn w f
s
® ®
®
= - - = - =-
Þ = - + = - + =
Work-energy theorem
Using the previous result
Be careful with the change of unit
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Motion with a varying forceEx 6.7
m
1v
Friction (k)
k 10.1 kg, 20 N/m, 1.5 m/s,
Find the maximum distance if 0 and if 0.47,
when the object is moving from 0 to the rightk k
m k v
d
x
m m
= = =
= =
=
When 0km=
y
x
n
mg
springF
frF
2
22 1 1
2 21 1
1Work done by the spring
21
Then by Work-Energy theorem; 2
1 1Therefore, 10.6 cm
2 2
W kd
W K K mv
mkd mv d v
k
=-
= - =-
- =- Þ = =
When 0km ¹
2 21
Work done by the friction force
Then by Work-Energy theorem;
1 10.086 m
2 2
fr k
k
W mgd
mgd kd mv d
m
m
=-
- - =- Þ =
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Ex 6.8 Motion on a curved path I
RRq
s Rq=
qdrl
y
x
Fr
q
T
F
mg
cosT q
sinT q
[ ]
0 0
0
0 0
0
0
The net force is zero ( remains in equilibrium)
sin 0, cos 0.
This leads to
tan
cos cos , and
Therefore,
tan cos ( ) sin
cos
(1 cos
x yF F T F T mg
F mg
F d Fdl F ds ds Rd
W mg Rd mgR d
mgR
mgR
q q
q
q q
q
q q q
q q q q q
q
q
= - = = - =
=
× = = =
= =
= -
= -
å å
ò ò
Q
rrl
)
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Ex 6.9 Motion on a curved path II (another way to compute the line integral)
In Ex 6.8, can be written as
ˆ ˆcos sin .
All forces acting on the object are
ˆ ˆ ˆ ˆ( sin ) cos , ( ) , .
Therefore,
( sin )( cos ) ( cos )( sin ) 0
sin
d
d ds i ds j
T T i T j w w j F F i
T d T ds T ds
w d w ds
F d
q q
q q
q q q q
q
= +
= - + = - =
× = - + =
× =-
×
r
r
r rr
rr
rr
rr
l
l
l
l
l
0
cos tan cos sin
sin (1 cos )w
F w
F ds w ds w ds
W w d w ds wR
W W
q q q q
q q
= = =
Þ
= × = - =- -
=-
ò òrrl
We get the same answer as in Ex 6.8
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Power
Average power
The time rate at which work is done.
av
WP
t
D=
D
(Instantaneous) power0
limt
W dWP
t dtD ®
D= =
D
Unit of powerwatt (W): 1W 1 J/s
horsepower (hp): 1 hp 746 W
=
=
Writing power in terms of F P F v= ×
r r