Download Physics 18 Spring 2011 Homework 3

Physics 18 Spring 2011
Homework 3 - Solutions
Wednesday February 2, 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, February
9th. 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. Astronauts in apparent weightlessness during their stay on the International Space
Station must carefully monitor their masses because significant loss of body mass is
known to cause serious medical problems. Give an example of how you might design
equipment to measure the mass of an astronaut on the orbiting space station.
————————————————————————————————————
Solution
There are all sorts of ways that you might measure the mass of an astronaut. One
way would be to put the astronaut on a spring with known spring constant, k, set
him oscillating, and measure the oscillation frequency, which depends on the mass.
Another way might be to attach him to that spring, and spin him in a circle. Then you
can measure the expansion of the spring, which is also equal to the centripetal force.
In this case, you can relate the astronaut’s mass to the expansion of the spring.
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2. A net force of (6.0 N) î − (3.0 N) ĵ acts on a 1.5 kg object. Find the acceleration ~a.
What is the magnitude of the acceleration, a?
————————————————————————————————————
Solution
Newton tells us that F~ = m~a, where F~ is the applied force, m is the mass of the object,
and ~a is the acceleration the mass experiences. So, the acceleration is just ~a = F~ /m.
So, we just find
6.0î − 3.0ĵ F~
=
= 4.0î − 2.0ĵ m/s2 .
~a =
m
1.5
√
√
p
This is our acceleration. The magnitude is just a = a2x + a2y = 4.02 + 2.02 = 20 =
4.47 m/s2 .
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3. Seat belts and air bags save lives by reducing the forces exerted on the driver and
passengers in an automobile collision. Cars are designed with a “crumple zone” in the
front of the car. In the event of an impact, the passenger compartment decelerates over
a distance of about 1 m as the front of the car crumples. An occupant restrained by
seat belts and air bags decelerates with the car. By contrast, an unrestrained occupant
keeps moving forward with no loss of speed (Newton’s first law!) until hitting the
dashboard or windshield. These are unyielding surfaces, and the unfortunate occupant
then decelerates over a distance of only about 5 mm.
(a) A 60 kg person is in a head-on collusion. The car’s speed at impact is 15 m/s.
Estimate the net force on the person if he or she is wearing a seat belt and if the
air bag deploys.
(b) Estimate the net force that ultimately stops the person if he or she is not restrained
by a seat belt or air bag.
(c) How do these two forces compare to the person’s weight?
————————————————————————————————————
Solution
For a constant acceleration we can relate the initial and final velocities as vf2 = vi2 +
v2
i
. Thus, the
2a∆x. Since the final velocity is zero, then the acceleration is a = − 2∆x
mv 2
force acting on the passenger is F = ma = − 2∆xi . We just need to plug in the different
distances for each case.
(a) When the air bags deploy and the seat belts hold, then the stopping distance is
∆x = 1, and so
60 × 152
mv 2
= −6750 N,
F =− i =−
2∆x
2×1
where the negative just means that it’s pushing the passenger back into the seat.
(b) If there are no restraints, then ∆x = 5 mm, or ∆x = 0.005 m, and so
F =−
mvi2
60 × 152
=−
= −1.35 × 106 N!
2∆x
2 × 0.005
This is much bigger than the restrained case!
(c) The passenger weighs W = mg = 60 × 9.8 = 588 N, and so the result from part
(a) is about 11.5 times bigger than their weight, while the result from part (b) is
about 2300 times their weight! It’s clear which is a more survivable impact!
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4. A 35.0-kg traffic light is supported by
two wires as in the figure. (a) Draw the
light’s free-body diagram and use it to
answer the following question qualitatively: Is the tension in wire 2 greater
than or less than the tension in wire 1?
(b) Verify your answer by applying Newton’s laws and solving for the two tensions.
————————————————————————————————————
Solution
(a) The free-body diagram is seen in the figure
to the right. You’ll notice that the vectors
do add to zero, which, as we know, is the
case if the light isn’t accelerating. That is,
y
T~1 + T~2 + F~g = 0.
From the diagram, it seems pretty clear
that the tension T~2 > T~1 .
(b) Now, let’s check the math. The sum of
the forces in each direction has to be zero.
Since gravity points down, it doesn’t contribute to the forces along the x direction.
So,
X
Fx = T1 cos 30◦ − T2 cos 60◦ = 0.
The two forces cancel out in the horizontal
direction. So, we can express T2 in terms
of T1 , which gives
√
√
3/2
cos 30◦
T1 =
T1 = 3T1 ,
T2 =
◦
cos 60
1/2
and so, indeed, T2 > T1 , as we guessed.
4
T2
T1
60
30
m
Fg
x
5. A 65-kg student weights himself by
standing on a force scale mounted on a
skateboard that is rolling down an incline,
as shown in the figure. Assume there is no
friction so that the force exerted by the
incline on the skateboard is normal to the
incline. What is the reading on the scale
if θ = 30◦ ?
————————————————————————————————————
Solution
The reading that the scale gives is equal
to the normal force, the force that the
scale exerts on the skater. We can draw
a free-body diagram, choosing our coordinates such that the x axis points along the
ramp, as in the figure to the right. Then,
the normal force points entirely along the
y direction. The sum of the forces in the y
direction has to be zero, since the skater is
accelerating along the ramp (i.e., in the x
direction). So,
X
Fy = Fn − Fg cos 30◦ = 0.
Since the gravitational force is Fg = mg,
then
√
√
3
3
◦
mg =
(65)(9.8) = 552 N.
Fn = mg cos 30 =
2
2
So, the scale reads 552 N.
5
y
Fn
m
x
30
Fg
6. A block of mass m slides across a frictionless floor and then up a frictionless ramp.
The angle of the ramp θ and the speed of
the block before it starts up the ramp is
v0 . The block will slide up to some height
h above the floor before stopping. Show
that h is independent of m and θ by deriving an expression for h in terms of v0 and g.
————————————————————————————————————
Solution
Since the floor is frictionless, the block slides along
the floor at a constant speed, v0 . Once it starts
going up the ramp, then it’s slowed by gravity
pulling it back down. We can draw a force diagram for the block, orienting our coordinate system along the ramp,. In this case, the normal
force points entirely along y. Suppose that the
block moves along the ramp a distance d, as in
the figure. Then, from the geometry, it is up a
height h = d sin θ. So, if we can figure out how
far it goes along the ramp, then we’ll know our
answer.
y
Fn
x
m
d
h
θ
θ
Fg
We can figure out how far along the ramp it goes using the kinematic equations. Since
we don’t have the time it takes to travel up the ramp, we can use the velocity-distance
equation,
2
2
vxf
= vxi
+ 2ax ∆x,
where ax is the acceleration along the x direction (which isn’t g!). When the block
reaches its maximum height, it stops moving, and so vf = 0. So, if it started with an
initial velocity v0 , and travels a distance ∆x = d, then solving for d gives
d=−
v02
.
2ax
What’s ax ? We can determine this from Newton’s laws. Since
Pthe block is accelerating
along x, and the normal force points entirely along y, then
Fx = −mg sin θ = max .
v02
Thus, ax = −g sin θ. So, d = 2g sin θ . Plugging in our expression for h gives
h = d sin θ =
v2
v02
sin θ = 0 ,
2g sin θ
2g
which is independent of the mass and angle, as advertised.
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7. A block of mass m is being lifted vertically by a uniform rope of mass M and length
L. The rope is being pulled upward by a force applied to its top end, and the rope
and block are accelerating upward with an acceleration of magnitude a. Show that
the tension in the rope at a distance x (where x < L) above the block is given by
(a + g) [m + (x/L) M ].
————————————————————————————————————
Solution
The system is seen in the figure to the
right. The total amount of mass below the
point x is the mass of the block, m, plus
the mass of the segment of rope of length
x. The trick is figuring out how much that
mass is. Since the mass is uniform, every meter of the rope has the same mass.
So, M/L is a constant number, and says
how much mass a unit length of the rope
carries. So, if we take this number, and
multiply it by the length, x, this this tells
us how much mass in in the segment. So,
x. Thus, the total mass bemsegment = M
L
low the point x is
M
L
x
m
T
(m+(M/L)x)
Fg
M
x.
L
Now that we know what the mass is, we can determine the tension from a free-body
diagram, as seen above. The sum of the forces along y is
X
M
Fy = T − Mtotal g = T − m + x g.
L
Mtotal = m +
But, the force is the mass times the acceleration,
and themass is the total mass of the
P
block and segment, Mtotal . So, setting
Fy = m + M
x a and solving for T gives
L
M
T = m + x (g + a) .
L
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8. The figure shows a 20-kg block sliding on
a 10-kg block. All surfaces are frictionless
and the pulley is massless and frictionless.
Find the acceleration of each block and
the tension in the string that connects the
blocks.
————————————————————————————————————
Solution
We can draw free-body diagrams
for each block, as seen to the left.
The diagram for the 20 kg block
is first, while the 10 kg is on the
right. The top block has three
forces acting on it: the tension
pulling it along x, the gravitational
force pulling it down, and the
normal force pushing it up along
y. The bottom block is a bit more
complicated, having four forces.
There is still the tension, the
gravitational force, and the normal
force, but the top block is sitting
on the bottom block, weighing it
down a bit. So, there’s an extra
force, F~20 .
y
y
Fn
Fn
20
T
10
x
T
m20
20
m10
Fg
20
F20
Fg
Let’s start with the top block. Since the acceleration is in the x direction,
m20 ax . So, working out the forces in the x direction gives
X
Fx = T − m20 g sin 20◦ = −m20 ax ,
P
Fx =
where we let the acceleration be negative, since the top block will likely slide backwards,
since it’s heavier. Working out the forces for the second block gives
X
Fx = T − m10 g sin 20◦ = +m10 ax ,
where the acceleration is positive because it’s opposite to that of the top block (if the top
block slides down, then the bottom one is sliding up, etc.). So, solving the top for the
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x
tension and substituting in to the second gives (m20 − m10 ) g sin 20◦ = (m20 + m10 ) ax .
Solving for ax gives
m20 − m10
ax =
g sin 20◦ .
m20 + m10
This is the acceleration. We can find the tension by plugging this back into our force
equation for, say, the bottom block.
m20 − m10
◦
T = m10 ax + m10 g sin 20 = m10
g sin 20◦ + m10 g sin 20◦ ,
m20 + m10
or,
T =2
m10 m20
g sin 20◦ .
m10 + m20
So, plugging in the numbers gives
20 − 10
m20 − m10
◦
g sin 20 =
(9.8) sin 20◦ = 1.12 m/s2 .
ax =
m20 + m10
20 + 10
So, the bottom block slides up at 1.12 m/s2 . The top block slides down at the same
acceleration. Furthermore,
m10 m20
20 × 10
◦
T =2
g sin 20 = 2
(9.8) sin 20◦ = 45 N.
m10 + m20
20 + 10
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9. A 2.0 kg block rests on a frictionless wedge
that has a 60◦ incline and an acceleration
~a to the right such that the mass remains
stationary to the wedge. (a) Draw the
free-body diagram of the block and use
it to determine the magnitude of the
acceleration. (b) What would happen
if the wedge were given an acceleration
larger than this value? Smaller than this
value?
————————————————————————————————————
Solution
(a) The ramp is accelerating to the right. This
pushes against the block leading to a normal force. So, there are two forces acting on the block, the normal force and
the gravitational force. Since the acceleration is in the x direction, let’s work out
the
P sum of the forces in the x direction,
Fx = ma.
X
Fx = Fn cos 30◦ = ma.
So, in order to determine the acceleration,
we need the normal force, which we can
determine from the forces along the y direction. Since there
P is no acceleration in
the y direction,
Fy = 0. Now,
mg
.
Fy = Fn sin 30 −mg = 0 ⇒ Fn =
sin 30◦
Plugging this in for our acceleration gives
X
◦
a=
y
Fn
30
x
Fg
Fn
cos 30◦ = g cot 30◦ = (9.8) × cot 30 = 17 m/s2 .
m
(b) The normal force comes from the acceleration. An acceleration greater than 17
m/s2 would provide a bigger normal force than needed to balance the gravitational
force. If this is the case, then, because there is a vertical component to the normal
force, which is bigger than the gravitational force, there is an acceleration upwards,
due to the forces not balancing. So, the block would slide up the ramp. Conversely,
if the acceleration was less than 17 m/s2 , then the normal force wouldn’t quite
balance the gravitational force, and the block would slide down the ramp.
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10. Elvis Presley has supposedly been sighted
numerous times since his death on August
16, 1977. The following is a chart of
what Elvis’s weight would be if here were
sighted on the surfaces of other objects
in our solar system. Use the chart to
determine: (a) Elvis’s mass on Earth, (b)
Elvis’s mass on Pluto, and (c) the free-fall
acceleration on Mars. (d) Compare the
free-fall acceleration on Pluto to the
free-fall acceleration on the moon.
Planet Elvis’s Weight (N)
Mercury
431
Venus
1031
Earth
1133
Mars
431
Jupiter
2880
Saturn
1222
Pluto
58
Moon
191
————————————————————————————————————
Solution
(a) The weight, W , of an object on a planet with gravitational acceleration g is just
W = mg. So, on Earth, Elvis has a mass m = W/g = 1133/9.8 = 116 kg.
(b) Because the mass is an intrinsic property of an object, it doesn’t change from
place to place. So, Elvis’s mass on Pluto is exactly the same as his mass on
Earth, m = 116 kg.
(c) The free-fall acceleration on Mars, gMars can be found from Elvis’s weight on Mars,
divided by his mass, gMars = WMars /m = 431/116 = 3.72 m/s2 .
(d) The free-fall acceleration on Pluto, gPluto , is again, Elvis’s weight on Pluto, divided
by his mass, gPluto = WPluto /m. The free-fall acceleration on the Moon is, by the
same reasoning, gMoon = WMoon /m. Solving the second for the mass, and plugging
it into the first gives
WPluto
gMoon .
gPluto =
WMoon
Plugging in the weights tells us that gPluto = (58/191)gMoon = 0.3gMoon . So, the
free-fall acceleration on Pluto is less than 1/3 that of the Moon!
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