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Slide 1 - Dacula High by yaofenji

VIEWS: 18 PAGES: 30

									           Home Work
p120: 21
p672: 1
p672: 2
p672: 4
p120: 22
p120: 23
p120: 24
p120: 25
p120: 27
p120: #21) Calculate the work done
when a 20 N force pushes a cart 3.5 m?

 List
               W = Fd
 F = 20 N
               W = 20(3.5)
 d = 3.5 m     W = 70 J
 W=?
p672: #1) Calculate the work done as a
100 N force pushes a crate 5 m along a
factory floor?


 List
                 W = Fd
 F = 100 N
                 W = 100(5)
 d=5m            W = 500 J
 W=?
p672: #2) Calculate the work done by a
person who carries a 50 N box vertically
3 m up a ladder?


 List
                  W = Fd
 F = 50 N
                  W = 50(3)
 d=3m             W = 150 J
 W=?
p672: #4) Calculate the power
expended by the person who carries
the 50 N box vertically 3 m up a ladder
in 3 seconds?

 List
                  P = Fd / t
 F = 50 N
                  P = 50(3) / 3
 d=3m             P = 50 W
 t= 3s
 P=?
p120: #22) Calculate the work done by lifting
a 500 N barbell 2.2 m above the floor. What
is the potential energy of the barbell when it
is lifted to this height?

 List               W = Fd

 Fg = 500 N         W = 500(2.2)
                    W = 1100 J
 d = 2.2 m
 W=?               PE = mgh = 500(2.2)
 PE = ?            PE = 1100 J
p120: #23) Calculate the power expended
when the barbell is lifted 2.2 m in 2 seconds.

 List
 W = 1100 J         P=W/t
                    P = 1100 / 2
 t=2s
                    P = 550 W
 P=?
p120: #24)a.) Calculate the work needed to
lift a 90 N block of ice a vertical distance of
3m. What PE does it have?

 List                W = Fd
 Fg = 90 N           W = 90(3) = 270 J
 d=3m                PE = mgh = 90(3)

 W=?                 PE = 270 J

 PE = ?
p120: #24)b.) When the same block of ice is raised
the same vertical distance by pushing it up a 5 m
long inclined plane, only 54N of force is needed.
Calculate the work needed to push the block up the
plane. What PE does it have?
 List                W = Fd
 F = 54 N            W = 54(5) = 270 J
 d=5m                PE = 90(3) = 270 J

 W=?
 PE = ?
p120: #25) Calculate the change in potential
energy of 8 million kg of water dropping 50m
over Niagara Falls.

List                  PE = mgh
m = 8 000 000 kg      PE = 8 000 000(9.8)(50)
d = 50 m
PE = ?                PE = 3 920 000 000 J
                      PE = 3.92 x 109
p120: #26) If 8 million kg of water flows over
Niagara Falls each second, what is the power
available at the bottom of the falls?
List
                      P = W/t
m = 8 000 000 kg
                      W = PE
d = 50 m
                     P = 3.92 x 109/1
PE = 3.92 x 109      P = 3.92 x 109 W
t = 1s
P=?
p120: #27)a) Calculate the kinetic energy of
a 3kg toy cart that moves at 4 m/s.


List             KE = ½ mv2
m = 3 kg         KE = ½ (3)(42) = ½ (3)(16)
v = 4 m/s
KE = ?           KE = 24 J
p120: #27)b) Calculate the kinetic energy of
the same cart at twice the speed.


List             KE = ½ mv2
m = 3 kg         KE = ½ (3)(82) = ½ (3)(64)
v = 8 m/s
KE = ?           KE = 96 J
12. How much PE does Tim, m = 60 kg,
gain when he climbs 3.5 m?

List:          PE = mgh
m = 60 kg      PE = 60(9.8)3.5
h = 3.5 m      PE = 2058 J
PE = ?
13. A 6.4 kg BB is lifted 2.1 m. What is
the change in PE?

List:            PE = mgh
m = 6.4 kg       PE = 6.4(9.8)2.1
h = 2.1 m        PE = 131.7 J
PE = ?
14. mary weighs 500 N and walks down
5.5 m. What is the change in PE?

List:          PE = mgh
mg = 500 N     PE = 500(-5.5)
h = -5.5 m     PE = -2750 J
PE = ?
15. A WL lifts a 180 kg barbell a height
of 1.95 m. What is the change in PE?

List:            PE = mgh
m = 180 kg       PE = 180(9.8)1.95
h = 1.95 m       PE = 3440 J
PE = ?
18. What mass must be lifted a height of
1 m to do 1 J of work?

List:           PE = mgh
m=?             1 = m(9.8)1
h=1m            m = 1/9.8 = .102 kg
PE = 1 J
W=1J
           Kinetic Energy
Kinetic Energy (KE) is the energy of motion.

                KE = ½ mv2
Example: What is the KE of a 3 kg toy
truck moving at 2 m/s?


  List
  m = 3 kg
              KE = ½ mv2 = ½(3)(22) = 6 J
  v = 2 m/s
  KE = ?
What is the KE of a 3 kg toy truck
moving twice that fast, or 4 m/s?

  List
  m = 3 kg
              KE = ½ mv2 = ½(3)(42) = 24 J
  v = 4 m/s
  KE = ?
  In a frictionless system, the kinetic
energy of a moving object is equal to the
 work needed to bring it to that speed.

                W = KE

              Fd = ½ mv2
Example: How much work is needed to
increase the speed of a 85 kg bicycle
from 22 m/s to 46 m/s?
List
               W = ΔKE
m = 85 kg
               W = ½ m(vf2 – vi2)
vi = 22 m/s
               W = ½ (85)(462 – 222)
v2 = 46 m/s
              W = ½ (85)(2116 – 484)
W=?
              W = ½ (85)(1632) = 69360J
 Class work:
 p. 672, 673:
   5, 6, 7, 8
Show all work
5. Calculate the KE of a 10 kg cart
moving at 4 m/s.

List:           KE = ½ mv2
                KE = ½ 10 (42)
m = 10 kg
                KE = 80 J
v = 4 m/s
KE = ?
6. Calculate the change in KE of a 10 kg
cart speeding up from 4 m/s to 10 m/s.

List:           KE = ½ mv2
                KE = ½ 10 (102 - 42)
m = 10 kg
                KE = 5(100 – 16)
vi = 4 m/s
                KE = 5(84) = 420 J
vf = 10 m/s
KE = ?        7. So it takes 420 J of work
              to change the KE by 420 J!
8. How much work is done to lift a 3 kg
crate up 2 m and throw it at 4 m/s?
List:       PE = mgh     KE = ½ mv2
m = 3 kg    PE = 3(9.8)2 KE = ½ 3 (42)

v = 4 m/s   KE = 58.8 J   KE = 24 J

h=2m
PE = ?      total energy = 82.8 J

KE = ?

								
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