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Different energies: • Kinetic/translation
Each energy is associated with a
- Gravitational potential• Heat energy• Electromagnetic energy
Each energy is associated with a “scalar”
which defines a state of a
system at a given time.
Kinetic EnergyKinetic Energy
- Electromagnetic energy• Strain or elastic energy Kinetic Energy
is associated with the state of motion
Kinetic EnergyKinetic Energy
2 1 2
KE
mv =^
Units of Joules: 1 J = kg·m
2 /s
2
v ) not
- KE doesn’t depend on which way something is moving
if i ’
h^
i^
di^
i
G v^
( here
2 v
G = v
G • v
=^
G^2 v^
or even if it’s changing direction
- KE is ALWAYS a positive scalar
How much is “Kinetic Energy”
Kinetic Energy
is associated with the state of motion
- KE depends on speed not• KE doesn’t depend on which way something is moving
G v^
( here v
2 =^
G^2 v^
)
- KE doesn t depend on which way something is moving
or even if it’s changing direction
1)^
Electron (e-) moving in Copper
me
= 9.
×^10
-^
kg and v ~ 1
×^10
6 m/s KE = 7
×^10
-^
J^
( ~ 4 eV)
2)^
Bullet traveling at 950 m/s (3100 ft/s). )^
g^
(^
)
m = 4.2 g and v ~ 950 m/s (3100 ft/s)
KE = 2000 J
3)^
Football Linebacker
m = 240 lbs and v ~ 18 mph (7 m/s)
KE = 2800 J
p^
(^
)
4)^
Aircraft Carrier Nimitz
m = 91,400 tons and v ~ 1 knot
KE = 10 MJ
How to find an
“alternate form” of Newton’s 2
nd^ Law that relates position and velocity.??
Work and Kinetic EnergyWork and Kinetic Energy^ How
to find an
alternate form
of Newton s 2
Law that relates position and velocity.??
Start in 1-D (e.g. Bead along wire
), we know …ˆ x
Fx
, net
= ma
= x^
m
dv dt
Fx , net (^
)^ ⎛ dx ⎜^ ⎝ dt
⎞ ⎟ =⎠
m
dv dt ⎛ ⎜^ ⎝
⎞ ⎟^ ⎠ ⎛ dx ⎜^ ⎝ dt
⎞ ⎟ =⎠
mv
dv dt
dx
d^
1 mv
2 (^
)
, (^
)^ ⎝^ dt
⎠^
dt ⎝^
⎠^ ⎝^ dt
⎠^
dt^ F
dx ∫^
=^
d^
1 mv
2 (^
)
∫^
=^ KE
KE
Fx , net
dx dt
=
d ^2
mv (^
) dt
Fx , net
dx ∫^
=^
d^
mv ( 2
)
∫^
=^ KE
− 2 KE
1
W
net^
≡^
Fx , net
dx ∫^
=^ KE
− 2 KE
= 1 Δ
KE
Fx , net
dx
= d^
1 mv^2
2 (^
)
W
net^
=^ Δ
KE
WorkWork-
-Kinetic Energy TheoremKinetic Energy Theorem
Work-Kinetic
change in the kinetic
net work done on
(^
)^
(^
Energy Theorem
change in the kineticenergy of an object
net work done on
the particle
(^
If^
is not a function of
, then
F^ net
x
G^
G
W
net
G F
net
G d
KE
No work is done on an object by a force unless there is aNo work is done on an object by a force unless there is aNo
work is done on an object by a force unless there is a
No work is done on an object by a force unless there is acomponent of the force along the objects line of motion.component of the force along the objects line of motion.
Positive and Negative WorkPositive
and Negative Work
Weight lifting: apply a FORCE up and DISPLACE thebarbell up…
both the force and displacement are in the +ydirection so work is
positive
On the downward motion the FORCE is still up and
the force is in the +y but the displacement is in –y direction so work is
negative
External Force acts on boxmoving rightward a distance dmoving rightward a distance d.Rank: Work done on box by F
QuestionQuestion
Two forces act on the box shown in the drawing, causing it tomove across the floor. The two force vectors are drawn toscale. Which force does more work on the box?scale. Which force does more work on the box?
F
1. F
1
2.^
F^2
3. They’re both zero (F
=F 1
4. They’re the same, but not zero
(F
=F 1
ExampleExample
Question 7-4:
The figure shows the values of a force
F ,
directed along an
x^ axis,
Q^
g^
,^
g^
,
that will act on a particle at the corresponding values of x. If the particlebegins at rest a x=0, what is the particle’s coordinate when it has (a)
the greatest speed (a)
the greatest speed
net^
net^
f^
i
W^
F^
dx^
KE
KE
KE
≡^
-^
= Δ
=^
−
G ∫
G^
Area under F-x plot is W.
(b) the minimum speed
Question:
The figure shows the values of a
force
F , directed along an
x^ axis, that
will act on a particle at the
pos. work -> KE increase
corresponding values of x. If theparticle begins at rest a x=0, what isthe particle’s coordinate when it has
Greatest KE here
(a)
the greatest speed (b) the minimum speed
neg work -> KE decrease
zero KE here
neg. work > KE decrease
Greatest speed at x=3 m
Minimum speed (zero) at x=6 m
Work done by Constant Gravitational Force
∫
Special Case:Special Case:
Wg
G Fg^
-^ d
G x
∫^
=^
G F • g^
G d
where
Fg^
=^
mg (^
−) ˆ j ( )
and
g
=^
m^ /
2 s
If an object is displaced upward (
Δ^ y positive), then the work done
by the
gravitational force
on the object is negative.
If an object is displaced downward (
Δy negative), then the work
done by the
gravitational force
on the object is positive.
If the only force acting on an object is Gravitational Force then What is the change in KE due to Gravitational Force
If^ the only force acting on an object is Gravitational Force then,
Wnet
=^
Wg
G Fg
-^
G d
(^
= Δ)
KE
=^
KE
− f^
KE
i
(^
)
If an object is displaced upward (
Δ^ y positive), the change in
Kinetic Energy is negative (it slows down). If an object is displaced downward (
Δy negative), the change in
Kinetic Energy is positive (it speeds up).
What work is needed to lift or lower an object?
In order to “lift” an object, we must apply an externalforce to counteract the gravitational force.
W
net^
≡^
W
ext^
=^
Δ KE
If^
(i.e.
),^
then
Δ KE
=
0
v^ f^
=^
vi^
W
= − g
W
ext
If an object is displaced upward (
Δ^ y positive), then the
work done by the
External force
on the object is positive
work done by the
External
force
on
the object is positive.
If an object is displaced downward (
Δy negative), then the
work done by the
External force
on the object is negative
work done by the
External
force
on
the object is negative.
Work due to Friction:
Special Case:Special Case:^ WORK due to friction is ALWAYS NEGATIVE
- Energy is transferred OUT - Kinetic energy decreases or
ΔKE < 0 (slow down)
Where did the energy go? THERMAL/Sound
gy g
