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Physics 18 Spring 2010
Midterm 1 Solutions
For the midterm, you may use one sheet of notes with whatever you want to put on it, front
and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please
ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR
FINAL ANSWERS! You have the full length of the class. If you attach any additional
scratch work, then make sure that your name is on every sheet of your work. Good luck!
1. A 75 kilogram base runner has an initial speed of 8 m/s (almost 18 miles per hour),
and he then slides to a stop over 4 m. What is the coefficient of sliding friction between
the runner and ground during the slide?
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Solution
The frictional force does work against the motion of the base runner, changing his
kinetic energy, W = ∆KE. The work done is the product of the frictional force Ff ,
and his stopping distance D, and so W = Ff D. Now, Ff = −µk FN = −µk mg, where
µk is the coefficient of sliding friction, and m is the runner’s mass. The negative sign
comes from the fact that the frictional force is resisting the runner’s motion. Thus,
W = −µk mgD.
The runner starts off with an initial speed vi , and then comes to a stop (vf = 0), so
∆KE = KEf − KEi = 21 mvf2 − 12 mvi2 = 0 − 21 mvi2 . Setting W = ∆KE gives
1 2
mv = µk mgD.
2 i
Solving for the coefficient gives
µk =
vi2
.
2gD
Plugging in the numbers gives
µk =
vi2
82
=
≈ 0.82.
2gD
2 × 9.8 × 4
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2. An accident victim with a broken leg is being
placed in traction. The patient wears a special
boot with a pulley attached to the sole. The foot
and boot together have a mass of 4.0 kg, and the
doctor has decided to hang a 6.0 kg mass from
the rope. The boot is suspended by the ropes and
does not touch the bed.
(a) Determine the amount of tension in the rope
by using Newton’s laws to analyze the hanging mass.
(b) The net traction force needs to pull straight
out on the leg. What is the proper angle θ
for the upper rope?
(c) What is the net traction force pulling the
leg?
Hint: If the pulleys are frictionless, which we will
assume, the tension in the rope is constant from
one end to the other.
————————————————————————————————————
Solution
(a) Since the tension is everywhere the same in the
wire, then the tension is due to the hanging mass,
only. We can draw the free-body diagram for the
system. From the sum of the forces we see that if
M is the mass of the hanging weight,
X
Fy = T − M g = 0,
where the net force is zero since the system isn’t
accelerating. So, T = M g. Plugging in the numbers gives T = M g = 6.0 × 9.8 = 58.8 N.
2
T
FG
(b) Now, in order for the traction force to be
entirely horizontal, we need the tension
forces to cancel with the weight of the boot.
We can draw the force diagram for the boot
in the figure to the right. In the vertical direction, the sum of the forces is (taking the
weight of the boot FG = mg, where m is
the mass of the boot)
X
Fy = T sin θ − T sin 15◦ − mg = 0.
y
T
Ftraction
θ
15
x
T
FG (foot)
Solving for the angle gives
−1
θ = sin
T sin 15◦ + mg
T
,
or since T = M g,
−1
θ = sin
m
sin 15 +
= sin−1 (0.26 + .66) = 67.7◦ .
M
◦
(c) Now that we know that the vertical components of the forces balance, then the
traction force is all horizontal. From the diagram, we see that (taking θ = 67.7◦ )
X
Fx = T cos 67.7◦ + T cos 15◦ − Ftraction = 0.
So,
Ftraction = T (cos 67.7◦ + cos 15◦ ) = 58.8 (1.35) = 79 N.
This is a force equivalent to hanging a bit less than four pounds directly from
your foot, so it’s not huge.
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3. Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 30 seconds to speed up from rest to its top speed of 1 rotation every 1.3
seconds. The astronaut is strapped into a seat 6.0 m from the axis.
(a) What is the astronaut’s tangential acceleration during the first 30 seconds?
(b) How many g’s of acceleration does the astronaut experience when the device is
rotating at top speed (each 9.8 m/s2 of acceleration is 1 g)?
————————————————————————————————————
Solution
. The
(a) The tangential acceleration is how fast the speed is changing, atang = ∆v
∆t
velocity is the total distance around the circle, 2πR = 2π × 6 = 12π, divided by
the time it takes to go around T = 1.3, so v = 12π/1.3 ≈ 29 m/s. It took the
centrifuge 30 seconds to build up to this speed, so
atang =
29
v
=
≈ 0.97 m/s2 .
t
30
(b) When the centrifuge is up to speed, there is no tangential acceleration (the speed
2
isn’t changing), but there is rotational acceleration, arot = vR . Plugging in numbers gives
292
v2
=
= 140 m/s2 .
arot =
R
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Dividing this answer by 9.8 gives arot = 140/9.8 = 14.3g, which is high but not totally unrealistic for launch and reentry. If astronauts can handle the accelerations
in the centrifuge, then they should be able to handle spaceflight.
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4. Let’s look the maximum safe heights that a person could jump from in two different
landing cases.
(a) First, suppose a person of mass m jumps from a height h, and lands with straight
legs such that they stop in a small distance d, cushioned only by the padding on
the bottom of their feet. Using conservation of energy, show that the force acting
.
back up on them from the ground is F = mgh
d
(b) Upon compression bones typical break when subjected to a force per area above
about 1.7 × 108 N/m2 . If the tibia near the ankle has a radius of about 1 cm,
estimate the maximum height from which a 70 kg person could just land without
breaking any bones if the padding on the soles of their feet is about 1 cm.
(c) Now, suppose the person lands at the end of the jump using the muscles in his
knees to help cushion the collision. Now his center of mass falls a height h before
his feet make contact with the ground, but then he bends down, dropping his
center of mass a further distance s (we can ignore the padding on his feet now,
since it’s small compared to how much he bends). Show that the required force
is now
h
.
F = mg 1 +
s
(d) If he bends down 0.5 meters, what is the maximum height that he could just land
from without breaking any bones? (Kids - don’t try this at home!)
————————————————————————————————————
Solution
(a) When he hits the ground the floor pushes back up on him with a force F , doing
work against the gravitational force. The work done is the force times the distance
over which it works (the thickness of the padding), W = −F d, where the minus is
because the force is resistive. The work done is also equal to (minus) the change
.
in potential energy, W = −∆P E = mgh. Thus, F d = mgh, or F = mgh
d
(b) The tibia could resist a force of F = 1.7×108 A = 1.7×108 (π(0.01)2 ) ≈ 5.34×104
Fd
N without breaking. The safe height is, then h = mg
= 53400×0.01
= 80 centimeters!
70×9.8
So, jumping from a height of less than a meter could break your bones, if you
land straight-legged!
(c) In this case the person moves a total distance h + s, but the resistive force only
acts for a distance s. So, the work done is W = −F s = −mg (h + s). Solving for
the force gives
mg
h
F =
(h + s) = mg 1 +
.
s
s
F
(d) Solving for the height and plugging in the numbers gives h = s mg
−1 =
53400
0.5 70×9.8
− 1 ≈ 38 meters! (Remember that this is the absolute edge of not
breaking your bone! To land safely you should jump from a much smaller distance.)
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Extra Credit Question!!
The following is worth 10 extra credit points!
where and σ are constants,
and r is the distance between the molecules. The
potential energy is plotted in
the figure to the right. The
vertical axis is in units of ,
while the horizontal axis is
in units of σ.
Energy
The potential energy between a pair of neutral
atoms or molecules is very
well-approximated by the
Lennard-Jones
Potential,
given by the expression
σ 12 σ 6
,
P E(r) = 4
−
r
r
Molcular Bond Energy
8
7
6
5
4
3
2
1
0
0.75
-1
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
-2
-3
-4
Distance
(a) Why does the potential energy approach zero as the distance gets bigger?
(b) At what separation distance, in terms of σ and , is the potential energy zero?
(c) At approximately what distance is the system in equilibrium? What is the potential energy at that distance? (Express your answers in terms of σ and .)
(d) How much energy would you need to add to the system at equilibrium in order
to break the molecular bonds holding it together? Why?
(e) How much energy is released in the breaking of those molecular bonds? Why?
Note - no calculation is needed to answer these problems!
————————————————————————————————————
Solution
(a) As the two molecules get further apart, the attractive force between them gets
weaker and weaker. When the are very far apart, they hardly interact at all they are basically free molecules. The potential energy of a free particle is zero,
since potential energy depends on the interaction between multiple particles.
(b) We can just read the value off from the graph. We see that the potential energy
crosses the x axis when x = 1, which means that r = σ. We can see this from the
equation, too: setting r = σ gives P E(σ) = 0.
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3.25
(c) The system is in equilibrium when the net force on it is zero. Since the force is
the slope of the potential energy graph, this happens when the slope is zero. The
potential energy graph has zero slope when it’s at it’s minimum point. Checking
the graph, we see that this happens right around x ≈ 1.15, or r ≈ 1.15σ. We could
d
(P E(r)) = 0, which gives r = 21/6 σ ≈ 1.12σ,
check the exact answer by finding dr
and so we were close on our guess. The energy at this distance can just be read
off the graph, giving y = −3, or P E = −3.
(d) In order to break the molecular bonds apart, we’d need to raise the energy to
zero. At equilibrium the energy is P E = −3, and so we’d need to add +3 units
of energy.
(e) There is no energy released in breaking these molecular bonds - we had to add
the energy to break these bonds. Energy is never released in the breaking of
bonds! One can obtain energy by breaking a less stable bond, then forming a
more stable bond. The more stable bond has a more negative potential energy
(a deeper potential “well”). The difference in energy between the initial and final
states is released to the environment. This is where the energy comes from in the
ATP reactions, and not by releasing energy from the breaking of bonds!
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