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Physics

PHY 303K

Rodenborn

Spring

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Version 001 – Test 3 Practice Test – rodenborn – (57770) 1
This print-out should have 20 questions. 002 10.0 points
Multiple-choice questions may continue on Ablkidonhepofampas
the next column or page – ﬁnd all choices shown in the ﬁgure below. The block is re-
before answering. leased from rest on the frictionless incline.
When the block reaches the ﬂat portion of
Acceleration of an Oscillator the table it begins to feel a frictional force
001 10.0 points with the table characterized by a coe▯cient
Amilamiaiasmied of kinetic frictiok µ .
0.85 m and period 2.3sec.
What is the maximum acceleration?
h
1. 1.00959 m/s2 µk
Which of the following statements about
2. 3.17171 m/s2 the energy and work of the system could
NOT be correct?
2
3. 6.34341 m/s correct
1. The sum of the potential energy and the
4. 0.369565 m/s2 kinetic energy of the block at any point along
2 the entire path of travel is conserved. correct
5. 14.5898 m/s
2. Along the incline, gravity does work on
6. 0.160681 m/s2 the block causing it to speed up.
Explanation:
3. The block’s gravitational potential en-
ergy loss is equal to kinetic energy gain as it
Let : A =0 .85 m and descends.
T =2 .3sec .
4. Along the incline, the potential energy of
For a simple harmonic oscillator, the dis- the system decreases from its initial value.
placement is
▯ ▯ 5. On the table, friction does work on the
2▯ block causing it to slow down.
x = Acos T t + ▯ ,
6. The normal force from the incline does no
so the acceleration is work on the block.
2 ▯ ▯2 ▯ ▯
d x 2▯ 2▯
a = dt2 = ▯A T cos T t + ▯ . 7. When the block reaches the bottom of
the incline, its kinetic energy reaches a maxi-
Since ▯1 < cos▯< 1, the maximum acceler- mum.
ation is
2 Explanation:
A = 4▯ A
max T2 The total mechanical energy of the block
2 (i.e.hemo fhepealngyadte
= 4▯ (0.85 m) kinetic energy) is not conserved. Mechani-
(2.3sec) cal energy is converted from potential energy
= 6.34341 m/s . and eventually is dissipated by work done by
friction.
Concepts of Energy Hewitt CP9 15 E13 Version 001 – Test 3 Practice Test – rodenborn – (57770) 2
▯
003 10.0 points =2 .2 dna
Which statement is wrong?
1. Adding the same amount of heat to two ▯ ▯
di▯erent objects will produce the same in- ▯T b T ▯bT =f8 .9 C ▯ 7.2 C
crease in temperature. correct ▯
=91 .7 C, so
2. When the same amount of heat produces
di▯erent changes in temperature in two sub-
Q gained,waterQ lost,brass
stances of the same mass, we say that they
have di▯erent speciﬁc heat capacities. m w p,w ▯T w m c b p,b▯T b
3. Each substance has its own characteristic
speciﬁc heat capacity.
c = m w p,w ▯T w
p,b mb▯T b
4. Temperature measures the average ki- ▯
netic energy of random motion, but not other (2.77 kg)(4186 J/kg · C)
= (0.52 kg)(91.7 C)
kinds of energy. ▯
▯ (2.2 C)
5. Di▯erent substances have di▯erent ther- ▯
= 534.969 J/kg · C .
mal properties due to di▯erences in the way
energy is stored internally in the substances.
Explanation: Power in a Car
The same amount of heat does not necessar- 005 10.0 points
ily produce the same increase in temperature
When an automobile moves with constant
in di▯erent objects. velocity, the power developed in a certain en-
gine is 113 hp.
Holt SF 10C 05 What total frictional force acts on the car
004 10.0 points at 35.4mi /h? 1 hp = 746 W and 1 m/s =
Brass is an alloy made from copper and zinc.
▯ 2.2374 mi/h.
A0 .52 kg brass sample at 98.9 Cirped
into 2.77 kg of water at 5.0 C. Correct answer: 5327.92 N.
If the equilibrium temperature is 7.2 C,
Explanation:
what is the speciﬁc heat capacity of brass?
The speciﬁc heat of water is 4186 J/kg · C.
▯ Let : P =113hp and
Correct answer: 534.969 J/kg · C.
v =35 .4mi /h.
Explanation:
Let : m b0 .52 kg ,
T =98 .9 C, W Fd
b P = = = Fv
m w2 .77 kg , t t
▯ P 113 hp
Tw=5 .0 C, F = v = 35.4mi /h = 5327.92 N .
Tf=7 .2 C, and
▯
cp,w=4186J /kg · C.
Falling Raindrops
006 10.0 points
▯T = T ▯ T =7 .2 C ▯ 5 C ▯ The force of air resistance on a raindrop is
w f w Version 001 – Test 3 Practice Test – rodenborn – (57770) 3
5.3 ▯ 10▯5 Nw ntali ril path is n times as long, the work done will be
n times as great. Clearly, the work by such
velocity of 4.9m /s. 2
The acceleration of gravity is 9.8m /s . afewlotlysezooerald
What is the mass of the raindrop? loop.
Conservative forces do work that depends
▯6
Correct answer: 5.40816 ▯ 10 kg. only on the end-points of the integral (and is
Explanation: completely independent of the path).
When the raindrop is falling with constant
velocity the sum of forces acting on it is zero. MandI: Strange Potential
008 10.0 points
Fnet =0= mg ▯ F air Consider a uniform conservative force on some
strange distant planet that is similar to the
mg = F air
F air gravitational force but pointing at an angle ▯
m = g to the vertical direction,
5.3 ▯ 10▯5 N
= 9.8m /s2 F = ma sin▯ ˆı ▯ ma cos▯ ˆ▯.
= 5.40816 ▯ 10 ▯6 kg y
P(x,y)
MandI: Conservative Force Concept
007 10.0 points
Frictional forces notoriously fail to satisfy the
x
criterion for a conservative force; iet-,t O(0,0)
tal work done by frictional forces does not
necessarily equal zero when integrated over a
▯
loop, ▯
f · dr ▯=0 .
f If you consider the planet that creates this
force and an object of mass m as one system,
Why don’t frictional forces satisfy the “con- what is the change in potential energy moving
servative” criterion? the object from O(0,0) to P(x,y)?
1. Frictional forces are always proportional 1. ▯U =+ max sin▯ ▯ may tan▯
to the force of gravity.
2. None of these 2. ▯U =+ max tan▯ ▯ may tan▯
3. ▯U =+ max cos▯ ▯ may sin▯
3. Frictional forces can only do zero or neg-
ative work. correct
4. ▯U = ▯max tan▯ + may tan▯
4. Frictional forces always act to decrease 5. ▯U = ▯max sin▯ + may cos▯ correct
the velocity v of an object.
6. ▯U = ▯max cos▯ + may tan▯
5. Frictional forces are never constant, but
are di▯erent at every point on an object’s 7. ▯U =+ max sin▯ ▯ may cos▯
path.
Explanation: 8. ▯U = ▯max cos▯ + may sin▯
Frictional forces are always opposite to the
displacement if there is one. Therefore if the 9. ▯U =+ max cos▯ ▯ may tan▯ Version 001 – Test 3 Practice Test – rodenborn – (57770) 4
10. ▯U = ▯max sin▯ + may tan▯ 4. PointS. correct
Explanation:
5. The particle remains stationary at point
A.
Let : Fx=+ ma sin▯,
Fy= ▯ma cos▯, Explanation:
Since the total energy is constant, the max-
Since the force is conservative, the workimum kinetic energy will occur when the po-
required to move the object from O to P is
tential energy is a minimum. Thus point S is
independent of the path taken. Consider a the correct answer.
straight horizontal path from O to (x,0), and
then a straight vertical path from (x,0) to P.
MandI: Potential Energy Graph
The work is 010 10.0 points
The following graph represents a hypothetical
W =+ F xx+ F yy so potential energy curve for a particle of mass
=+ max sin▯ ▯ ma y cos▯. m.
U(r)
The change is potential is the negative of
the work done by the force or
3U 0
▯U = ▯W 2U 0
= ▯max sin▯ + ma y cos▯.
U 0
r
O r
Complicated Potential 02 0 2r0
009 10.0 points If the particle is released from rest at posi-
Use the potential orpoitiss.op tion 0 ,whatwillbeitsspeed▯▯▯ at position
shown below to answer the following questio2r0?
Acsmfoi A and ▯
moves in the potential U(x).Suppose the 4U
1. ▯▯▯ = 0 correct
mechanical energy of the system is conserved. m
▯
A U(x) U 0
T 2. ▯▯▯ = 2m
x ▯
U 0
3. ▯▯▯ =
▯ 6m
2U 0
Z 4. ▯▯▯ =
S m
▯
5. ▯▯▯ = 8U 0
At which position(s) will the kinetic energy m
▯
of the particle have its maximum value? U 0
6. ▯▯▯ = 8m
1. PointT. ▯
U 0
7. ▯▯▯ =
2. PointZ. ▯ 4m
U 0
8. ▯▯▯ =
3. PointsZ and S. m Version 001 – Test 3 Practice Test – rodenborn – (57770) 5
=2(6 .67259 ▯ 10▯11Nm /kg )2
▯
9. ▯▯▯ = 6U 0 (6.77 ▯ 104 kg)
m ▯ 8 2
(2.99792 ▯ 10 m/s)
Explanation: = 0.0100524 m .
The total energy of the particle is con-
served. So the change of the potential en-
ergy is converted into the kinetic energy of Technically speaking, in a region where
the particle, which gives
gravity is extremely intense, Newton’s me-
1 2 chanics cannot be used. Rather, one needs to
mv =3 U ▯ 0 0 apply the “general theory of relativity” devel-
2 ▯
4U 0 oped by Albert Einstein.
v = . It turns out that the theory of relativity
m gives the same expression for this limiting
MandI: Black Hole Radius radius, referred to as the “Schwarzschild ra-
dius”.
011 10.0 points
Ablackholeisanobjectsoheavythatneither
matter nor even light can escape the inﬂuence Potential Graph 03
012 10.0 points
of its gravitational ﬁeld. Since no light can The graph is a potential energy diagram for
escape from it, it appears black. Suppose a
body with the same mass as the Earth 6 .77 ▯ animes satof
24 position x.
10 kg is packed into a small uniform sphere.
Based on Newtonian mechanics, determine
the radius 0 at which this body would appear
to be a black hole. B
Hint: The escape speed at the sur-
face of the body must be equal to the x
speed of light c =2 .99792 ▯ 10 m/s. The U
universal gravitational constant is G = A
6.67259 ▯ 10▯11Nm /kg .2
x
0 x 2x 3x 4x 5x
Correct answer: 0.0100524 m.

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