Download Physics 130 - University of North Dakota

10/4 Potential Energy Intro
Text: Chapter 6 Energy
HW 10/4 “Potential Energy with Friction”
due Wednesday 10/9 (to be posted soon)
Potential Energy and Projectile Motion
Discuss “Block up a Ramp”
Discuss “2nd Law vs. Energy”
Exam 2 Thursday, 10/17
5-7 Wit 116
6-8 Wit 114 (only if needed)
Example: Energy in 2D
Energy
vi,y = 2m/s
m = 3kg
3.0m
6J
KEi
“Work”
90J
96J
KEf
vf = ?
WE,B =Fnet
= mg
Let’s see…
Find KEi (1/2mvi2)
Find “Work” (Fnet y)
Add to Kei to get KEf
Find vf from KEf
KEi = 1/2mvi2 = 6 Joules
Work = mgy = 90 Joules
KEf = 96J = 1/2mvf2
vf = 8.0m/s Oh Baby!
Energy
Always draw “Buckets”
For “Work” the net force must be constant in Fnetx
x is the “Displacement,” as usual
Always consider Energy as an alternative to the 2nd law
Energy is a “Scalar” not a “Vector” (See Ch. 1)
For Friday, think about this example in the case where there
is a horizontal component to the motion. (projectile motion)
Energy in 2D
Always draw “Buckets”
For “Work” the net force must be constant in Fnetx
x is the “Displacement,” as usual
Always consider Energy as an alternative to the 2nd law
Energy is a “Scalar” not a “Vector” (See Ch. 1)
What if Fnet and x point in different directions?
Example: Energy in 2D
Let’s just add a
Energy
horizontal
vi,y = 2m/s
Let’s see…
component to the
m = 3kg
velocity.
Find KEi (1/2mvi2)
vi,x = 6m/s Find “Work” (Fnet y)
The y-component
Add to Kei to get KEf
solution does not
change.
Find vf from KEf
3.0m
WE,B =Fnet KEi = 1/2mvi2 = 6 Joules
96J
= mg Work = mgy = 90 Joules
6J
“Work”
1/ mv 2
90J
KE
=
96J
=
f
2
f
v
=
6m/s
i,x
KEi
KEf
vf = 8.0m/s Oh Baby!
vf = 10m/s
vf,y = 8m/s
Example: Energy in 2D
Let’s just add a
Energy
horizontal
vi,y = 2m/s vi = 40 m/s
component to the
m = 3kg
Watch what happens
velocity.
when we ignore the
v
=
6m/s
i,x
The y-component
vectors!
solution does not
change.
1/ mv 2 = 60J
KE
=
i
2
i
3.0m
150J WE,B =Fnet Work = mgy = 90J ???
60J
= mg KEf = 150J = 1/2mvf2
“Work”
vf = 10m/s !!!!!
90J
vi,x = 6m/s
KEi
KEf
vf,y = 8m/s
vf = 10m/s
Potential Energy
Since this worked,
Energy
we choose to
vi,y = 2m/s vi = 40 m/s
define “Potential
m = 3kg
Watch what happens
Energy” as..
vi,x = 6m/s when we ignore the
vectors!
PEg = mgy
PEg = mgy
3.0m
96J
60J
KEi
“Work”
90J
WE,B =Fnet
= mg
KEf
vf,y = 8m/s
vi,x = 6m/s
KEi = 1/2mvi2 = 60J
Work = mgy = 90J ???
KEf = 150J = 1/2mvf2
vf = 10m/s !!!!!
vf = 10m/s
Watch Out for +- Signs!!!
PE in the previous case is negative, corresponding to a drop in
PE and a rise in KE.
So- a drop in PE will mean a rise in KE
and a rise in PE will mean a drop in KE.
KE = - PE
Here the minus signs mean gain or loss, not direction.