Transcript Physics 130 - University of North Dakota

10/3 Energy Intro
Text: Chapter 6 Energy
HW 10/3 “2nd Law vs. Energy” due Monday
10/7
Tomorrow, Potential Energy and Projectile
Motion
Exam 2 Thursday, 10/17
5-7 Wit 116
6-8 Wit 114 (only if needed)
Use 5kg and 30N in “Block on a Ramp” but
try it with all combinations of numbers!
Energy Introduction
The equation I never let you use:
vi2 + 2ax = vf2
Multiply by the mass of the object, m:
mvi2 + 2max = mvf2
Divide by 2:
1/
2 + max = 1/ mv 2
mv
2
i
2
f
Note that ma = Fnet :(2nd Law)
1/
2+F
1/ mv 2 Defines Kinetic Energy!
mv
x
=
2
i
net
2
f
Kinetic Energy and “Work”
1/
2+F
1/ mv 2 Defines Kinetic Energy!
mv
x
=
2
i
net
2
f
KEi + “Work” = KEf Shorthand for “Work Energy Relation”
This is not a vector equation, it’s a “Bucket” equation!
“Work”
KEi
KEf
Think of Energy as a “fluid”
Think of “Work” as the
“change in Energy” or the
amount of fluid I pour into (or
out of) a Bucket
Example: 2nd law vs. Energy
vi = 2m/s
m = 3kg
Find the final
velocity for a ball
moving up at 2m/s
when it is 3m below
3.0m
its starting point.
2nd Law
Let’s see….
find acceleration
find time to top
find y to top
WE,B =Fnet find total y
= mg find vf
Wish for an easier way!!!
vf = ?
Example: 2nd law vs. Energy
Benefits:
Don’t need
highest point!
(y = 3.0m)
Energy
vi = 2m/s
m = 3kg
No acceleration
calculation!
No vectors!
6J
KEi
“Work”
90J
3.0m
96J
KEf
vf = ?
Let’s see…
Find KEi (1/2mvi2)
Find “Work” (Fnet y)
Add to Kei to get KEf
Find vf from KEf
WE,B =Fnet
= mg
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)