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PHYS 218
sec. 517-520
Review
Chap. 6
Work and Kinetic Energy
Constant force
Straight-line displacement
Force and displacement are in the same direction.
Work
r
F
r
s
Displacement vector
W = Fs,
r
r
F : magnitude of F, s : magnitude of s
Unit of work
1 joule = (1 Newton)´ (1 meter)
or 1J = 1N ×m
Work
Constant force in direction of straight-line displacement
W = Fs
Constant force, straight-line displacement
r r
W = F ×s = Fs cos f
r
r
F : force vector,
s: displacement vector,
f : angle between the two vectors
Varying x-component of force, straight-line displacement
W=
ò
x2
x1
Fx dx
Work done on a curved path (general definition of work)
W=
ò
P2
P1
F cos f dl =
ò
P2
P1
FPdl = ò
P2
P1
r r
F ×dl
Evaluation of Work
Ex 6.1
r
r
Constant force F = (160 N) iˆ - (40 N) ˆj, displacement s = (14 m) iˆ + (11m) ˆj.
r r
Then W = F ×s = (160 N)´ (14 m) + (- 40 N)´ (11m) = 1800 J
Ex 6.2
displacement = 20 m, total weight = 14, 700 N,
constant force exerted by the tractor (FT ) = 5000 N, (angle f = 36.9o )
friction force ( f ) = 3500 N
work done by the tractor
n
WT = FT s cos f = (5000 N)(20 m) cos 36.9o = 80, 000 J = 80 kJ
FT
f
f
w
work done by the friction force
W f = fs cos180o = (3500 N)(0 m) (- 1) = - 70 kJ
r
s
work done by the normal force = work done by the weight = 0
(Q angle = 90o )
total work W = WT + W f + Wn + Ww = 10 kJ
Fx ( x)
Work done by a varying force, straight-line motion
F2x
F1x
m
m
r
s
x1
F2x
x2
F1x
x2
x1
F2x
To calculate the work,
Fbx
Fax
divide the displacement into small segments D xa , etc
F is nearly constant for this small segment, and then
F1x
x1
x2
W = FaxD xa + FbxD xb + L =
ò
x2
x1
Fx dx
D xa D xb
Work is the area under the curve between the initial and final positions in the
graph of force as a function of position.
Evaluation of Work: Stretched spring
Hooke's law: F = - kx, k : spring constant
x
Fx = kx
Restoring force
Fx = - kx
Work done by the force which stretches the spring ( = kx)
when the elongation goes from 0 to a maximum value X is
W=
ò
0
X
kx dx =
1 2
kX
2
Generally, from x = x1 to x = x2 ,
W=
1 2 1 2
kx2 - kx1
2
2
(work done by the spring is -
1 2
kX )
2
Evaluation of Work
W=
ò
x2
x1
Fx dx
W=
x1
Varying x-component of force,
straight-line displacement
In general, W =
ò
x2
ò
P2
x2
Fx dx = Fx ò dx = Fx (x2 - x1 ) = Fx s
x1
When Fx is constant
r r
F ×dl ,
P1
so the work is obtained by calculating the above line integral,
which is defined on a given path.
Kinetic energy and work-energy theorem
Kinetic energy (K)
K=
1 2
mv
2
When v = 0, K = 0
Work-Energy Theorem
The work done by the net force on a particle equals the change in the
particle’s kinetic energy
Wtot = K2 - K1 = D K
This 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
Proof of the work-energy theorem
Constant force in direction of straight-line displacement
v1
v2
m
r
F
m
x1
r
s
x2
For a constant force (acceleration a),
v22 - v12
v = v + 2as Þ a =
2s
v22 - v12
1
1
Þ
W = Fs = mas = m
s = mv22 - mv12 = K 2 - K1
2s
2
2
2
2
2
1
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.
Proof of the work-energy theorem
Varying force, straight-line motion
Start from the definition of acceleration,
dv
dv dx dvx
dv
ax = x = x
=
vx = vx x
dt
dx dt
dx
dx
Then,
Wtot =
ò
x2
x1
Fx dx =
ò
x2
x1
max dx =
ò
x2
x1
dv
mvx x dx =
dx
1 2 1 2
mv2 - mv1
2
2
The work-energy theorem is valid.
Eliminate time dependence since W is
an integral over x.
ò
v2
v1
mvx dvx
=
\
At x = x1, v = v1 and at x = x2 , v = v2
Examples
Ex 6.3
Same as Ex 6.2: suppose that the initial speed v1 is 2 m/s. What is the
final speed?
We already know that Wtot = 10 kJ. To know the final speed, we use the work-energy theorem.
Wtot = K 2 - K1
To compute kinetic energy, we need to know the mass. Since the total weight is 14,700 N,
w 14, 700 N
=
= 1500 kg
g
9.8 m/s 2
1
1
2
Initial kinetic energy K1 = mv12 = (1500 kg )(2 m/s) = 3000 J
2
2
Then K 2 = K1 + Wtot ,
m=
K 2 = 3000 J + 10, 000 J = 13, 000 J
Therefore,
v2 =
2K2
=
m
2´ 13, 000 J
= 4.2 m/s
1500 kg
Use Work-Energy Theorem to calculate speed.
Ex 6.4
Forces on a hammerhead
Falling hammerhead
y
hammerhead
Point 1
f = 60 N
3m
Point 2 7.4 cm
I-beam
w = mg
v
Point 3
Total work done on the hammerhead: Wtot1® 2 = ( w - f )s12 = (mg - f ) s12 = 5700 J.
Work-Energy Theorem gives Wtot1® 2 = K 2 - K1 = K 2 =
1 2
mv2
2
( Q v1 = 0)
Therefore,
v2 =
2Wtot
=
m
2´ (5700 J)
= 7.55 m/s
200 kg
Speed of the hammerhead at point 2
Ex 6.4 (cont’d)
Forces on a hammerhead
Hammerhead pushing I-beam
i.e., Point 2 g Point 1
Using the previous result
y
n
Total work done on the hammerhead:
Wtot2® 3 = (w - f - n)s23 = K 3 - K 2 = - Wtot1® 2
Work-energy theorem
f = 60 N
Þ
w = mg
Wtot1® 2
5700 J
n = w- f +
= (1960 N)- (60 N )+
= 79, 000 N
s23
0.074 m
Be careful with the change of unit
Ex 6.7
Motion with a varying force
v1
k
m
m = 0.1 kg,
k = 20 N/m, v1 = 1.5 m/s,
Find the maximum distance d if mk = 0 and if mk = 0.47,
when the object is moving from x = 0 to the right
When mk = 0
Friction (mk)
1 2
kd
2
Then by Work-Energy theorem; W = K 2 - K1 = -
y
Therefore, -
n
1 2
mv1
2
1 2
1
m
kd = - mv12 Þ d = v1
= 10.6 cm
2
2
k
When mk ¹ 0
Fspring
Ffr
Work done by the spring W = -
x
Work done by the friction force W fr = - mk mgd
mg
Then by Work-Energy theorem;
1
1
- mk mgd - kd 2 = - mv12 Þ d = 0.086 m
2
2
Ex 6.8
Motion on a curved path I
The net force is zero ( Q remains in equilibrium)
q
R
R
å
r
dl
r
F
Fx = F - T sin q = 0,
s = Rq
y
W=
ò
q0
mg tan q cos q( Rd q) =
0
T
q
= mgR (1- cos q0 )
T sin q
mg
x
and ds = Rd q
ò
q0
0
= mgR [- cos q ]00
T cos q
F
Fy = T cos q - mg = 0.
This leads to
F = mg tan q
r r
F ×dl = Fdl cos q = F cos qds,
Therefore,
q
q
å
mgR sin qd q
Ex 6.9
Motion on a curved path II (another way to compute the line integral)
r
In Ex 6.8, dl can be written as
r
dl = ds cos q iˆ + ds sin q ˆj.
All forces acting on the object are
r
r
r
ˆ
ˆ
ˆ
T = (- T sin q) i + T cos q j , w = (- w) j , F = F iˆ.
Therefore,
r r
T ×dl = (- T sin q)(ds cos q) + (T cos q)(ds sin q) = 0
r r
w ×dl = - w sin qds
r r
F ×dl = F cos qds = w tan q cos qds = w sin qds
Þ
r r
Ww = ò w ×dl = ò - w sin qds = - wR(1- cos q0 )
WF = - Ww
We get the same answer as in Ex 6.8
Power
The time rate at which work is done.
Average power
Pav =
(Instantaneous) power
Unit of power
Writing power in terms
of F
DW
Dt
D W dW
=
D t® 0 D t
dt
P = lim
watt (W): 1W = 1 J/s
horsepower (hp): 1 hp = 746 W
r r
P = F ×v