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  k  C
s
s
2) U  k  C
s
s
3) U  k x 2  C
s
s
4)
5)
1
U s   ks x2  C
2
1
U s  ks x2  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? 2
Chapter 7 Clickers
1
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? 4
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? 1
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?
Initial
Final
1)
A
B
2)
B
C
3)
A
C
4)
C
A
Initial
Final
1)
A
B
2)
B
C
3)
A
C
Q7.3.b: skip
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?
Chapter 7 Clickers
2
4)
C
A
Q 7.3.c
An ideal mass-spring system is oscillating with angular frequency
  k m . More energy is put into the mass-spring system. What is now true?
s
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?
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  E
= 4.2 J/K on a per-gram basis.
thermal / T
Chapter 7 Clickers
1) 0 J
2) 2000 J
3) 3000 J
4) 5000 J
5) 7000 J
1) 0.0006 K
2) 0.6 K
3) 1.7 K
4) 1667 K
5) Insufficient information
3
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.
SKIP
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
Chapter 7 Clickers
4
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 E
sys
 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 E  W  Q .
sys
ext
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) Mc2  K  E
chemical  Ethermal
4) Mc2  K  E
E
U
chemical
thermal
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)?
Chapter 7 Clickers
5
1) Mc2
2) K
3) Echemical
4) E
thermal
Q7.8.h What is true about the change in E
inside the chosen system (the horse)?
chemical
1) E
because the horse provides chemical energy to go up the hill
chemical  0
2) Echemical  0 because the horse’s temperature doesn’t change
3) E
because the horse provides chemical energy to go up the hill
chemical  0
Put it all together for system = horse: E
chemical   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 E  W  Q ?
sys
ext
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 E
sys
 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
Eair  0
4) The new terms are U
and E  0
air
grav   Mgh
5) The new terms are U
and Eair  0
grav   Mgh
3) The only new term is
Put it all together for system = Universe: E
chemical  Mgh  Eair  0
Chapter 7 Clickers
6
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:
We see that E
air
Echemical  Mgh  Q
 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.
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
Chapter 7 Clickers
7
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
Chapter 7 Clickers
8