Transcript 10/7 Potential Energy

10/7 Potential Energy
Text: Chapter 6 Energy
HW 10/7 “Potential Energy with Friction”
due Thursday 10/10 None due Wed
Potential Energy and Projectile Motion
Exam 2 Thursday, 10/17
5-7 Wit 116
6-8 Wit 114 (only if needed)
Please send email if other time needed
Potential Energy
Our energy ideas so far:
Consider each force in Fnet
W is special. KE lost to
the work done by gravity
as an object rises is
completely returned to KE
as it falls.
KEi + Fnetx = KEf
W
N
f
T
non-contact
contact
contact
contact
v=0
PEg KE
KE = - PE
We say “Mechanical Energy
is Conserved”
PEg KE
PEg KE
5m
KE
v= 0 m= 1kg
0
+
50
PE
=
E
50
PEg = mg h ( = WE,B h)
(g  10N/kg)
Now, release the ball.
The “Energy Bucket” Method
0
5m
KE
v= 0 m= 1kg
+
PE
=
0
50
50
10
40
50
20
30
50
E
If we pull W out of Fnet and include it in the “total
energy” then we can talk about “energy conservation.”
40
0
10
50
5m
KE
v= 0 m= 1kg
+
PE
=
0
50
50
10
40
50
20
30
50
Here Fothers = 0 so total energy is conserved!
KE = 40J = 1/2mv2
v = 8.9m/s
40
0
10
50
E
PEg,i KEi
37
0
PEg,f KEf
2.5m
The initial velocity is 4m/s. Find the velocity with which Carl
(70 kg) impacts the water.
Use buckets and pick PE = 0 at water.
PEg,i = mgh = 70(10)2.5 = 1750J
KEi = 1/2mvi2 = 1/2(70)42 = 560J
KEf = 2310J = 1/2mvf2
vf = 8.1m/s
Note: Compare to previous method!