Download Important Notes: 1. Energy (U = mgh, K = 1/2mv2) is ALWAYS

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AP Physics
Chapter 8: Potential Energy
by Brenda Chen
Background/Summary
This chapter takes our basic knowledge of energy a step further by introducing nonisolated and isolated
systems. Nonisolated systems can have energy cross boundaries, a concept called conservation of energy.
Isolated systems have a total energy that is constant. We are also introduced to power.
Important Notes:
1. Energy (U = mgh, K =
1/2mv2) is ALWAYS
conserved! (Remember,
heat is not a form of
energy)
2. A nonisolated system
has other forces acting on
it. An isolated system
does not; all the terms on
its right side will equal
zero.
3. Work is added to the
conservation of energy
equation if the system is
nonisolated. Work is a
method of transferring
energy to a system by
applying a force to a
system so that a point of
application undergoes a
displacement.
4. A friction force
transforms kinetic energy
in a system to internal
energy. The two are
inversely proportional.
5. Power takes into
account the time interval
it takes to get something
done. There is
instantaneous power,
dE/dt, and an average
power, W/Δt. The unit
for power is Watts.
Common Diagrams: Pendulum
Lcosθ
Important Formulae: 𝑈! + 𝐾! = 𝑈! + 𝐾! θ L L - Lcosθ
Block on Table
𝐸!"!#$%!!"!#!$%
− 𝛥𝐸!"!"#$%& + 𝛴𝑊!"!!"#$"%!&
= 𝐸!"!#$%!!"#$% 𝐸!"#!!"#$!% = 𝐾 + 𝑈 𝛥𝐸 = 𝑊 = !𝐹Ÿ 𝑑𝑟 𝛥𝐸!"#$%"&' = 𝑓! 𝑑 𝑑𝐸
𝑃=
𝑑𝑡
𝑃 = 𝐹Ÿ 𝑣 1
𝑈! = 𝑘𝑥 ! 2
Hanging Mass
m1
m2
Strategy:
1. Identify problem as using an Energy analysis.
2. Define your system: Identify initial and final positions
for all objects.
3. Select zero reference points for all potential energies.
4. If any type of friction is present, mechanical energy will
not be constant – account for friction using
ΔEint = Ffrictionx
5. Write an equation with all energies present, and solve for
the unknown.
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AP Physics
Problem 1 (Easy):
Lena is at the Pasadena Ice Rink. Her devious friend gives her a little push so that Lena is sliding at an
initial speed of 3.00 m/s. The coefficient of kinetic friction between her ice skates and the ice is .140. How
far does she travel until she finally reaches a stop?
Solutions:
3m/s
µ = .140
Remember to think of the five steps of our strategy. In this case, we are sure that this is an energy analysis
problem, and we won’t have to assign heights because they are on the same y-axis. For that reason, we also
won’t have to include the potential energy in the equation because there are staying on the same y-axis.
We will, however, have to add Einternal to the equation because they gave us a coefficient of friction.
𝛴𝐸!"!#!$% = 𝛴𝐸!"#$%
𝐾! − 𝐸!"# = 𝐾!
!
!
𝑚𝑣 ! − 𝑓! 𝑑 = 𝐾!
0
The final kinetic energy is zero because Lena eventually comes to a stop.
𝐹! = 𝜇𝐹! = 𝜇𝑚𝑔
1
𝑚𝑣 ! = 𝜇𝑚𝑔𝑥
2
𝑣!
3!
𝑥=
=
2𝜇𝑔
2 . 14 9.8
𝑥 = 3.27 𝑚
Lab: AP Review Sheets
AP Physics
Problem 2 (Medium):
Hana is trying to hypnotize her sister. Her 8 cm pendulum is a attached to a mass of 10 g. If the speed of
the particle is 2.5 m/s at its lowest point, at what angle is the pendulum released at? What is the tension T
at the lowest point?
Solutions:
FT
Fg
1cosθ
θ
1m
v = 2.5 m/s
10g
1-1cosθ
We know that we need to use the L-Lcosθ to find where the mass is at its lowest point. Once we
established that, we can then use our conservation of energy equation.
𝑈!" + 𝐾! = 𝑈!" + 𝐾!
0
0
The initial kinetic energy and final potential energy cancel out because the pendulum is released from rest,
and at its lowest point, we consider the height to be zero.
1
𝑚𝑣 !
2
1
. 010 9.8 1 − 1𝑐𝑜𝑠𝜃 = (.010)(2.5)!
2
1
!
2 (.010)(2.5) = 1 − 1𝑐𝑜𝑠𝜃
. 010 9.8
−.681 = 1𝑐𝑜𝑠𝜃
𝑚𝑔 𝐿 − 𝐿𝑐𝑜𝑠𝜃 =
𝜃 = 47.1
Now we need to do a force analysis to figure out the tension. Because the pendulum travels in a circular
path, we need to use centripetal force.
𝑚𝑣 !
𝑟
𝑚𝑣 !
𝐹! − 𝐹! =
𝑟
𝑚( 2𝑔 𝐿 − 𝐿𝑐𝑜𝑠𝜃 )!
𝑇=
+ 𝑚𝑔
𝐿
𝐹!"# =
𝑇=
(.01) (2)(9.8)(1 − 1𝑐𝑜𝑠47.1)!
+ (.01)(9.8)
1
𝑇 = .161 𝑁
Lab: AP Review Sheets
AP Physics
Problem 3 (Hard):
A marshmallow gun that requires a special type of dense marshmallow has a launching mechanism that
uses a spring with unknown k. When compressed only .10 m from the barrel end, the gun will launch the
heavy marshmallow 2 meters above the end of the barrel. The marshmallow, which is 14 g, sits on top of
the vertically oriented gun. What is the spring constant? The equilibrium position of the spring with the
heavy marshmallow resting on it? The marshmallow’s speed as it moves through this position?
Solutions:
2m
Marshmallow (14 g)
.10m
Marshmallow Gun
To find the spring constant, we should first assume that there is no air friction. Then we write out the
conservation of energy equation (Be careful sure to use the potential energy equation for a spring)
𝑈!!!"!#!$% + 𝑈!!!"!#!$% + 𝐾! = 𝑈!!!"#$! + 𝑈!!!"#$% + 𝐾!
1 !
1
1
1
𝑘𝑥 + 𝑚𝑔ℎ + 𝑚𝑣 ! = 𝑘𝑥 ! + 𝑚𝑔ℎ + 𝑚𝑣 !
2
2
2
2
1 !
𝑘𝑥 = 𝑚𝑔ℎ
2
The Ug-initial, Ki, Us-final, and Ug-final cancelled out because our starting point is at h = 0, the marshmallow is
starting from rest, and the final speed is also 0.
𝑘=
2𝑚𝑔ℎ
2 . 014 9.8 . 1 + 2
=
= 3.29 𝑁/𝑚
!
𝑥
. 1!
For finding the equilibrium position, we need to use a force analysis. The right side of the equation is zero
because the forces need to be equal for the system to be in equilibrium.
𝐹!"# = 0
𝐹!"#$%& − 𝐹! = 0
𝐹!"#$%& = −𝑚𝑔 = −𝑘𝑥
𝑚𝑔 (.014)(9.8)
𝑥=
=
= .042𝑚
𝑘
(3.29
Now to find the speed of the marshmallow as it moves through this position, we would again need to do an
energy analysis.
𝑈!!!"!#!$% + 𝑈!!!"!#!$% + 𝐾! = 𝑈!!!"#$% + 𝑈!!!"#$% + 𝐾!
1 !
1
1
1
𝑘𝑥 + 𝑚𝑔ℎ + 𝑚𝑣 ! = 𝑘𝑥 ! + 𝑚𝑔ℎ + 𝑚𝑣 !
2
2
2
2
1 ! 1 !
1
!
𝑘𝑥 = 𝑘𝑥 + 𝑚𝑔ℎ + 𝑚𝑣
2
2
2
1
1
1
!
!
(3.29)(−.1) = (3.29) −.042 + (.014)(9.8)(2 − .12) + (.014)𝑣 !
2
2
2
37.3 = 𝑣 ! = 6.10 𝑚/𝑠