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1
PHYSICS FOR ENGINEERS.
A-EXAMS 2006-2007.
A1 A2 A3 A4
2
A1A student raises one end of a board slowly, where a block lies. The block starts to move when the angle is 30º. The static friction coefficient between the block and the board surface should be:
30º
1. a)
b)
c)
d)
e) None of above
33
32
22
23
An object falls freely from rest on a planet without atmosphere where the gravity acceleration is 10 m s-2. Its velocity when it crashes against the ground is 5 m s-1, therefore its initial height was:
2. a)
b)
c)
d)
e) None of above
m 00.2
m .500
m .251
m 40.2
3. The angle formed by the vectors jiA
jiB
22 and is:
a)
b)
c)
d)
e) None of above
º45
rad 6
º90
rad 6
“An object at rest stays at rest unless acted on by an external force”. This is:
4.a)
b)
c)
d)
e) None of above
Newton’s third law
Law of inertia
Law of motion along a curved path
The work-energy theorem
A punctual mass follows a circular trajectory with constant speed. As for its acceleration, it is true that
a)
b)
c)
d)
e) None of above
The punctual mass is not accelerated
Its tangential acceleration is positive
The modulus of its normal acceleration is constant
There is no normal acceleration5.
A particle whose mass is (1.00±0.01)10-2 kg is moving along a straight path at (1.00±0.10) m s-1. Its momentum is
6. a)
b)
c)
d)
e) None of above
m/skg 1011.000.1 2- m/skg 0.01.10
m/skg 0.01.110
m/skg 100.110.1 4-
AGR
FOR
3
A1A spring initially at rest, attached to a mass m, is stretched to distance x. As for the mass m, it is true that:
7.
a)
b)
c)
d)
e) None of above
The potential energy of m only varies in case the mass hangs vertically on the spring
The potential energy of m varies if the mass hang vertically on the spring, but even if the spring and the mass lie on a horizontal surface.
The potential energy of m never changes, only its kinetic energy undergoes some variation.
Neither the potential energy of m nor its kinetic energy undergo any variation.
Two blocks of masses m1 and m2 lie on a horizontal table. On m1 we apply a horizontal force F0 as shown in the picture. The kinetic friction coefficients for both masses are 1 y 2, respectively. Using the numerical values given below, answer the following questions:
m1
m2
F0
Assuming the force F0 is big enough to move the set of two blocks, find its acceleration.
b)
c) Find the force exerted by the first block (m1) on the second one (m2) and the force exerted by the second one on the first one.
1 = 0.075 2 = 0.040 m1 = 8 kgF0 = 2,50 kp m2 = 6 kg
Separately draw the free body diagram for the set of two blocks, for m1 and for m2. a)
PROBLEM
GRADING:
PROBLEM: 6 POINTS
QUESTIONS: 4 POINTS
EACH CORRECT ANSWER: +0.500
EACH WRONG ANSWER: -0.125
A harmonic oscillator obeys the equation ty 90.9cos 5 where
every quantity is given in S.I. units. The frequency and the period are:
8. 4.5 Hz and 0.22 s a) 0.5 Hz and 2.0 s b)
6.3 Hz and 0.16 s c) 5 Hz and 9.90 s d)
None of abovee)
4
A1A student raises one end of a board slowly, where a block lies. The block starts to move when the angle is 30º. The static friction coefficient between the block and the board surface should be:
30º
1. a)
b)
c)
d)
e) None of above
33
32
22
23
An object falls freely from rest on a planet without atmosphere where the gravity acceleration is 10 m s-2. Its velocity when it crashes against the ground is 5 m s-1, therefore its initial height was:
2. a)
b)
c)
d)
e) None of above
m 00.2
m .500
m .251
m 40.2
3. The angle formed by the vectors jiA
jiB
22 and is:
a)
b)
c)
d)
e) None of above
º45
rad 6
º90
rad 6
“An object at rest stays at rest unless acted on by an external force”. This is:
4.a)
b)
c)
d)
e) None of above
Newton’s third law
Law of inertia
Law of motion along a curved path
The work-energy theorem
A punctual mass follows a circular trajectory with constant speed. As for its acceleration, it is true that
a)
b)
c)
d)
e) None of above
The punctual mass is not accelerated
Its tangential acceleration is positive
The modulus of its normal acceleration is constant
There is no normal acceleration5.
A particle whose mass is (1.00±0.01)10-2 kg is moving along a straight path at (1.00±0.10) m s-1. Its momentum is
6. a)
b)
c)
d)
e) None of above
m/skg 1011.000.1 2- m/skg 0.01.10
m/skg 0.01.110
m/skg 100.110.1 4-
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A1.
SOLUTION
5
A1A spring initially at rest, attached to a mass m, is stretched to distance x. As for the mass m, it is true that:
7.
a)
b)
c)
d)
e) None of above
The potential energy of m only varies in case the mass hangs vertically on the spring
The potential energy of m varies if the mass hang vertically on the spring, but even if the spring and the mass lie on a horizontal surface.
The potential energy of m never changes, only its kinetic energy undergoes some variation.
Neither the potential energy of m nor its kinetic energy undergo any variation.
Two blocks of masses m1 and m2 lie on a horizontal table. On m1 we apply a horizontal force F0 as shown in the picture. The kinetic friction coefficients for both masses are 1 y 2, respectively. Using the numerical values given below, answer the following questions:
m1
m2
F0
Assuming the force F0 is big enough to move the set of two blocks, find its acceleration.
b)
c) Find the force exerted by the first block (m1) on the second one (m2) and the force exerted by the second one on the first one.
1 = 0.075 2 = 0.040 m1 = 8 kgF0 = 2,50 kp m2 = 6 kg
Separately draw the free body diagram for the set of two blocks, for m1 and for m2. a)
PROBLEM
GRADING:
PROBLEM: 6 POINTS
QUESTIONS: 4 POINTS
EACH CORRECT ANSWER: +0.500
EACH WRONG ANSWER: -0.125
A harmonic oscillator obeys the equation ty 90.9cos 5 where
every quantity is given in S.I. units. The frequency and the period are:
8. 4.5 Hz and 0.22 s a) 0.5 Hz and 2.0 s b)
6.3 Hz and 0.16 s c) 5 Hz and 9.90 s d)
None of above
SOLUTION(CONTINUED)
e)
6
A1m1 m2
F0
F21
FR1
gmNFR 11111
gmNFR 22222
F0
F12
FR2
N2
FR1 FR2
ammFFF RR 21210
N1 N2
amm 21 Newton’s 2nd law
21
22110
mm
gmgmFa
Friction forces
Newton’s 2nd law
amFFF R 11210
amFFF R 11021
1 = 0.075 2 = 0.040 m1 = 8 kgF0 = 2,50 kp m2 = 6 kg
Numerical result a = 1.162 m/s2
N 324.91110 amgmF
Newton’s 2nd law
amFF R 2212
amFF R 2212 N 324.9222 amgm
am1
am2
F12 and F21 have to be equal (action and reaction)
N1
SOLUTION(CONTINUED)
7
A2a) The sphere rolls from the beginning on the horizontal surface.
b) The velocity of the center of mass remains constant, the angular velocity increases.
c) The velocity of the center of mass drops, the angular velocity remains null.
d) The sphere slides, but cannot roll because the friction is not zero.
e) None of the above.
P2. A skater on ice has her arms extended while she spins at 2 rps. When she cross her arms on her body, her moment of inertia halves. As a result, we’ll see that:
a) Her angular velocity halves.
b) Her angular velocity duplicates.
c) Her angular velocity reduces to a fourth.
d) Her angular velocity increases four times.
e) Her angular velocity remains constant, however her angular momentum duplicates.
P3. The figure depicts a ring of radious R and mass M, where the mass is homogenously shared out along its rim. The momenta of inertia about the different axis depicted in the figure are:
a)
b)
c)
d)
e) None of the aboveX
Y
Z222
2
1
2
1 MRIMRIMRI XXYYZZ
222
4
1
4
1
2
1MRIMRIMRI XXYYZZ
222
2
1
2
1
4
1MRIMRIMRI XXYYZZ
222
2
1
2
3 MRIMRIMRI XXYYZZ
P4. The variation of the angular momentum with time quantifies:
a) The sum of external forces
b) The sum of the momenta of inertia
c) The sum of the torques of external forces
d) The variation of the angular velocity
e) None of the above
Warning: to answer questions P1 to P5 you have to use the answer form, only the answers including there will be considered. For the choosen right answer, use the symbol . To invalidate a previously marked cell, use the symbol
P1. A homogeneous sphere is thrown on a rough horizontal floor. Its initial center of mass velocity is v, meanwhile its initial angular velocity is zero. What happens is:
X X
For questions P1 to P5, each correct answer adds +1; each wrong answer takes off -0.25
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A2E
QUESTIONS AGR
FOR
SURNAMES
FIRST NAME
8
A2P5. A torque of 10 Nm is acting on a spinning disc, which turns around a normal fixed axis passing through its simmetry center. If the rotational work is 5.24 J, then we can say that the angle gone over is (within ±1º):
a) 90º
b) 30º
c) 60º
d) 45º
e) 180º
a b c d e
P1
P2
P3
P4
P5
ANSWER FORM
A B
LL
A B
A thin rigid rod of lenght 2L = 80 cm is composed by two sections, each of lenght L, made on different materials whose linear densities are A = 2 kg/m and B = 5 kg/m.
The rod is hung horizontally from both ends on two threads A and B, as shown in the figure. Find:
a) The tension supported by each thread (1.5 p)
c) Assume the thread B is cut. Find the angular acceleration and the angular velocity of the rod when it reaches the vertical position (2 p).
b) The moment of inertia about end A (1.5 p)
PROBLEM
9
A2A B
0xLL
A B
CMx
mg
AT BT
a) Both sections are homogeneous and have the same lenght, so the position of both center of masses will be
2/LxA 2/3LxB
Center of mass of the whole rod:
Mass of each section:Lm AA
Lm BB
BA
BBAACM mm
xmxmx
BA
BACM mm
LmLmx
2/32/
LL
LLLLx
BA
BACM
2/32/
2
3 Lx
BA
BACM
Total mass: Lm BA
Y
02 LTxmg BCMA L
xmgT CM
B 2
0 BAy TTmgF
L
xmg CM
21 BA TmgT
BA
BACM
L
x
3
4
1
2
BA
BABAB gLT
3
4
1 gLT BAB 3
4
1
BA
BABAA gLT
3
4
11 gLT BAA 3
4
1
PROBLEM SOLUTION
10
A2A
LL
A B
0x
A B
L
L
B
L
A
xx2
3
0
3
3
3
333 8
3
1
3
1LLL BA
73
3
BAA
LI
mg
CMx
90
90sin ACMA Imgx
'AT
cos
A
CM
I
mgx
Lm BA
2
3 Lx
BA
BACM
cos
2
3
7
33
gLL
L BABA
BA
BA
L
g
BA
BA
2
cos3
7
3
b) Moment of inertia about A
mg
dx
x
dxdm
L
L
B
L
AA dmxdmxI
2
2
0
2
L
L
B
L
A dxxdxx
2
2
0
2
c) Angular acceleration and angular velocity
PROBLEM SOLUTION
11
A3The pipe shown in figure 1 is used to supply water to a reservoir. It has two open tubes, each in different parts of the pipe having cross sections S1 and S2 (see numerical values below).
1. The difference of height between the water level in both open tubes is h (see numerical value below). Find the volume flux and the mass flux in the pipe.
2. There is a pump P (figure 2) which takes out from the reservoir M kg of water per minute and carries them up to the height H at c0 m/s (see numerical values below). Find the power of the pump (neglect friction losses)
S1
h
1 2S2
H
0c
P
Figure 1
Figure 2
PROBLEM
S 1 (cm2) = 350
S 2 (cm2) = 180h (m) = 1
M (kg/min) = 4500c 0 (m/s) = 9,6
H (m) = 5
Gravity acceleration g = 9.8 m/s2; density of water = 1.00 g/cm3
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A3E.
Warning: to answer questions P1 to P5 you have to use the answer form, only the answers including there will be considered. For the choosen right answer, use the symbol . To invalidate a previously marked cell, use the symbolX X
For questions P1 to P5, each correct answer adds +1; each wrong answer takes off -0.25
E
12
A3Bernoulli’s law
S1
y1
h
1 2
1211 2
1gycP
S2
y2
2222 2
1gycP
c1c2
z2
z1
11 gzPP atm
22 gzPP atm
21
2221 2
1ccPP
2121 zzgPP
ghPP 21
1
221 S
Scc
21
222
22221 2
1
S
SccPP
Continuity equation
gh
212
21
2
SS
ghc
212
221
2
SS
ghSVV
212
221
2
SS
ghSmm
PROBLEM SOLUTION
13
A3H
0c
P
S
0
SSS yc
gg
P 2
2
1
020
0
2
1yc
gg
P
PH
Sy
0y0
SP yycg
H 0202
1
PHgmW
PHgdt
dm
dt
dWW mHgW P
kg/s 60
1
(s) 60
(kg) M
Mm PHgMW
60
1
Bernoulli’s LawPressures in S and 0 are the same The surface of the water is at rest
H
Hcg
H P 202
1
Work done by the pump raising a water mass m to a height H
Power:
The pump raises M kg a minute:
Figure 2
PROBLEM SOLUTION (CONTINUED)
2. There is a pump P (figure 2) which takes out from the reservoir M kg of water per minute and carries them up to the height H at c0 m/s (see numerical values below). Find the power of the pump (neglect friction losses)
14
A3 (kg/m3) = 1000 1000 1000 1000g (m/s2) = 9,8 9,8 9,8 9,8
h (m) = 0,25 1 0,5 0,4S1 (m
2) = 3,50E-02 3,50E-02 2,80E-02 2,80E-02
S2 (m2) = 2,00E-02 1,80E-02 1,50E-02 2,00E-02
P1-P2 (Pa) = 2450 9800 4900 3920
c2 (m/s) = 2,70 5,16 3,71 4,00
c1 (m/s) = 1,54 2,65 1,99 2,86
V (m3/s) = 5,39E-02 9,29E-02 5,56E-02 8,00E-02m (kg/s) = 53,95 92,92 55,61 80,02
g (m/s2) = 9,8 9,8 9,8 9,8M (kg/min) = 3000 4500 5000 800
c0 (m/s) = 8 9,6 10 5
H (m) = 4 5 2 8
HP (m) = 7,27 9,70 7,10 9,28
W (watt) = 3560,00 7131,00 5800,00 1212,00
MODEL E
PROBLEM SOLUTION (CONTINUED)
15
A3PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A3E. E
P1. A student blows up the tyres of his car in a place where the atmospheric pressure is 980 mb. When the gauge indicates 2.2 kp/cm2, the absolute pressure inside the tyres is:
a) 284200 Pa b) 3136 mb c) 3.05 kp/cm2 e) None of aboved) 3.45 kp/cm2
1 2 3
P2. Three identical glasses are filled up with water. The glass 1 contains just water; a cube of ice floats in the glass 2 (ice density 0.89 g/cm3), and a piece of wood floats in the glass 3 (wood density 0,75 g/cm3). We call W1, W2 y W3 the weights of the three glasses including their contents, then it is true that:
b) Ordered from more to less weight , the order is W1 > W2 > W3
c) Ordered from more to less weight , the order is W3 > W2 > W1
d) The weight of the three glasses is the same.
e) The weight of the glass containing just water is smaller than the weights of the other two glasses.
a) We cannot say which glass is more weighted without knowing the weights of the ice and the piece of wood.
C
1 m25 m2
1 m0.5 m
1 m
A
B
CP
P3. The initial level of the water in the system shown in the figure is indicated by A. The cylinder C is held in a fixed position (it cannot move). If we fill out the vertical thin tube up to the level B, what happens with the pressure PC exerted by the liquid against the cylinder C is:
a) Pressure PC multiplies by 3 its initial value.
b) Pressure PC keeps nearly exactly its initial value because the vertical tube is thin.
c) Pressure PC multiplies by 5 the its inital value.
d) Pressure PC duplicates.
e) None of the above.
QUESTIONS
16
A3F1
F2
F3
P4. The arrows depicted in the schemes F1, F2 and F3 indicate the velocities of the particles of a fluid moving to the right in a pipe. We can say about that:
b) The three velocity schemes are characteristic of turbulent flow
a) F1 represents laminar flow and F3 turbulent flow
c) The three velocity schemes are characteristic of laminar flow
d) The schemes F2 and F3 represent laminar flow; F1 represents turbulent flow
e) None of above
A
B C
P5. The figure shows a siphon. We try to transvase liquid from the left–hand to the right-hand vessel. The neccesary condition for the siphon to work is:
e) The P5 heading is wrong, because it is impossible that a siphon setup such as the depicted one could work.
a) It is necesary that h << d.
b) It is necesary that d.= 0
d) The pressure in A must be smaller than the pressure in B.
c) That siphon only works if the liquid is less dense than the water.
a b c d e
P1
P2
P3
P4
P5
ANSWER FORM
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A3E. E
QUESTIONS
17
A3gaugeatmabs PPP
mb 920atmP bar 920.0 bar
Pa 105 Pa 92000
2kp/cm 0.2gaugeP 2-4 m N/10 8.90.2 Pa 196000
A kp/cm2 mb bar PaPatm 0,9 882 0,882 88200Pgauge 2 1960 1,96 196000
Pabs 2,9 2842 2,842 284200
B kp/cm2 mb bar PaPatm 1 980 0,98 98000Pgauge 2,2 2156 2,156 215600
Pabs 3,2 3136 3,136 313600
C kp/cm2 mb bar PaPatm 0,95 931 0,931 93100Pgauge 2,1 2058 2,058 205800
Pabs 3,05 2989 2,989 298900
D kp/cm2 mb bar PaPatm 1,05 1029 1,029 102900Pgauge 2,4 2352 2,352 235200
Pabs 3,45 3381 3,381 338100
QUESTION 1 SOLUTIONEvery floating body drives out of the glas a volume of liquid equal to its weight. Therefore, the weights of the three glasses are the same.
QUESTION 2 SOLUTION
QUESTION 3 SOLUTION
The lower part of the cylinder C lies initially 0.5 m below the free surface of the water (level A). When the vertical thin tube is fillep up to level B (1 m besides level A), the new deep of the lower part of cylinder C is three times the initial one, so the pressure in C multiplies 3 times.
QUESTION 4 SOLUTION
See theory.
QUESTION 5 SOLUTION
See the siphon problem.
18
A410 mol of an ideal gas (initial temperatue 20 ºC) expand reversibly following the law p = aV, where p, V are pressure and volume, whereas a is a constant. The final volume after the expansion is twice the initial one. Gasses constant R = 8.314 J/(Kmol).
1. Plot the expansion process in a p-V diagram (1 p).
PROBLEM
A
2. Find the final temperature (1.5 p).
3. Find the work in the expansion (1.5 p).
P1. About the phase diagram depicted aside, it can be said that:
a) 1 is the critical point; 2 is the triple point; AB is a sublimation process.
b) 1 is the triple point; 2 is the boiling point; AB is a melting process.
c) 1 is the melting point; 2 is the boiling point; AB is an evaporation process.
d) 1 is the triple point; 2 is the critical point; AB is a sublimation process.
e) None of the above.
QUESTIONS
T
P
T
P
1
2
AB
P2. An ideal gas expands reversibly without changing its temperature. The gas absorbs 200 Kcal. The internal energy variation in this process is:
a) -200 kJ b) +200 Kcal c) -200 Kcal e) Zerod) +200 KJ
ANSWER FORM
a b c d eP1P2P3P4P5P6
PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A4E.
Warning: to answer questions P1 to P6 you have to use the answer form, only the answers including there will be considered. For the choosen right answer, use the symbol . To invalidate a previously marked cell, use the symbolX X
For questions P1 to P6, each correct answer adds +1; each wrong answer takes off -0.25
19
A4QUESTIONS
a) The internal energy variation of the gas is positive if the cycle is described is clockwise direction.
b) The entropy variation of the gas once the cycle is completed is positive if any of the cycle steps is irreversible.
c) The heat absorbed by the gas is always the same than the work done by the gas.
d) The internal energy variation of the gas once the cycle is completed is not well-defined, it depends on the way.
e) None of the above.
P3. An ideal gas complete a thermodynamics cycle composed by two adiabatics, one isochoric and one isobar. Indicate the correct option.
P4. A 20-ton iceberg fall of from a coastal glacier. The temperature of the ice is -5º C. The specific heat of the ice is 0.50 Kcal/kgK, ans its latent heat is 80 Kcal/kg. The energy needed to melt a half of the mass of the iceberg is:
a) 8105 Kcal b) 1.6105 Kcal c) 82.5104 Kcal d) 25105 Kcal e) None of the above
aQ
bQ
W
aT
bT
P5. About the picture on the right, W is work, and about the temperatures, Ta > Tb. The energies transferred are Qa = 60 KJ and Qb= 50 kJ. Using those data, we can say that the picture is a representation of:
a) A heat engine, whose yield is 20%
b) A refrigerator, whose COP is 5.
c) A heat engine, whose yield is 6/5.
d) A refrigerator, whose COP is 5/6.
e) None of the above
P6. A system absorbs 300 Kcal from a source at 300 K during a reversible process. The variation of entropy of the universe, once the process is completed, is:
a) Zero b) +1 Kcal/K c) -1 Kcal/K e) None of the aboved) +1 KJ/K
APHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A4E.
20
A4PROBLEM
PHYSICS FOR ENGINEERS. 2006-2007. EXAM A4. A
p
V
1
2
V1 V2=2V1
aVp 11 aVp
22 aVp
nR
VpT 11
1
nR
VpT 22
2
We know T1 = 20º C = 293 K
We must find a relationship to calculate T2 from T1 and the volume rate V2/V1.
22
11
2
1
Vp
Vp
T
T
2
1
2
1
V
V
p
p
2
2
1
V
V
12 mVV
21
m1
22 TmT
(in this case m = 2)
C º899K 11729322 21
22 TmT
Expansion work:
2
1
2
1
2
1
2 2
V
V
V
V
V
V
Va
dVVapdVW 21
221
22 1
2
2Vm
aVV
aW
nR
aVT
21
1 a
nRTV 12
1
12 1
2Tm
nR
J 36540932 12 2
314.810 2
W
10 mol of an ideal gas (initial temperatue 20 ºC) expand reversibly following the law p = aV, where p, V are pressure and volume, whereas a is a constant. The final volume after the expansion is twice the initial one. Gasses constant R = 8.314 J/(Kmol).
1. Plot the expansion process in a p-V diagram (1 p).
2. Find the final temperature (1.5 p).
3. Find the work in the expansion (1.5 p).