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PHYS 207.013
Chapter 8
potential energy
conservation of energy
Instructor: Dr. Bianco
TAs: Joey Betz; Lily Padlow
University of Delaware - Spring 2021
Energy
- kinetic energy
- work
-Potential energy -conservation of energy
H&R CH8 potential energy - conservation of energy
Instantaneous Work
H&R CH8 potential energy - conservation of energy
Instantaneous Work
H&R CH8 potential energy - conservation of energy
WAB=∫ABFdx
W_{AB} = \int_A^B F dx
[N m]
Fdx=ma dx=mdtdvdxv=dtdx=>dx=v dt
F dx = ma ~dx = m \frac{dv}{dt} dx \\
v = \frac{dx}{dt} => dx = v ~dt\\
WAB=∫ABFdx=∫ABmdtdvvdt=∫ABmv dv
W_{AB} = \int_A^B F dx = \int_A^B m \frac{dv}{dt} v {dt} = \int_A^B m v ~dv
=m∫ABv dv=21mv2∣AB
= m \int_A^B v ~dv = \frac{1}{2}m v^2 |_A^B
Instantaneous Work
H&R CH8 potential energy - conservation of energy
K=21mv2
K = \frac{1}{2} m v^2
kinetic energy
=m∫ABv dv=21mvB2−21mvA2
= m \int_A^B v ~dv = \frac{1}{2} mv_B^2 - \frac{1}{2}m v_A^2
WAB=
W_{AB} =
WAB=∫ABFdx
W_{AB} = \int_A^B F dx
[N m]
Instantaneous Work
H&R CH8 potential energy - conservation of energy
K=21mv2
K = \frac{1}{2} m v^2
kinetic energy
=m∫ABv dv=21mvB2−21mvA2=KB−KA
= m \int_A^B v ~dv = \frac{1}{2} mv_B^2 - \frac{1}{2}m v_A^2 = K_B - K_A
WAB=
W_{AB} =
WAB=∫ABFdx
W_{AB} = \int_A^B F dx
[N m]
Instantaneous Work
H&R CH8 potential energy - conservation of energy
=m∫ABv dv=21mvB2−21mvA2=KB−KA
= m \int_A^B v ~dv = \frac{1}{2} mv_B^2 - \frac{1}{2}m v_A^2 = K_B - K_A
K=21mv2
K = \frac{1}{2} m v^2
kinetic energy
WAB=
W_{AB} =
WAB=∫ABFdx
W_{AB} = \int_A^B F dx
[N m]
WAB=KB−KA
W_{AB} = K_B - K_A
work-energy theorem
H&R CH8 potential energy - conservation of energy
WAB=KB−KA
W_{AB} = K_B - K_A
A
B
h
WAB=KB−KA
W_{AB} = K_B - K_A
WAB=F⋅d
W_{AB} = \vec{F}\cdot\vec{d}
W and K in 1D
v
H&R CH8 potential energy - conservation of energy
WAB=KB−KA
W_{AB} = K_B - K_A
A
B
h
WAB=KB−KA
W_{AB} = K_B - K_A
WAB=F⋅d
W_{AB} = \vec{F}\cdot\vec{d}
mg
m\vec{g}
WAB=mghcos(180o)=−mgh
W_{AB} = mgh\cos{(180^o)}= -mgh
W and K in 1D
lift m from A to B
v
H&R CH8 potential energy - conservation of energy
WAB=KB−KA
W_{AB} = K_B - K_A
A
B
h
WAB=KB−KA
W_{AB} = K_B - K_A
WAB=F⋅d
W_{AB} = \vec{F}\cdot\vec{d}
mg
m\vec{g}
WAB=mghcos(180o)=−mgh
W_{AB} = mgh\cos{(180^o)}= -mgh
WAB=F⋅d
W_{AB} = \vec{F}\cdot\vec{d}
ΔK=mgh
\Delta K = mgh
W and K in 1D
W and K in 1D
H&R CH8 potential energy - conservation of energy
A
B
h
mg
m\vec{g}
Wup=−mghWdown=mgh
W_{up} = -mgh\\
W_{down} = mgh
Wrowndtrip=0
W_{rowndtrip} = 0
W and K in 3D
H&R CH8 potential energy - conservation of energy
WAB=∫ABFdr
W_{AB} = \int_A^B F dr
dr
d\vec{r}
dr
d\vec{r}
W and K in 3D
H&R CH8 potential energy - conservation of energy
WAB=∫ABFdr
W_{AB} = \int_A^B F dr
dr
d\vec{r}
dr
d\vec{r}
F
F
F=Fxi^+Fyj^+Fzk^
\vec{F} = F_x \hat{i} +F_y \hat{j}+F_z \hat{k}
dr=dxi^+dyj^+dzk^
\vec{dr} = dx \hat{i} +dy \hat{j}+dz \hat{k}
W and K in 3D
H&R CH8 potential energy - conservation of energy
WAB=∫ABFdr
W_{AB} = \int_A^B F dr
dr
d\vec{r}
dr
d\vec{r}
F
F
F=Fxi^+Fyj^+Fzk^
\vec{F} = F_x \hat{i} +F_y \hat{j}+F_z \hat{k}
dr=dxi^+dyj^+dzk^
\vec{dr} = dx \hat{i} +dy \hat{j}+dz \hat{k}
At every step in the path:
dW=Fx⋅dx+Fy⋅dy+Fz⋅dz
dW = Fx \cdot dx+ Fy \cdot dy+ Fz \cdot dz
W and K in 3D
H&R CH8 potential energy - conservation of energy
WAB=∫ABFdr
W_{AB} = \int_A^B F dr
dr
d\vec{r}
dr
d\vec{r}
F
F
F=Fxi^+Fyj^+Fzk^
\vec{F} = F_x \hat{i} +F_y \hat{j}+F_z \hat{k}
dr=dxi^+dyj^+dzk^
\vec{dr} = dx \hat{i} +dy \hat{j}+dz \hat{k}
At every step in the path:
∫ABdW=∫ABFx⋅dx+∫ABFy⋅dy+∫ABFz⋅dz
\int_A^BdW = \int_A^BF_x \cdot dx+ \int_A^BF_y \cdot dy+ \int_A^BF_z \cdot dz
each piece is the 1D prob we did before
W and K in 3D
H&R CH8 potential energy - conservation of energy
WAB=∫ABFdr
W_{AB} = \int_A^B F dr
dr
d\vec{r}
dr
d\vec{r}
F
F
F=Fxi^+Fyj^+Fzk^
\vec{F} = F_x \hat{i} +F_y \hat{j}+F_z \hat{k}
dr=dxi^+dyj^+dzk^
\vec{dr} = dx \hat{i} +dy \hat{j}+dz \hat{k}
At every step in the path:
∫ABdW=21(vBx2−vAx2)+21(vBy2−vAy2)+21(vBz2−vAz2)
\int_A^BdW = \frac{1}{2}(v_{Bx}^2 - v_{Ax}^2) + \frac{1}{2}(v_{By}^2 - v_{Ay}^2) + \frac{1}{2}(v_{Bz}^2 - v_{Az}^2)
W and K in 3D
H&R CH8 potential energy - conservation of energy
WAB=∫ABFdr
W_{AB} = \int_A^B F dr
dr
d\vec{r}
dr
d\vec{r}
F
F
F=Fxi^+Fyj^+Fzk^
\vec{F} = F_x \hat{i} +F_y \hat{j}+F_z \hat{k}
dr=dxi^+dyj^+dzk^
\vec{dr} = dx \hat{i} +dy \hat{j}+dz \hat{k}
At every step in the path:
∫ABdW=21(vBx2−vAx2)+21(vBy2−vAy2)+21(vBz2−vAz2)=21mvA2−21mvB2=ΔK
\int_A^BdW = \frac{1}{2}(v_{Bx}^2 - v_{Ax}^2) + \frac{1}{2}(v_{By}^2 - v_{Ay}^2) + \frac{1}{2}(v_{Bz}^2 - v_{Az}^2) \\
= \frac{1}{2}mv^2_A - \frac{1}{2}mv^2_B = \Delta K
W and K in 3D
slightly more practical
H&R CH8 potential energy - conservation of energy
dr
d\vec{r}
dr
d\vec{r}
F
F
F=mg=Fyj^
\vec{F} = m \vec{g} = F_y \hat{j}
∫ABdW=∫ABF⋅dr=∫ABFy⋅dy=∫ABmgdy=−mgh
\int_A^BdW = \int_A^BF \cdot dr = \int_A^BFy \cdot dy = \\
\int_A^B m g dy = -mgh
h
W and K in 3D
slightly more practical
H&R CH8 potential energy - conservation of energy
dr
d\vec{r}
dr
d\vec{r}
F
F
F=mg=Fyj^
\vec{F} = m \vec{g} = F_y \hat{j}
h
only force is g: acts along y
W and K in 3D
slightly more practical
H&R CH8 potential energy - conservation of energy
dr
d\vec{r}
dr
d\vec{r}
F
F
F=mg=Fyj^
\vec{F} = m \vec{g} = F_y \hat{j}
∫ABdW=∫ABF⋅dr=∫ABFy⋅dy=∫ABmgdy=−mgh
\int_A^BdW = \int_A^BF \cdot dr = \int_A^BFy \cdot dy = \\
\int_A^B m g dy = -mgh
h
But this does not depend on the crazy path I took!!
W and K in 3D
slightly more practical
H&R CH8 potential energy - conservation of energy
F
F=mg=Fyj^
\vec{F} = m \vec{g} = F_y \hat{j}
∫ABdW=∫ABF⋅dr=∫ABFy⋅dy=∫ABmgdy=−mgh
\int_A^BdW = \int_A^BF \cdot dr = \int_A^BFy \cdot dy = \\
\int_A^B m g dy = -mgh
h
But this does not depend on the crazy path I took!!
W and K in 3D
slightly more practical
H&R CH8 potential energy - conservation of energy
F
F=mg=Fyj^
\vec{F} = m \vec{g} = F_y \hat{j}
∫ABdW=∫ABF⋅dr=∫ABFy⋅dy=∫ABmgdy=−mgh
\int_A^BdW = \int_A^BF \cdot dr = \int_A^BFy \cdot dy = \\
\int_A^B m g dy = -mgh
h
But this does not depend on the crazy path I took!!
Instantaneous Work
H&R CH8 potential energy - conservation of energy
If the Work done by a Force does not depend on the path taken that force is said to be conservative
Conservative Forces
H&R CH8 potential energy - conservation of energy
If the Work done by a Force does not depend on the path taken that force is said to be conservative
- e.g. Gravity is conservative
can you think of a non conservative force and another non conservative force?
Conservative Forces
H&R CH8 potential energy - conservation of energy
If the Work done by a Force does not depend on the path taken that force is said to be conservative
- e.g. Gravity is conservative
can you think of a non conservative force and another non conservative force?
CONSERVATIVE: Gravity - Spring
NON-CONSERVATIVE: Friction
Conservation of Mechanical Energy
H&R CH8 potential energy - conservation of energy
A
A
B
h
mg
m\vec{g}
W=−mgh=−mg(yB−yA)
W = -mgh = -mg (y_B-y_A)
lift m from A to B
W=KB−KA
W = K_B - K_A
v
Conservation of Mechanical Energy
H&R CH8 potential energy - conservation of energy
A
A
B
h
mg
m\vec{g}
W=−mgh=−mg(yB−yA)
W = -mgh = -mg (y_B-y_A)
lift m from A to B
−mgyB+mgyA=KB−KA
-mgy_B + mgy_A = K_B - K_A
W=KB−KA
W = K_B - K_A
Conservation of Mechanical Energy
H&R CH8 potential energy - conservation of energy
A
A
B
h
mg
m\vec{g}
W=−mgh=−mg(yB−yA)
W = -mgh = -mg (y_B-y_A)
lift m from A to B
−mgyB+mgyA=KB−KA
-mgy_B + mgy_A = K_B - K_A
W=KB−KA
W = K_B - K_A
KA+mgyA=KB+mgyB
K_A + mgy_A = K_B + mgy_B
Conservation of Mechanical Energy
H&R CH8 potential energy - conservation of energy
A
A
B
h
mg
m\vec{g}
lift m from A to B
KA+mgyA=KB+mgyB
K_A + mgy_A = K_B + mgy_B
The sum of potential energy and kinetic energy remains the same
when the force is conservative
call mgy Potential Energy
use the letter U to refer to it
the 0 point of the potential gravitational energy
H&R CH8 potential energy - conservation of energy
A
A
this point is arbitrary: the only thing that matters is the distance h - just be consistent!
A
B
0
h
U=mgh
U = mgh
the 0 point of the potential gravitational energy
H&R CH8 potential energy - conservation of energy
A
A
this point is arbitrary: the only thing that matters is the distance h - just be consistent!
A
B
-h
0
U=mgh
U = mgh
the 0 point of the potential gravitational energy
H&R CH8 potential energy - conservation of energy
A
A
this point is arbitrary: the only thing that matters is the distance h - just be consistent!
A
B
h
2h
U=mgh
U = mgh
0
Conservation of Elastic Energy
Uspring=21kx2
U_{spring} = \frac{1}{2}kx^2
loop-the-loop problem
H&R CH8 potential energy - conservation of energy
loop-the-loop problem
H&R CH8 potential energy - conservation of energy
will a ball released from the top left of this track make it around the loop?
H&R CH8 potential energy - conservation of energy
A
B
C
D
E
loop-the-loop problem
H&R CH8 potential energy - conservation of energy
UA+KA=UB+KB=UC+KC=UD+KD=UE+KE
U_A + K_A = U_B+K_B=U_C+K_C =U_D+K_D =U_E+K_E
A
C
D
E
starting at h0:mgh0=mgy+21mv2
\mathrm{starting~at}~h_0:\\
mgh_0 = mgy + \frac{1}{2}mv^2
everywhere
B
loop-the-loop problem
H&R CH8 potential energy - conservation of energy
UA+KA=UB+KB=UC+KC=UD+KD=UE+KE
U_A + K_A = U_B+K_B=U_C+K_C =U_D+K_D =U_E+K_E
A
B
C
D
E
mgh=mgy+21mv2
mgh = mgy + \frac{1}{2}mv^2
everywhere
rvD2=a>=g
\frac{v^2_D}{r} = a >= g
loop-the-loop problem
H&R CH8 potential energy - conservation of energy
UA+KA=UB+KB=UC+KC=UD+KD=UE+KE
U_A + K_A = U_B+K_B=U_C+K_C =U_D+K_D =U_E+K_E
A
B
C
D
E
mgh=mgy+21mv2
mgh = mgy + \frac{1}{2}mv^2
rvD2=a>=g
\frac{v^2_D}{r} = a >= g
v2=2g(h−y)
v^2 = 2g(h -y)
loop-the-loop problem
H&R CH8 potential energy - conservation of energy
UA+KA=UB+KB=UC+KC=UD+KD=UE+KE
U_A + K_A = U_B+K_B=U_C+K_C =U_D+K_D =U_E+K_E
A
B
C
D
E
mgh=mgy+21mv2
mgh = mgy + \frac{1}{2}mv^2
rvD2=a>=g
\frac{v^2_D}{r} = a >= g
v2=2g(h−y)
v^2 = 2g(h -y)
vD2=2g(h−2r)>=gr
v_D^2 = 2g(h -2r) >= gr
loop-the-loop problem
H&R CH8 potential energy - conservation of energy
UA+KA=UB+KB=UC+KC=UD+KD=UE+KE
U_A + K_A = U_B+K_B=U_C+K_C =U_D+K_D =U_E+K_E
A
B
C
D
E
mgh=mgy+21mv2
mgh = mgy + \frac{1}{2}mv^2
rvD2=a>=g
\frac{v^2_D}{r} = a >= g
v2=2g(h−y)
v^2 = 2g(h -y)
h>=21r+2r=25r
h >= \frac{1}{2}r + 2r = \frac{5}{2}r
pendulum problem
H&R CH8 potential energy - conservation of energy
what can you immediately say about the height a pendulum will reach on the opposite side of the initial release point?
pendulum problem
H&R CH8 potential energy - conservation of energy
what can you immediately say about the height a pendulum will reach on the opposite side of the initial release point?
how high can it go?
pendulum problem
H&R CH8 potential energy - conservation of energy
what can you immediately say about the height a pendulum will reach on the opposite side of the initial release point?
how high can it go?
why will it eventually stop?
Conservation of Energy
EM=K+U is conserved
EM = K + U~ \mathrm{is~conserved}
Conservation of Energy
ΔU=−∫xixfF(x)dx
\Delta U =- \int_{x_i}^{x_f} F(x) dx
F(x)=−dxdU(x)
F(x) = -\frac{dU(x)}{dx}
EM=K+U is conserved
EM = K + U~ \mathrm{is~conserved}
Conservation of Energy
slope U < 0
=> F > 0
Turning point: K = 0
U=E
particle changes direction
Emec=5J
E_{mec} = 5J
EM=K+U EM is conserved
EM = K + U~ \\
EM~\mathrm{is~conserved}
if the system is isolated and only contains conservative forces
Conservation of Energy
slope U < 0
=> F > 0
Turning point: K = 0
U=E
particle changes direction
Emec=3J
E_{mec} = 3J
EM=K+U EM is conserved
EM = K + U~ \\
EM~\mathrm{is~conserved}
if the system is isolated and only contains conservative forces
Conservation of Energy
Neutral Equilibrium: U + Emec
K = 0
dU/dx = F = 0
Emec=4J
E_{mec} = 4J
EM=K+U EM is conserved
EM = K + U~ \\
EM~\mathrm{is~conserved}
if the system is isolated and only contains conservative forces
Conservation of Energy
K(x)=0
K(d+dx)>0
K(d-dx)>0
Unstable equilibrium: U=E
K=0
particle would start moving if displaced
Emec=3J
E_{mec} = 3J
K(x±dx)=0
K(x\pm dx) \neq 0
EM=K+U EM is conserved
EM = K + U~ \\
EM~\mathrm{is~conserved}
if the system is isolated and only contains conservative forces
Conservation of Energy
U = Emec
U(x+/- dx) > Emec
Stable equilibrium:
U = E
particle cannot move without violating K>0
Emec=1J
E_{mec} = 1J
U(x±dx)>E
U(x\pm dx) > E
EM=K+U EM is conserved
EM = K + U~ \\
EM~\mathrm{is~conserved}
if the system is isolated and only contains conservative forces
Non conservative forces
ΔEmec=−fkdFd=ΔEmec+ΔEther
\Delta E_{mec} =- f_k d\\
Fd = \Delta E_{mec} + \Delta E_{ther}
exercise: think about how the examples we saw earlier, pendulum, loop-the-loop would change if we included non-conservative forces in the problem
KEY POINTS:
- Gravitational potential energy:
- Hooks potential energy
-
The potential energy relates to the force
-
in absence of non conservative forces, for an isolated system, the mechanical energy is conserved
Us=kx
U_s = kx
ΔU=−∫xixfF(x)dx
\Delta U =- \int_{x_i}^{x_f} F(x) dx
potential energy and
conservation of energy
Emec=U+K=const
E_{mec} = U + K = \mathrm{const}
H&R CH7 kinetic energy and work
Ug=mgh
U_g = mgh
PHYS 207.013 Chapter 8 potential energy conservation of energy Instructor: Dr. Bianco TAs: Joey Betz; Lily Padlow University of Delaware - Spring 2021 https://slides.com/federicabianco/phys20713_8
phys207 potential energy conservation of energy
By federica bianco
phys207 potential energy conservation of energy
potential energy and conservation of energy
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