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2AP Physics C
What is Energy?
"It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount." -Richard Feynman "Lectures on Physics"
4AP Physics C
Matter and Energy
The combination of matter and energy makes up the universe. Matter is substance and energy is the mover of substance.
-Paul Hewitt
5AP Physics C
Definition
Energy is an abstract quantity with the ability to effect physical change in matter.
Energy is the substance from which all things in the Universe are made up.
6AP Physics C
Forms of Energy
Two broad categories of energy:
Potential: Stored Energy
Chemical – Stored in chemical bonds
Mechanical – Stored in objects by tension
Nuclear – Stored in nucleus of atoms
Gravitational – Stored based on height and weight
Electrical – Stored in batteries
Kinetic: Energy due to motion
Radiant – electromagnetic
Thermal - Heat
Motion – Moving objects
Sound – motion of air and other media by sound waves
7AP Physics C
Conservation of Energy
Energy Conservation during freefall: 2 2 2f i f iv v g y y
2 2 2 2f i f iv v gy gy
2 22 2f f i iv gy v gy
2 2 2 21 12 22 2
2 f f i i f f i i
mv gy v gy mv mgy mv mgy
222
2 2
ms
mmv units kg kg m Nm
s
m mmgy units kg m kg m Nm
s s
AP Physics C 8
Calculus Approach
Lets look at the freefall of a mass, m, from N2L and calculus:
dvF ma m mg
dt
y
dv dy dv dvv
dt dt dy dy
y
dvmv mg
dy
ymv dv mgdy
f f
i i
v y
yv ym v dv mg dy
2 2 212
1 1
2 2f f
ii
v y
f f i f iyvmv mv mv mgy mgy mfy
2 21 1
2 2f f i imv mgy mv mgy
AP Physics C 11
Gravitational Potential Energy
Similar concept to kinetic energy except that the orientation is vertical and the force is gravity. Think KE which has not been created.
M
d
mg
13AP Physics C
Problem 1
A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he releases it. Use energy to find:
a. The ball’s maximum height above the ground.
b. The ball’s speed as it passes the window on its way down.
c. The speed of impact on the ground.
Analysis: Use Kf + Uf = Ki + Ui to solve for y1, v2 and v3.
2 20 0 1 1
1 1
2 2mv mgy mv mgy
a. At y1, v1 = 0; at y0, v0 = 10 and y0 = 0
21
1
110 0 0
2100
5.12 9.8
m mgy
y m
AP Physics C 14
Problem 1 con’t
2 20 2
1 10 0
2 2v g v g
b. At y0, v0 = 10; at y2=0, find v2
2 20 2v v
2 0 10v v
Speed up = speed down.c. At y0 = 0, v0 = 10; At y3 = -20, find v3.
2 23
1 110 0 20
2 2g v g
23
12 10 20 9.8 22.2
2m
sv
AP Physics C 15
Energy Bar Charts
Bar chart modeling a rock thrown upward and returning to same elevation:
AP Physics C 16
Zero of Potential Energy
f f
i i
U y
U yU mg dy
Our earlier calculus based derivation of The following expression represented the change in potential energy:
2 21 1
2 2f f i imv mgy mv mgy
The initial position was considered to be zero and resulted in U = mgyf
U will vary based on where the zero potential energy level is placed.
ΔU will always be the same regardless of the location of the zero level.
17AP Physics C
Non-Freefall Ug
Horizontal surface: no change in gravitational potential energy
Slanted/undulating surface: define s-axis parallel to surface of movement.
snet ss
dvF ma m
dt
s s snet ss
dv dv dvdsF m m mv
dt ds dt ds
sin ss
dvmg mv
ds
sin s smg ds mv dv
s smgdy mv dv
2 21 1
2 2f f i imv mgy mv mgy
18AP Physics C
Ballistic Pedulum
A 10 g bullet is fired into a 1200 g wood block hanging from a 150 cm long string. The bullet embeds itself into the block and the block swings out to an angle of 40°. What was the speed of the bullet?
Analysis: Two part problem: The impact of the bullet with the block is inelastic. Momentum is conserved. After the collision the block swings as a pendulum. The sum of the kinetic and potential energies before and after do not change as the block swings to its largest angle.
AP Physics C 19
Ballistic Pendulum con’t
1 0 0( )w B x w x B xw Bm m v m v m v
0 1W B
x xBB
m mv v
m
Collision/Momentum calculations:
If we can calculate v1x from swing/energy relationships, we can calculate the speed of the bullet.
2
2 2
2
1 1
1
2
1
2
W B W B
W B W B
m m v m m gy
m m v m m gy
2
2 1 1
1 1
2 20 gy v gy
V2 = 0 and dividing by (mW + mB) gives:
or 1 22v gy
1 2 9.8 0.351 2.62 msv
20AP Physics C
Restoring Forces
Restoring ForceElasticEquilibrium Length, L0
Displacement, Δs Δs = L - L0
Fsp = k ΔsSpring constant, k
AP Physics C 22
Hooke’s Law Problem
You need to make a spring scale for measuring mass. You want each 1.0 cm length along the scale to correspond to a mass difference of 100 g. What should be the value of the spring constant?
/ (0.100 kg)(9.8 N/m)/(0.010 m) 98 N/m k mg x
sp . F k x mg
AP Physics C 23
Elastic Potential Energy
Is the force applied to the ball constant?
Describe the mechanical energy in both situations; before and after.
AP Physics C 24
Elastic Potential Energy con’t
N2L for the ball is:
net ss
dvF ma m
dt
By Hooke’s law, (Fnet)s = -k(s – se), substituting gives:
e
dvm k s s
dt
Using the chain rule:
s s ss
dv dv dvdsv
dt ds dt ds
Substituting gives: s s emv dv k s s ds
Integrating from initial to final conditions:
2 21 1( )
2 2s f
i i
v s
s f i ev smv ds mv mv k s s ds
AP Physics C 25
Elastic Potential Energy con’t
2 21 1( )
2 2s f
i i
v s
s f i ev smv ds mv mv k s s ds
From previous slide:
2 21 1( )
2 2f
i
s
e f isk s s ds k s k s
Substituting and rewriting gives:
2 22 21 1 1 1
2 2 2 2f f i imv k s mv k s
½ mv2 is obviously the kinetic energy.
21
2k s is the elastic potential energy, Us
AP Physics C 26
Elastic Potential Energy Problem
How far must you stretch a spring with k = 1000 N/m to store 200 J of energy?
21s 2 ( ) U k s
s2 / 2(200 J) / 1000 N/m 0.632 m s U k
Elastic potential energy is defined as:
Solving for ∆s:
AP Physics C 27
Elastic Collisions
Perfectly Elastic Collision: A collision in which mechanical energy is conserved.
Must conserve momentum and mechanical energy
Not possible where friction is involved
If only one object, m1, initially moving and all motion is along a line:
fx fx ixm v m v m v 1 2 11 2 1
fx fx ixm v m v m v 2 2 2
1 2 11 2 1
1 1 1
2 2 2
Solving the first equation for (vfx)1 and substituting into the second gives:
ix fx fx ix
mm v v m v m v
m
2 1
22 22
1 2 11 21
AP Physics C 28
Elastic Collisions con’t
ix fx fx ix
mm v v m v m v
m
2 1
22 22
1 2 11 21
Squaring and simplifying gives:
fx fx ix
mv v v
m
2
2 2 11
1 2 0
There are two solutions, (vfx)2 = 0 which is trivial and:
fx ix
mv v
m m
1
2 11 2
2
Substituting this into the momentum equation gives
fx ix fx ix
m m mv v and v v
m m m m
1 2 1
1 1 2 11 2 1 2
2
These allow us to compute the final velocity of each object in terms of the initial velocity of m1 and the relative masses of each object
AP Physics C 29
Elastic Collision Problem
A 50 g marble moving at 2.0 m/s strikes a 20 g marble at rest. What is the speed of each marble immediately after the collision?
1 2f 1 i 1
1 2
1f 2 i 1
1 2
50 g 20 g( ) ( ) (2.0 m/s) 0.86 m/s
50 g 20 g
2 2(50 g)( ) ( ) (2.0 m/s) 2.9 m/s
50 g 20 g
x x
x x
m mv v
m m
mv v
m m
Analysis: Expect that v1 will decrease and v2 will significantly increase. Laws conservation of Momentum and ME will be observed.
AP Physics C 31
Energy Diagrams
Energy Diagram: A graph showing a system’s potential energy and total energy as a function of position.