Transcript Document

Conservation of energy
Contents:
•Definition Conservation of Energy
•Sample problem 1
•Sample problem 2
•Whiteboards
Conservation of Energy
Energy is neither created nor
destroyed, it just moves around.
The total energy of a closed system
remains constant.
Total Energy before = Total Energy After
(Pendulum of death)
TOC
Conservation of Energy
Total Energy before = Total Energy After
Comes from = Goes to
Assets = Expenditures
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
Speeds you up work
Slows you down work
TOC
Example 1
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
v = 4.5 m/s
250 kg
What is its velocity
at the bottom?
1.75 m
0 + (250 kg)(9.8 N/kg)(1.75 m) + 1/2(250 kg)(4.5 m/s)2 + 0 =
0 + 0 + 1/2(250 kg)v2 + 0
TOC
Example 2
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
v = 6.2 m/s
890 kg
What is its velocity
after the puddle?
3.6 m
(Puddle - Exerts 3200 N
of retarding force)
1/
2(890
kg)(6.2 m/s)2 = (3200 N)(3.6 m) + 1/2(890 kg)v2
TOC
Whiteboards:
Conservation of Energy
1|2|3|4|5|6|7
TOC
15 kg
vi = 0
What speed at the bottom?
h = 2.15 m
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
0 + mgh + 0 = 0 + 0 + 1/2mv2
(15 kg)(9.8 N/kg)(2.15 m) = 1/2(15 kg)v2
6.5 m/s
W
15 kg
vi = 5.8 m/s
What speed at the bottom?
h = 2.15 m
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
0 + mgh + 1/2mv2 = 0 + 0 + 1/2mv2
(15 kg)(9.8 N/kg)(2.15 m) + 1/2(15 kg)(5.8 m/s)2= 1/2(15 kg)v2
8.7 m/s
W
What final velocity?
vi = 4.6 m/s
350 kg
Pushes with 53 N for 35 m
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
Fd + 0 + 1/2mv2 = 0 + 0 + 1/2mv2
(53 N)(35 m) + 1/2(350 kg)(4.6 m/s)2 = 1/2(350 kg)v2
5.6 m/s
W
v = 3.68 m/s
2.34 kg
The hammer pushes in the
nail 3.50 mm. (.00350 m).
What force did it exert on
the nail
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
0 + 0 + 1/2mv2 = Fd + 0
1/ (2.34
2
4530 N
+
0
kg)(3.68 m/s)2 = F(.0035 m)
W
3.2 m
The 125 kg pile driver
drives the piling in .25 m
with one stroke. What
force does it exert?
(careful what you use as the
change in height)
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
0 + mgh + 0 = Fd + 0
+
0
(125 kg)(9.8 N/kg)(3.2 m + .25 m) = F(.25 m)
17,000 N
W
What speed at the top?
vi = 8.6 m/s
15 kg
h = 4.25 m
h = 2.15 m
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
0 + mgh + 1/2mv2 = 0 +mgh + 1/2mv2
(15 kg)(9.8 N/kg)(2.15 m) + 1/2(15 kg)(8.6 m/s)2 =(15 kg)(9.8 N/kg)(4.25 m) 1/2(15 kg)v2
5.7 m/s
W
15 kg
vi = 5.8 m/s
Brakes for 4.5 m. Final
speed after is 4.0 m/s.
What force brakes?
h = 2.15 m
Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2
0 + mgh + 1/2mv2
= Fd + 0
+ 1/2mv2
(15 kg)(9.8 N/kg)(2.15 m) + 1/2(15 kg)(5.8 m/s)2 = F(4.5 m) + 1/2(15 kg)(4.0 m/s)2
99.6 N
W