Transcript 投影片 1 - National Cheng Kung University

6. Work, Energy & Power
1.
2.
3.
4.
Work
Forces that Vary
Kinetic Energy
Power
Does the work of climbing a mountain
depend on the route chosen?
No
Direct application of Newton’s law can be infeasible.
What’s the speed of
these skiers at the
bottom of the slope?
simple
complicated
Energy conservation to
the rescue (Chap 7).
6.1. Work
Work W done on an object by a constant force F is
W  F rF
rF = displacement along direction of F.
 Fr r
Fr = force along direction of r .
Note: F need not be a net force.
W   Joule  1 N  m

W  F cos x
 F  x F
Example 6.1. Pushing a Car
The man pushes with a force of 650 N, moving the car 4.3 m.
How much work he does?
W  F x   650 N   4.3 m  2.8 kJ
Example 6.2. Pulling a Suitcase
Woman exerts 60 N force on suitcase, pulling at 35 angle to the horizontal.
How much work is done if the suitcase is moved 45 m on a level floor?
W  Fx x  F cos  x
  60 N   cos 35  45 m
 2.2 kJ
 F xF
xF = x cos
x
 F x cos 
Work & the Scalar Product
Scalar = quantity specified by a single number
that is the same in every coordinate system.
Scalar has no direction.
B
Scalar (dot) product of vectors A & B :
A  B  A B cos 

BA = B cos
A
A  B  A B cos  A BA
 AB B
is a scalar
 Ax Bx  Ay By
2-D
 Ax Bx  Ay By  Az Bz
3-D
W  F r cos 
W  F  r
 = angle between F & r
Work is a scalar.
Example 6.3. Tugboat
Tug boat pushes a cruiser with force F = ( 1.2, 2.3 ) MN,
displacing the ship by r = ( 380, 460 ) m.
(a) Find the work done by the tugboat.
(b) Find the angle between F & r.
W  F  r  Fx x  Fy y
 1.2 MN  380 m   2.3 MN   460 m
 1510 MJ
 F r cos
F
Fx2  Fy2 
r 
 x    y 
  cos 1
2
W
F r
1.2 MN    2.3 MN 
2
2
 cos1

 2.59 MN
2
380 m   460 m
2
1510 MJ
 2.59 MN   597 m 
2
 12 
 597 m
6.2. Forces that Vary
N
N
i 1
i 1
W   Wi   F  X i  x
W  lim
N 
N
 F  X  x
i 1
W   F  x d x
x2
x1
i
x2  x1
N
 1
X i  x1   i   x
 2
1
X 1  x1  x
2
1 x x

X N  x1   N   2 1
2 N

1
 x2  x
2
x 
Tactics 6.1. Integrating
inverse
Derivative  Indefinite Integral  Antiderivative 
dg
 f
dx

g   xn d x
Example:
d xn
 n x n 1
dx
Since

x2
x1
x dx  g
n
x2
x1

g   f dx
dg
 xn
dx
we have
 g  x2   g  x1 
x n 1
g
n 1
x2n 1 x1n 1


n 1 n 1
See Appendix A for integral table.
x n 1
 x dx  n  1
n
Stretching a Spring
W   F  x d x   k x d x
x
x
0
0
1
 k x2
2

1 2
kx
2
x
x n 1
 x dx  n  1
n
0
Example 6.4. Bungee Jumping
Bungee cord is 20 m long with k = 11 N/m.
At lowest point, cord length is doubled.
(a)
How much work is done on cord?
(b)
How does work done in the last meter compare with that done in the 1st meter?
W12 
W
(a)
1
k  x22  x12 
2
1
2
2
11 N / m   20 m    0 m  
2
 2.2 kJ
(b) 1st meter
W
1
2
2
11 N / m  1 m    0 m    5.5 J
2
Last meter
W
1
2
2
11 N / m   20 m   19 m    214 J
2
Example 6.5. Rough Sliding
Workers pushing a 180 kg trunk across a level floor encounter a 10 m region where floor
becomes increasingly rough.
There, k = 0 + a x2, with 0 = 0.17, a = 0.0062 m2 & x is the distance into the rough part.
How much work does it take to push the trunk across the region?
x n 1
 x dx  n  1
F  x   k m g
n
W   F  x d x
x2
x1
W 
x2
x1

0
 ax2 
1

3
m g d x  m g  0 x  a x 
3


x2
x1

1
1
 

 m g  0 x2  a x23    0 x1  a x13  
3
3
 


1
3

 180 kg   9.8 m / s 2   0.17 10 m    0.0062 m2  10 m  
3


 6.6 kJ
Force & Work in 2- & 3- D
W   F r   d r
r2
Line integral
r1
x2
  F dx
1-D
x1

r2

r2
r1
r1
F
d x  Fy d y 
F
d x  Fy d y  Fz d z 
x
x
2-D
3-D
Work Done Against Gravity
W  m g  y
Only vertical displacement
requires work against gravity
W=mgh
GOT IT? 6.2.
3 forces have magnitudes in N that are numerically equal to
(a) x,
(b) x2,
(c) x,
where x is the position in meters.
Each force moves an object from x = 0 to x = 1 m.
Note that each force has the same values at the end points, namely, 0 N & 1 N.
Which force does the most work?
Which does the least?
(b)
(c)
6.3. Kinetic Energy
Wnet   Fnet d x
Wnet  
v2
v1
 m
dv
dx
d x   m dv
dt
dt
m v dv  1 m v 2
2
v2

v1
  m v dv 
1
1
m v22  m v12
2
2
Kinetic energy:
K
1
m v2
2
K is relative (depends on reference frame).
K is a scalar.
Work-energy theorem:
K  Wnet
1
m v2
2
Example 6.6. Passing Zone
A1400 kg car enters a passing zone & accelerates from 70 to 95 km/h.
(a)
How much work is done on the car?
(b)
If the car then brakes to stop, how much work is done on it?
Wnet  K 
(a)
Wnet 
1
1
m v22  m v12
2
2
1
2
2
2
1400 kg   95 km / h    70 km / h    2,887,500 kg  km / h 
2
2
 1000 m 
 2887500 
 kg
3600
s


b)
Wnet 
 222.8 kJ
1
2
2
2
1400 kg   0    95 km / h    6,317,500 kg  km / h 
2
2
 1000 m 
 6,317,500 
 kg
3600
s


 487.5 kJ
GOT IT? 6.3.
For each situation, tell whether the net work done on a soccer ball is
positive,
negative,
or zero.
Justify your answer using the work-energy theorem.
(a) You carry the ball to the field, walking at constant speed.
zero (K=0)
(b) You kick the stationary ball, starting it flying through the air.
positive (K>0)
(c)
negative (K<0)
The ball rolls along the filed, gradually coming to a halt.
Energy Units
[energy] = [work]
CGS:
= J (SI)
1 erg  1 g cm / s 2
 107 J
Other energy units:
eV (electron-volt):
used in nuclear, atomic, molecular, solid state physics.
cal (calorie), BTU (British Thermal Unit): used in thermodynamics.
kW-h (kilowatt-hours): used in engineering.
See Appendix C
6.4. Power
Average power:
(Instantaneous) power:
P
W
t
 power   watt
W  J / s
W
dW

t  0  t
dt
P  lim
Example 6.7. Climbing Mount Washington
A 55 kg hiker makes the vertical rise of 1300 m in 2 h.
A 1500 kg car takes ½ h to go there.
Neglecting loss to friction, what is the average power output for each.
mgh
P
t
Hiker:
Car:
 55 kg   9.8 m / s 2  1300 m 
P
 2h  3600 s / h 
1500 kg   9.8 m / s 2  1300 m 
P
 0.5 h  3600 s / h 
 97 W
 11 kW
P
W
t

W  lim
t  0
W  P t
for constant power P
 P t   P dt
t2
general case
t1
Example 6.8. Yankee Stadium
Each of the 500 floodlights at Yankee stadium uses 1.0 kW power.
How much do they cost for a 4 h night game, if electricity costs 9.5 ₵ / kW-h ?
W  P t
 500  1.0 kW   4h$ 0.095 / kW  h  $190
Energy and Society
2008: 4.71020 J or 1.51013 W
Power & Velocity
P
dW  F  d r
dW
dr
 F
dt
dt
P  Fv
Example 6.9. Bicycling
Fair
v
Riding a 14 kg bicycle at a steady 18 km/h (5.0 m/s),
you experience a 30 N force from air resistance.
If you mass 68 kg, what power must you supply
Fair
(a) on level ground.
v
(b) up a 5 slope.
(a)
(b)
Fg
P  Fair v  30N 5.0 m / s   150 W
P    Fair  Fg  v
  Fair  m g sin   v
 30 N  14 kg  68 kg   9.8 m / s 2   sin 5   50 m / s 
 500 W