Q7.1.a
Which of the following
expressions satisfies the
requirement that
Fx   dU s / dx   k s x where x is
measured from the equilibrium
position (C is a constant)?
1) U s  k s  C
2) U s  k s  C
3) U s  k s x 2  C
1
4) U s   k s x 2  C
2
1
5) U s  k s x 2  C
2
Q7.1.b
A horizontal spring has a mass attached which can move with negligible friction.
You stretch the spring and release the mass from rest. For the resulting motion,
which of the following statements is TRUE?
1) When the spring is (momentarily) fully compressed, K has its largest value.
2) When the spring (momentarily) has its relaxed length, U has its largest value.
3) When the spring (momentarily) has its relaxed length, K has its smallest value.
4) When K is large, U is small, and vice versa.
5) When K is large, U is large, and vice versa.
Q7.1.c
A horizontal spring with stiffness 3 N/m has a relaxed length of 25 cm (0.25 m). A
mass of 50 grams (0.050 kg) is attached and you stretch the spring to a total length
of 29 cm (0.29 m). The mass is then released from rest and moves with negligible
friction. What is the kinetic energy of the mass at the moment when the spring
returns through the position where its length is its relaxed length of 25 cm?
1) 2.4e-3 J
2) 4.8e-3 J
3) 6e-2 J
4) 1.26e-1 J
5) 1.5 J
Q7.2.a
y-axis: energy; x-axis: separation
Which graph correctly shows U for two interacting electrons?
5: none of the above
Q7.2.b
y-axis: energy; x-axis: separation
Which graph correctly shows U for two atoms in a diatomic molecule?
5: none of the above
Q7.2.c (review)
y-axis: energy; x-axis: separation
Which graph correctly shows U for the Moon and the Earth?
5: none of the above
Q7.3.a:
Consider a block falling onto a vertical spring.
States:
A: Block 0.8 m above floor
B: Block just touching top of spring
C: Block 0.3 m above floor
To find the speed of the block just before it
hits the spring, what should we pick for initial
and final states?
1)
Initial Final
A
B
2)
B
C
3)
A
C
4)
C
A
Q7.3.b:
Consider a block falling onto a vertical spring.
States:
A: Block 0.8 m above floor
B: Block just touching top of spring
C: Block 0.3 m above floor
To find the maximum compression of the
spring, what should we pick for initial and
final states?
1)
Initial Final
A
B
2)
B
C
3)
A
C
4)
C
A
Q 7.3.c
An ideal mass-spring system is oscillating with angular frequency
  k s m . More energy is put into the mass-spring system. What is
now true?
1) The frequency of the oscillations increases
2) The frequency of the oscillations decreases
3) The frequency increases and amplitude increases
4) The frequency is unchanged and the amplitude increases
5) The frequency decreases and amplitude decreases
Q7.4.a
Two lead bricks moving in the +x and –x directions, each with kinetic energy K,
smash into each other and come to a stop. What happened to the energy?
1) The kinetic energy changed into rest energy.
2) The kinetic energy changed into thermal energy.
3) The total energy of the system decreased by an amount 2K.
4) Since the blocks were moving in opposite directions, the initial kinetic
energy of the system was zero, so there was no change in energy.
Q7.4.b
You have a pot containing 1000 grams of water
over a fire, and you are also stirring the water
with a paddle. Take the water as the system.
Because the fire is at a higher temperature than
the water, there is energy transfer due to a
temperature difference (microscopic work) Q =
5000 J to the water due to the fire. At the same
time, there is external work W = 2000 J done by
you. What is the increase in the thermal energy of
the water?
1) 0 J
2) 2000 J
3) 3000 J
4) 5000 J
5) 7000 J
Q7.4.c
The thermal energy of the 1000 grams of water
increased 7000 J. What was the temperature
increase in Kelvins of the water? The heat
capacity of water is C  Ethermal / T = 4.2 J/K
on a per-gram basis.
1) 0.0006 K
2) 0.6 K
3) 1.7 K
4) 1667 K
5) Insufficient
information
Q7.5.a
Which of the following statements is correct?
1) Q and Ethermal are the same thing.
2) Q and Ethermal are not the same thing, but they are always equal.
3) Ethermal can be nonzero even if Q is zero.
4) Q and Ethermal are both always positive.
Q7.8.a: A ball travels straight up, coming to a stop after it has risen
a distance h. Which equation (Ef = Ei + W) applies to the system
of the ball alone?
1)
0 = 1/2 mvi**2 – mgh
2) 0 + mgh = 1/2 mvi**2 + 0
3) 1/2 mvi**2 = – mgh
4)
0 = 1/2 mvi**2 + mgh
The following questions are a series based on 7.X.44, the analysis of a
horse running up a hill, for two different choices of system.
Q7.8.b. A horse of mass M gallops with constant speed v up a long hill of height h
and horizontal extent d. Choose the horse as the system to analyze. Start from the
energy principle, Esys  Wext  Q . First, work on right side of equation.
What objects in the surroundings exert forces on our chosen system, the horse?
1) Earth (gravitational)
2) Earth (gravitational) and ground (electric; interatomic)
3) Earth (gravitational), ground (electric; interatomic), and air (electric; air
resistance)
4) Earth (gravitational), ground (electric; interatomic), air (electric; air resistance),
and horse’s hooves (electric; interatomic)
Q7.8.c
The horse’s hooves don’t slip on the rocky ground, so the work done
by the ground on the horse is
1) W > 0 because the force points upward
2) W > 0 because work is a positive quantity
3) W = 0 because there is no displacement of the force
4) W < 0 because the hooves move downward
5) W < 0 because the hill doesn’t speed up the horse
Q7.8.d.The displacement of the horse is < d, h, 0 >, and the force of the Earth on the
horse is < 0, -Mg, 0 >. The work done by the Earth on the horse is
1) < 0, -Mgh, 0 >
2) |< 0, -Mgh, 0 >| =
3) -Mgh
4) < d, h-Mg, 0 >
02  (  Mgh)2  02
= Mgh
Q7.8.e. The term “Q” in the energy principle Esys  Wext  Q is microscopic work done
on the system due to a temperature difference between the system and the
surroundings. When the horse started running, its temperature quickly rose, and its
temperature is quite a bit higher than the surroundings. Which of the following is
true?
1) Q < 0 because there is energy transfer from the horse to the surrounding air
2) Q = 0 because air is a thermal insulator
3) Q > 0 because the horse is hotter than the surrounding air
The body temperature of the horse quickly rises until the rate of energy transfer
(|dQ/dt|, joules per second) to the surroundings (microscopic work done by the
horse on the surroundings) is high enough to keep the body temperature constant,
no longer increasing.
We now know a lot about the right side of Esys  Wext  Q , which is
(0 due to ground) + (-Mgh due to Earth) + (-|Q| from horse to air)
= -Mgh-|Q|
Next let’s look at the left side of Esys  Wext  Q .
Q7.8.f. Which of the following correctly represents the energy of the chosen system
(the horse)?
1) Mc 2  K
2) Mc 2  K  U grav
3) Mc 2  K  Echemical  Ethermal
4) Mc 2  K  Echemical  Ethermal  U grav
Q7.8.g Which of the following energy terms change during the run at constant speed
and constant body temperature (constant because there is energy transfer |Q| to the
surrounding air)?
1)
2)
3)
4)
Mc2
K
Echemical
E thermal
Q7.8.h What is true about the change in Echemical inside the chosen system (the
horse)?
1) Echemical  0 because the horse provides chemical energy to go up the hill
2) Echemical  0 because the horse’s temperature doesn’t change
3) Echemical  0 because the horse provides chemical energy to go up the hill
Put it all together for system = horse: Echemical  Mgh  Q
The horse’s store of chemical energy decreases. The energy principle relates this
energy decrease to the energy transfers out of the system (negative work done by
Earth, energy transfer from hot horse to cool air).
Q7.8.i. Instead, choose the Universe as the system.
What is different about the right side of Esys  Wext  Q ?
1) It’s the same; the energy principle is fundamental.
2) The right side is zero; there are no surroundings.
3) We have to add potential energy to the right side.
Q7.8.j. What is different about the left side of Esys  Wext  Q if we choose the
Universe as the system?
1) The only new term is U grav   Mgh
2) The only new term is U grav   Mgh
3) The only new term is Eair  0
4) The new terms are U grav   Mgh and Eair  0
5) The new terms are U grav   Mgh and Eair  0
Put it all together for system = Universe: Echemical  Mgh  Eair  0
The gravitational potential energy of the Universe increases, and the thermal energy
of the air increases. These increases are “paid for” out of the horse’s store of
chemical energy.
Compare with system = horse: Echemical  Mgh  Q
We see that Eair  Q , which makes sense; the energy transfer from horse to air in
the form of microscopic work is equal to the energy increase of the air.
Answers to the horse questions:
Q7.8b 3
Q7.8c 3
Q7.8d 3
Q7.8e 1
Q7.8f 3
Q7.8g 3
Q7.8h 3
Q7.8i 2
Q7.8j 4
The following questions go through a chain of reasoning about air
resistance, step by step.
Q7.10.a
A single falling coffee filter quickly reaches a constant terminal speed
of 1 m/s. At this speed, what do we know about the air resistance force?
1) Fair > Fgrav
2) Fair = Fgrav
3) Fair < Fgrav
4) not enough information
Q7.10.b
Three nested coffee filters fall, quickly reaching a terminal speed
of ~ 1.7 m/s. If the air resistance force on a single falling coffee filter
was F1, how large is the air resistance force on the three falling filters?
1) F1
2) 3*F1
3) (1/3)*F1
4) not enough information
Q7.10.c: How does the air resistance force depend on an object's speed?
1 filter v1 = 1 m/s Fair = 1*mg
3 filters v3 = 1.7 m/s Fair = 3*mg
1) Fair = C*v
(C is a constant)
2) Fair = C*(1/v)
3) Fair = C*v**(1/2)
4) Fair = C*v**2
5) Fair = C*v**(1/3)
6) Fair = C*v**3
Q7.10.d: If the cross-sectional area of a coffee filter were doubled,
how would you expect the air resistance force to change?
1) Fair would be larger
2) Fair would stay the same
3) Fair would be smaller