Physics 18 Spring 2011 Homework 6 - Solutions ... - Faculty

Physics 18 Spring 2011
Homework 6 - Solutions
Wednesday February 23, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish
them. The homework is due at the beginning of class on Wednesday, March 2nd. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. The metabolic rate is defined as the rate at which the body uses chemical energy to
sustain life functions. The average metabolic rate has been found to be proportional to
the total skin surface area of the body. The surface area for a 5-ft, 10-in male weighing
175 lb is about 2.0 m 2 , and for a 5-ft, 4-in female weighing 110 lb it is approximately
1.5 m 2 . There is about a 1 percent change in surface are for every three pounds above
or below the weights quoted here and a 1 percent change for every inch above or below
the heights quoted.
(a) Estimate your average metabolic rate over the course of a day using the following
guide for metabolic rates (per square meter of skin area) for various physical
activities: sleeping, 40 W/m 2 ; sitting, 60 W/m 2 ; walking 160 W/m 2 ; moderate
physical activity, 175 W/m 2 ; and moderate aerobic exercise, 300 W/m 2 . How do
your results compare to the power of a 100 W light bulb?
(b) Express your average metabolic rate in terms of kcal/day (1 kcal = 4.19 kJ). (A
kcal is the “food calorie” used by nutritionists.)
(c) An estimate used by nutritionists is that each day the “average person” must eat
roughly 12-15 kcal of food for each pound of body weight to maintain his or her
weight. From the calculations in Part (b), are these estimates plausible?
————————————————————————————————————
Solution
(a) The total energy required for all the activities is
E = Esleeping + Esitting + Ewalking + Emodphys + Emodaer.
We just need to figure out the energies for each case. We know the power (per
unit area) for each, so E = AP ∆t, where A is the surface area of the body, and
∆t is the amount of time spend doing the activity. Thus, assuming that you sleep
for 8 hours, walk for 2 hours, sit for 8 hours, engage in moderate activity for 5
hours, and engage in moderate aerobic activity for 1 hour, we have
∆Esleeping = 2.0 × 40 × 8 × 3600 = 2.3 MJ
∆Esitting = 2.0 × 60 × 8 × 3600 = 3.5 MJ
∆Ewalking = 2.0 × 160 × 2 × 3600 = 2.3 MJ
∆Emodphys = 2.0 × 175 × 5 × 3600 = 6.3 MJ
∆Emodaer = 2.0 × 300 × 1 × 3600 = 2.2 MJ
1

Thus, altogether, we have ∆E = 16.6 = 17 MJ. This is the energy over the course
of a day. So, the power is P = ∆E 17×106 = = 200 W, which is twice that of a
∆t 24×3600
100 W light bulb.
(b) If 1 kcal is 4190 J, then P = 17×106
4190 day
= 4060 kcal/day.
(c) For a 175 lb person, then 4060/175 = 23 kcal/day/lb, which is higher than these
estimates. However, these calculations are all approximate, based on the average
metabolic rates, etc. So, to be within a factor of 2 is pretty good.
2

2. Assume that your maximum metabolic rate (the maximum rate at which your body
uses its chemical energy) is 1500 W (about 2.7 hp). Assuming a 40 percent efficiency
for the conversion of chemical energy into mechanical energy, estimate the following:
(a) The shortest time you could run up four flights of stairs if each flight is 3.5 m
high,
(b) the shortest time you could climb the Empire State Building (102 stories high)
using your Part (a) result. Comment on the feasibility of you actually achieving
your Part (b) result.
————————————————————————————————————
Solution
The metabolic rate is just the power, P = ˙ W , where, as always, a dot indicates a
derivative with respect to time. So, if the efficiency is ɛ, then the amount of time it
takes to do work ∆W may be found by
ɛP = dW
dt
≈ ∆W
∆t
⇒ ∆t = ∆W
ɛP .
Now, in climbing a height h, the work done is ∆W = mgh. So,
∆t = mgh
ɛP
80 × 10 4
= h =
0.4 × 1500 3 h,
where we have assumed a mass of 80 kg. Now, we can answer the questions.
(a) Here the height of each floor is 3.5 meters, so ∆t = 4 × 4 × 3.5 ≈ 19 seconds.
3
(b) Now, we have to run up 102 flights of stairs, each flight having a height of 3.5
meters, for a total height of 102 × 3.5 = 357 m. This gives a time ∆t = 4 × 357 =
3
476 seconds, or about 8 minutes. It’s extremely unlikely that you could maintain
your maximum metabolic rate at full speed for eight minutes!
3

3. A 3.0 kg block slides along a frictionless
horizontal surface with a speed of 7.0
m/s. After sliding a distance of 2.0 m,
the block makes a smooth transition to
a frictionless ramp inclined at an angle
of 40 ◦ to the horizontal. What distance
along the ramp does the block slide before
coming momentarily to rest?
————————————————————————————————————
Solution
The block starts off with only kinetic energy, KE = 1
2 mv2 . Because the horizontal
surface is frictionless, the block has the same energy at the ramp. After the block
starts sliding up the ramp its kinetic energy changes to potential energy. Eventually,
all of the energy becomes potential energy, and the block reaches a maximum height
h, with P E = mgh. Energy conservation gives the height h = KE v2 = . Now, the
mg 2g
distance along the ramp is given in terms of the height the block reaches as d = h
sin θ ,
where θ = 40 ◦ is the angle of inclination of the ramp. Thus, we finally solve for the
distance
d = v2
2g sin θ =
7 2
2 × 10 × sin (40 ◦ )
4
= 3.8 m.

634
4. A girl of mass m is taking a picnic lunch to her grandmother. She ties a rope of length
R to a tree branch over a creek and starts to swing from rest at a point that is a
distance R/2 lower than the branch. What is the maximum breaking tension for the
rope if it is not to break and drop the girl into the creek?
Chapter 7
46 •• A ————————————————————————————————————
girl of mass m is taking a picnic lunch to her grandmother. She ties a
rope of length R to a tree branch over a creek and starts to swing from rest at a
Solution
point that is a distance R/2 lower than the branch. What is the minimum breaking
tension for the rope if it is not to break and drop the girl into the creek?
The girl is shown to the right. At the bottom
of her swing, we can determine the
tension in the rope using Newton’s laws.
Recalling that the net force is the centripetal
force we have
�
Fy = T − mg = mv2
R ,
Picture the Problem Let the system
consist of the girl and the earth and let
Ug = 0 at the lowest point in the girl’s
swing. We can apply conservation of
mechanical energy to the system to
relate the girl’s speed v to R. The force
diagram shows the forces acting on the
girl at the low point of her swing.
Applying Newton’s 2 nd law to her will
allow us to establish the relationship
between the tension T and her speed.
and so the tension is T = mg + mv2 . Now,
2
we just need to know what the velocity is at
the bottom of the path. This we can determine
from energy conservation. At the top
of the path the girl has a height R/2 above
the bottom. So, she has a potential energy
P E = mgh = mgR
. At the bottom, all
2
this energy is changed into kinetic, 1
2mv2 ,
giving a speed
Apply ∑ radial = radial ma
v = � gR.
F to the girl
So, plugging back into the tension gives
at her lowest point and solve for T:
Apply conservation of mechanical
energy to the system to obtain:
Substituting for Kf and Ui yields:
T = mg + mv2
R
R
T r
1
2
R
g 0 = U
r r
F = mg
2
v
T − mg = m
R
= mg + mgR = 2mg.
and R
2
v
T = mg + m
(1)
R
This is the tension for which all the forces balance. Beyond this, there will be a net
force down, and the rope will break.
W
= ΔK
+ ΔU
= 0
ext
or, because Ki = Uf = 0,
K
f
−U
i
= 0
2
2 R v
mv − mg 0 ⇒ = g
2 R
1
2
=
Substitute for v 2 /R in equation (1) T = mg + mg = 2mg
5
g

5. The 2.0 kg block in the figure slides down a
frictionless curved ramp, starting from rest
at a height of 3.0 m. The block then slides
9.0 m on a rough horizontal surface before
coming to rest.
(a) What is the speed of the block at the
bottom of the ramp?
(b) What is the energy dissipated by the
friction?
(c) What is the coefficient of kinetic friction
between the block and the horizontal
surface?
————————————————————————————————————
Solution
(a) The block starts at a height h, and so has an initial potential energy P E = mgh.
After it has reached the bottom of the ramp, all of its potential energy is now
kinetic, KE = 1
2mv2 . The conservation of energy gives v = √ 2gh. So, the block
is moving at a speed v = √ 2gh = √ 2 × 10 × 3 = 7.7 m/s.
(b) At the bottom of the ramp, before the rough surface, the block only has kinetic
energy, KE = 1
2mv2 = 1
2 × 2 × 7.72 = 60 J. At the end of the surface, the block
isn’t moving, and so it has zero energy. Thus, the energy dissipated by the friction
is Ef = 60 J.
(c) Recalling that the energy dissipated by friction is equal to the work done by
friction, Ef = Wf = µkmgd, where d is the distance the block moves along. So,
µk = Ef
, or mgd
µk = Ef
mgd =
6
60
2 × 10 × 9
= 0.33.

6. A small object of mass m moves in a horizontal circle of radius r on a rough table. It
is attached to a horizontal spring fixed at the center of the circle. The speed of the
object is initially v0. After completing one full trip around the circle, the speed of the
object is 0.5v0.
(a) Find the energy dissipated by friction during that one revolution in terms of m,
v0, and r.
(b) What is the coefficient of kinetic friction?
(c) How many more revolutions will the the object make before coming to rest?
————————————————————————————————————
Solution
(a) Initially, the object was moving at a speed v0, and so it had an initial kinetic
energy KEi = 1
2mv2 0. After running around the circle it is moving at half the
speed, and so has a final kinetic energy KEf = 1
2m (0.5v0) 2 = 1
8mv2 0. So, the
difference in energy is ∆KE = � �
1 1 2 − mv 2 8 0 = 3
8mv2 0, and is the energy lost to
friction. Thus,
∆E = 3
8 mv2 0.
(b) Now this energy is equal to the work done by friction, W = Ff∆ℓ, where Ff =
µkmg is the force due to friction, and ∆ℓ = 2πr is the total distance the object
travels in one orbit. Thus, W = 2πµkmgr = ∆E. Plugging in for ∆E = 3
8 mv2 0
gives
µk = 3v2 0
16πgr .
(c) The object has an energy of only KE = 1
8 mv2 0 left. This energy can do a work
against friction of W = µkmg∆ℓ = 3mv2 0 ∆ℓ. Setting this equal to the kinetic
16πr
energy gives for ∆ℓ
∆ℓ = 1
8 mv2 0 × 16πr
3mv2 =
0
1
(2πr) .
3
So, the object will only make 1/3 of a revolution before stopping.
7

7. If a black hole and a “normal” star orbit each other, gases from the normal star falling
into the black hole can have their temperatures increased by millions of degrees due to
frictional heating. When the gases are heated that much, they begin to radiate light in
the X-ray region of the electromagnetic spectrum (high-energy light photons). Cygnus
X-1, the second strongest known X-ray source in the sky, is thought to be one such
binary system; it radiates at an estimated power of 4 × 10 31 W. If we assume that 1.0
percent of the in-falling mass escapes as X ray energy, at what rate is the black hole
gaining mass?
————————————————————————————————————
Solution
If only one percent of the in-falling mass is released, then the energy released is E =
, then
0.01mc 2 . Since the power is dE
dt
P = dE
dt
= 0.01c2 dm
dt .
So, the rate at which the black hole is gaining mass is just dm,
or dt
Plugging in the values gives
dm
dt
dm
dt
P
= = 100P .
0.01c2 c2 4 × 1031
= 100P = 100 ×
c2 9 × 1016 = 4.4 × 1016 kg/s.
So, the black hole is eating lots of mass every second!
8

8. A large nuclear power plant produces 1000 MW of electrical power by nuclear fission.
(a) By how many kilograms does the mass of the nuclear fuel decrease in one year?
(Assume an efficiency of 33 percent for a nuclear power plant.)
(b) In a coal-burning power plant, each kilogram of coal releases 31 MJ of thermal
energy when burned. How many kilograms of coal are needed each year for a
1000 MW coal-burning power plant? (Assume an efficiency of 38 percent for a
coal-burning power plant.)
————————————————————————————————————
Solution
(a) The plant has a power of P = 1000 MW, which is 10 9 J/s. So, if one year is about
π × 10 7 seconds, then the total energy produced by the plant is
E = 10 9 × π × 10 7 = π × 10 16 J.
Now, Einstein told us that E = mc 2 . So, if the energy is released due to a change
in mass, then ∆m = E
c 2 . If the efficiency is ɛ, then the total change in mass is
Plugging in the numbers gives
∆m = E
=
ɛc2 ∆m = E
.
ɛc2 π × 10 16
0.33 × 9 × 10 1 6
≈ 1.1 kg.
(b) In this case, the coal releases a much smaller amount of energy, at a rate r = 31
MJ. In order to release the same amount of energy as the nuclear plant (π × 10 16
J), at an efficiency of 38%, we would need to burn
∆m = E
ɛr =
π × 10 16
0.38 × 31 × 10 6 ≈ 2.7 × 109 kg.
So, you can see that the nuclear energy requires much less mass!
9

9. In particle physics, the potential energy associated with a pair of quarks bound together
by the strong nuclear force is in one particular theoretical model written as the following
function: U (r) = − (α/r) + kr, where k and α are positive constants, and r is the
distance of separation between the two quarks.
(a) Sketch the general shape of the potential-energy function.
(b) What is a general form for the force each quark exerts on the other?
(c) At the two extremes of very small and very large values of r, what does the force
simplify to?
————————————————————————————————————
(a) The general shape of the potential
energy function is seen in the figure
to the right.
(b) The force is (minus) the slope of the
potential energy, so
F = − d
dr U(r).
Working out the derivatives gives
Solution
-5
-10
-15
-20
-25
F (r) = − d
�
−
dr
α
�
+ kr
r
Potential energy
15
10
5
Potential Energy Graph
0
0 0.5 1 1.5 2 2.5
= − α
− k.
r2 (c) For very small values of r, the first term in the force becomes important, and so
we get a large attractive force
F → − α
r 2 for r → 0,
while for very big values of r, the first term dies away, and so the force becomes
a constant attractive force
F → −k for r → ∞.
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Distance

10. In one model of a person jogging, the energy expended is assumed to go into accelerating
and decelerating the feet and the lower portions of the legs. If the jogging speed is v,
then the maximum speed of the foot and lower leg is about 2v. (From the moment
a foot leaves the ground, to the moment it next contacts the ground, the foot travels
nearly twice as far as the torso, so it must be going, on average, nearly twice as fast as
the torso.) If the mass of the foot and lower portion of a leg is m, the energy needed to
accelerate the foot and lower portion of a leg from rest to speed 2v is 1
2 m (2v)2 = 2mv 2 ,
and the same energy is needed to decelerate this mass back to rest for the next stride.
Assume that the mass of the foot and lower portion of a man’s leg is 5.0 kg and that he
jogs at a speed of 3.0 m/s with 1.0 m between one footfall and the next. The energy he
must provide to each leg in each 2.0 m of travel is 2mv 2 , so the energy he must provide
to both legs during each second of jogging is 6mv 2 . Calculate the rate of the man’s
energy expenditure using this model, assuming that his muscles have an efficiency of
20 percent.
————————————————————————————————————
Solution
The rate of the man’s energy expenditure is just the power, which is P = ∆E . Because
∆t
his muscles have an efficiency of only 20%, much of the energy is wasted, and so he has
to work a lot harder to provide the energy to run. Thus, if the power being expended
to run is P , then he has to expend power at a rate P/0.2 = 5P , in order to get out a
power P . So, he has to expend energy
P = 5 ∆E
∆t
= 56mv2
∆t = 30mv2 ,
where we have taken ∆t = 1 second, since we’re told that the energy is 6mv 2 per
second. Plugging in values gives, taking m = 10 kg for both legs,
P = 30mv 2 = 30 × 10 × 3 2 = 30 × 10 × 9 = 2700 W.
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