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Physics 18 Fall 2009
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 typical laboratory centrifuge rotates at 4000 RPM. Test tubes have to be placed into
a centrifuge very carefully because of the very large accelerations.
(a) What is the acceleration at the end of a test tube that is 10 cm from the axis of
rotation?
(b) For comparison, what is the magnitude of the acceleration a test tube would
experience if dropped from a height of 1.0 m and stopped in a 1.0 ms (10−3
seconds) encounter with a hard floor?
————————————————————————————————————
Solution
(a) Since it’s spinning, the test tube is subjected to a centripetal acceleration acent =
v 2 /r, where v is the rotational velocity and r is the rotational distance. Now,
we know the rotational frequency, f = 4000 rotations/min ÷ 60 sec/min ≈ 67
rotations per second. This means that the period is T = 1/f = 1/67 = 0.015
seconds. Now, the period is the time it takes to go around one orbit in the
centrifuge, and is the total distance around (2πr) divided by how fast it’s going
around (v). So,
2πr
,
T =
v
which means that v = 2πr/T . Thus, the acceleration is
acent
v2
(2π)2
=
=
r.
r
T2
Plugging in the numbers gives acent =
1790g!
(2π)2
.0152
× 0.1 = 1.75 × 104 m/s2 , which is
(b) If we drop a test tube, it experiences a big acceleration when it hits the ground.
Since it goes from some velocity vdrop to zero over a time ∆t, then the acceleration
is a = vdrop /∆t. What’s the initial speed? Since it’s dropped from a height −h,
under the constant acceleration due to gravity then vf2 = vi2 − 2g(−h) = 0 + 2gh,
since the tube is assumed
to have dropped
√ from√rest. The final velocity it gets
√
2gh
is just vf = vdrop = 2gh. Thus, a = ∆t
= 2×9.8×1
= 4400 m/s2 , which is
10−3
about 450g. So, the centrifuge subjects the tube to accelerations even greater
than hitting the ground!
1
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. To assess a patient who is suspected of having heart disease, the physician must examine
the cardiac function when (a) the patient is at rest with a heart beating at a normal
pace, and (b) when under stress, for example, after exercise. To simulate the conditions
of stress, the patient exercises by walking a treadmill to increase heart beat and sustain
high levels of cardiac stress.
Suppose a 50 kg patient exercises on the treadmill, angled at 30◦ from the ground,
exerting a constant force of 500 N up the slope of the treadmill while running a constant
velocity of 4 m/s along the treadmill for 5 minutes. The coefficient of friction of the
treadmill is µs = 0.45. Determine the work done by the patient.
————————————————————————————————————
Solution
The work, W , done by the patient is given
by
y
W = Fnet d cos θ,
F
where Fnet is the net force on the patient,
N
and d is the total distance that she runs.
Because the net force is along the direction
that she’s moving (she’s accelerating up
along the treadmill), the angle is zero, and
so cos θ = 1. Thus, we just find W = Fnet d.
P
So, in order to proceed, we need to know
the total force acting on the patient.
We can draw the force diagram seen to
f
the right, orienting our coordinate system along the treadmill to make life eas30
ier. Here F~N is the normal force, F~P is
the constant force exerted by the patient,
F~G = mg is her weight, and F~f is the fricG
tional force.
Let’s look at the forces along the x and y directions. Doing so gives
P
◦
P Fx = −Ff − mg sin 30 +◦ FP = Fnet
Fy =
FN − mg cos 30
= 0.
x
F
F
F
Solving the y components gives FN = mg cos 30◦ . Recalling that the frictional force
Ff = µs FN gives Ff = µs mg cos 30◦ . Plugging this back into the x component expression we have the net force, Fnet = FP − mg (sin 30◦ + µs cos 30◦ ).
Now, we need the distance. Since the patient ran at a speed v for a time t, the
“distance” she ran along the treadmill is d = vt = 4 m/s × 5 min × 60 seconds = 1200
meters. Thus,
◦
W = Fnet d = [FP − mg (sin30◦ + µs cos 30
)] vt
√
3
1
= 500 − 50 × 9.8 × 2 + 0.45 × 2
× 1200
4
=
7.69 × 10 J.
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4. Suppose you eat a Snickers bar that contains 280 Calories.
(a) If one Calorie is 4187 joules, then how much energy is contained in the Snickers
bar?
(b) Digestion requires some energy, and so only about 85% of the food energy is
directly available for other uses (like living). How much energy is this?
(c) Suppose you want to work off the candy by climbing stairs. Because the efficiency
of the muscles is relatively low, only about 20% of the food energy is available for
conversion to mechanical energy. If you have a mass of 80 kg, and a single step is
15 centimeters tall, how many steps would you be able to climb from the energy
in part (b) available to you?
(d) How many steps would you need to climb in order to work off all the Calories
from the candy?
————————————————————————————————————
Solution
(a) Since one Calorie is 4187 joules, then the Snickers bar contains 280 × 4187 =
1.2 × 106 J.
(b) Since only 85% is available for other uses, then we have available 0.85×1.2×106 =
1.02 × 106 J to us.
(c) Due to the low efficiency, only 20% of the energy can be used for mechanical work.
So, we have only 0.2 × 1.02 × 106 = 2.04 × 105 J that we can use from the candy
bar. The energy that we gain in going up a height h is just mgh. This potential
energy that we gain came at the expense of the food energy that we’re burning.
So, Efood = mgh, which says that we can climb a height
h=
2.04 × 105
Efood
=
= 260 meters.
mg
80 × 9.8
If one step is 15 centimeters tall, then it would take 260/0.15 = 1733 stairs to
completely burn off the energy available from the candy! Fortunately, other life
processes are burning the Calories, too!
(d) The total energy in the candy was found in part (a). To completely work off all
this energy would require climbing to a height of
h=
Efood
1.2 × 106
=
= 1531 meters,
mg
80 × 9.8
which is 10,200 steps!! You would have to climb very nearly a mile up!
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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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