TIME OF COMPLETION_______________

Transcription

TIME
OF
COMPLETION_______________
NAME_______SOLUTION______________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1111, Exam 2
Version 1
Total Weight: 100 points
Section 1
November 2, 2005
1.
Check your examination for completeness prior to starting. There are a total of nine (9)
problems on seven (7) pages.
2.
Authorized references include your calculator with calculator handbook, and the Reference
Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are five (5) multiple choice and four (4) calculation problems. Work all multiple choice
problems and 4 calculation problems. Show all work; partial credit will be given for correct work
shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
6:00 p.m.
7:15 p.m
PROBLEM
POINTS
1-5
20
6
20
7
20
8
20
9
20
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR
PARTIAL CREDIT.
1. The quantity ½ kx2 is
a. The kinetic energy of the object.
b. The elastic potential energy of the object.
(4)
c. The work done on the object by the force.
d. The power supplied to the object by the force.
2. The total mechanical energy of the system
a. Is equally divided between kinetic energy and potential energy.
b. Is either all potential or all kinetic energy.
(4)
c. Can never be negative.
d. Is constant if there are no nonconservative forces.
3. Two objects collide and stick together. Kinetic energy
a. Is definitely conserved.
(4)
b. Is definitely not conserved.
c. Could be conserved or not conserved depending on the masses.
d. Is conserved only if there is no friction present.
4. Angular velocity is expressed in units of
a. Meters per second.
b. Radians per second.
(4)
c. Omegas per second.
d. Arcs per second.
5.
Negative work done on an object by an applied force implies that
a. The kinetic energy of the object increases.
(4)
b. The applied force is not constant.
c. The applied force is perpendicular to the displacement.
d. The applied force is directed opposite to the displacement.
6. An automobile engine slows down from 4500 rpm to 1200 rpm in 2.50 s.
Given: 4500 rpm
1200 rpm
t = 2.50 s
a. Express these angular velocities in terms of rad/s.
(4500 rev/min)(2 rad/rev)(1 min/60s) = 471 rad/s
(1200 rev/min)(2 rad/rev)(1 min/60s) = 126 rad/s
b. Calculate the angular acceleration of the engine, assumed constant.
t
(t = ((126 rad/s) – (471 rad/s))/(2.50 s) = -138 rad/s2
c. The total number of revolutions the engine makes in this time.
)
) = ()
) = (126 rad/s471 rad/s138 rad/s2) = 746 rad = 119 rev
7. A 1.00-kg box is initially at rest on a horizontal road. The coefficient of kinetic friction between
the box and the road is 0.200. The box is pushed a distance of 3.00 m by a force of 30.0 N applied to
the box at an angle of 25.0o below the horizontal. (Hint: the magnitude of the normal force that the
road exerts on the box is n = A sin + mg)
a. Find the work done by the applied force.
WA = (30.0 N)(3.00 m)cos(25.0o) = 81.6 J
b. Find the energy dissipated by friction.
fk = k n = k(A sin + mg) = (0.200)((30.0 N) sin(25.0o) + (1.00 kg)(9.81 m/s2)) = 4.50 N
Wfk = (4.50 N)(3.00 m)cos(180o) = -13.5 J
c. Find the change in the kinetic energy of the box.
Wnet K
Wnet = 81.6 N + (-13.5 N) = 68.1 J
K = 68.1 J
Bonus (5 points): Find the speed of the box after it has traveled 3.00 m.
K = Kf – Ki = Kf – 0 = Kf = ½ mV2
½ mV2K
V2K)/m
VK)/m)1/2
V = 11.7 m/s
8. A 2.00-kg block is pushed against a spring that has a spring constant of 500 N/m, compressing the
spring by 20.0 cm. The block is then released, and the spring projects it along a frictionless
horizontal surface. Right before a frictionless incline of angle 45o, there is another 3.00-kg block.
Two blocks collide and stick together.
(20)
a. What is the speed of the 2.00-kg block the moment it looses the contact
with the spring?
Ei = m g yi + ½ mVi2 + ½ kxi2 = 0 + 0 + ½ kxi2 = ½ kxi2
Ef = m g yf + ½ mVf2 + ½ kxf2 = 0 + ½ mVf2 + 0 = ½ mVf2
Ei = Ei
½ kxi2 = ½ mVf2
kxi2 = mVf2
Vf2 = kxi2 /m
Vf = (kxi2 /m)1/2
Vf = ((500 N/m)(0.200 m)2 /(2.00 kg))1/2 = 3.16 m/s
b. What is the speed of combined mass right after the collision?
m1V1i + m2V2i = (m1 + m2 )Vf
Vf = (m1V1i) / (m1 + m2 ) = (2.00 kg)(3.16 m/s) / (2.00 kg + 3.00 kg) = 1.26 m/s
c. How far up the incline do the blocks travel before momentarily coming to
rest?
Ei = m g yi + ½ mVi2 + ½ kxi2 = 0 + ½ mVi2 + 0 = ½ mVi2
Ef = m g yf + ½ mVf2 + ½ kxf2 = m g yf + 0 + 0 = m g yf
Ei = Ei
½ mVi2 = m g yf
Vi2 = 2 g yf
yf = Vi2 /(2 g) = 0.0816 m

L = yf / sin(45.0o) = 0.115 m

9. A small car of mass 1.20 Mg (1.20 x 103 kg) traveling east at 60.0 km/h collides at an intersection
with a truck of mass 3.00 Mg traveling west at 40.0 km/h.
a. The car and truck stick together. Find the velocity of the wreckage just after the collision.
m1V1i + m2V2i = (m1 + m2 )Vf
Vf = (m1V1i + m2V2i)/ (m1 + m2 ) = ((1.20 x 103 kg)(60.0 km/h) + (3.00 x 103 kg)(-40.0
km/h))/ (1.20 x 103 kg + 3.00 x 103 kg) = -11.4 km/h = -3.17 m/s
b. Calculate the decrease in kinetic energy that results from the collision.
Ki = ½ m1V1i2 + ½ m2V2i2 = ½ (1.20 x 103 kg) (16.7 m/s)2 + ½ (3.00 x 103 kg) (-11.1 m/s)2 = 352000
J = 352 kJ
Kf = ½ (m1 + m2)Vf2 = ½ (4.20 x 103 kg) (-3.17 m/s)2 = 21000 J = 21.0 kJ
K = 21.0 kJ – 352 kJ = -330 kJ
c. What happens to the “lost” kinetic energy?
The kinetic energy was transformed into the energy of sound waves, energy of deformation, internal
energy, etc.
TIME
OF
NAME___SOLUTION__________________________
PHYS 1111, Exam 2
Version 1
Section 1
November 6, 2006
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
2.
3.
4.
5.
There are six (6) multiple choice and four (4) calculation problems. Work all calculation
problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work
shown.
6.
7.
Start:
Stop:
6:00 p.m.
7:15 p.m
PROBLEM
POINTS
1-6
20
7
20
8
20
9
20
10
20
TOTAL
100
CREDIT
PERCENTAGE
PARTIAL CREDIT.
1. What is the SI unit of momentum?
a. N∙m.
b. N/s.
(4)
c. N∙s.
d. N/m.
2. Which of the following is an accurate statement?
a. The momentum of a projectile is constant.
b. The momentum of a moving object is constant.
(4)
c. If an object is acted on by a non-zero net external force, its momentum will not remain
constant.
d. If the kinetic energy of an object is doubled, its momentum will also double.
3. A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is
near the outer edge, and the girl is closer to the center. Who has the greater angular
displacement?
a. The boy.
b. The girl.
(4)
c. Both have the same non-zero angular displacement.
d. Both have zero angular displacement.
4. Rolling without slipping depends on
a. Kinetic friction between the rolling object and the ground.
b. Static friction between the rolling object and the ground.
(4)
c. Tension between the rolling object and the ground.
d. The force of gravity between the rolling object and the Earth.
5. If the net work done on an object is zero, then the object's kinetic energy
a. Decreases.
b. Remains the same.
(4)
c. Increases.
d. Is zero.
6. A ball falls from the top of a building, through the air (air friction is present), to the ground
below. How does the kinetic energy (K) just before striking the ground compare to the potential
energy (U) at the top of the building?
a. K is equal to U.
b. K is greater than U.
(4)
c. K is less than U.
d. It is impossible to tell.
7.
An automobile engine slows down from 4500 rpm to 1200 rpm in 2.50 s.
a. Express the initial and final angular velocities in rad/s.
0 = (4500 rev/min)(2 rad/rev)(1 min/60 s) = 471 rad/s
 = (1200 rev/min)(2 rad/rev)(1 min/60 s) = 126 rad/s
b. Calculate engine’s angular acceleration, assumed constant, and
 = 0 +  t
 = ( - 0) /t
 = (126 rad/s – 471 rad/s) / (2.50 s) = - 138 rad/s2
c. the total number of revolutions the engine makes in this time.

 = 02+ 2 ()
() = ( - 02)/ (2
() = ((126 rad/s)2 - (471 rad/s)2)/(2138 rad/s2 = 746 rad = 119 rev
8. A 6.0-kg block is pushed 8.0 m up a frictionless 37º inclined plane by a horizontal force of 75 N.
If the initial speed of the block is 2.2 m s up the plane calculate
a. The initial kinetic energy of the block;
Ki = ½ m Vi 2 = ½ (6.00 kg) (2.20 m/s) 2 = 14.5 J
b. The work done by the 75-N force;
WA = A d cos () = (75.0 N)(8.00 m)cos (37.0o) = 479 J
c. The work done by gravity;
Ww = w d cos (127.0o) = (m g) d cos (127.0o) = (6.00 kg) (9.81 m/s2) (8.00 m) cos (127.0o) = -283 J
d. The work done by the normal force;
Wn = n d cos (90o) = 0
e. The final kinetic energy of the block.
Wnet = 479 J + (-283 J) + 0 = 196 J
Wnet = Kf - Ki
Kf = Ki + Wnet = 14.5 J + 196 J = 211 J
9. A bullet is fired vertically into a 1.40-kg block of wood at rest directly above it. If the bullet
has a mass of 29.0 g and a speed of 510 m s , how high will the block rise after the bullet
becomes embedded in it?
Inelastic collision:
m1 V1i + m2 V2i = (m1 + m2)Vf
Vf = (m1 V1i + m2 V2i)/(m1 + m2) = [(0.029 kg)(510 m/s) + (1.40 kg)(0)]/(0.029 g + 1.40 kg) = 10.3
m/s
Upward motion (total energy is conserved):
m g yi + ½ m Vi 2 = m g yf + ½ m Vf 2
½ Vi 2 = g yf
yf = ½ Vi 2 / g = ½ (10.3 m/s)2 / (9.81 m/s2) = 5.41 m
10. Tarzan swings on a 30.0-m long vine initially inclined at an angle of 37.0o with the vertical.
What is his speed at the bottom of the swing
a. If he starts from rest?
m g yi + ½ m Vi 2 = m g yf + ½ m Vf 2
m g yi = m g yf + ½ m Vf 2
g yi = g yf + ½ Vf 2
Vf 2 = 2 (g yi - g yf)
Vf = [2 (g yi - g yf)]1/2
yi = L ( 1- cos ()) = 6.04 m
yf = 0
Vf = 10.9 m/s
b. If he pushes off with a speed of 4.00 m/s?
m g yi + ½ m Vi 2 = m g yf + ½ m Vf 2
g yi + ½ Vi 2 = g yf + ½ Vf 2
Vf 2 = 2 (g yi - g yf) + ½ Vi 2
Vf = [2 (g yi - g yf) + ½ Vi 2 ]1/2
yi = L ( 1- cos ()) = 6.04 m
yf = 0
Vf = 11.2 m/s
TIME
OF
NAME____SOLUTION_________________________
PHYS 1111, Exam 2
Version 1
Section 1
October 29, 2007
1.
2.
3.
4.
5.
shown.
6.
7.
Start:
Stop:
6:00 p.m.
7:15 p.m
PROBLEM
POINTS
1-6
20
7
20
8
20
9
20
10
20
CREDIT
TOTAL
100
PERCENTAGE
PARTIAL CREDIT.
A mass is attached to one end of a string. The other end is attached to a rigid support. The mass
is released at A and swings in a vertical arc to points B, C, and D, as shown below. Neglect air
resistance.
1. At what point does the mass have the most gravitational potential energy?
(A) A.
(B) B.
(C) C.
(D) D.
(E) None of the given points.
2. At what point does the mass have the most kinetic energy?
(A) A.
(B) B.
(C) C.
(D) D.
3. At what point does the mass have its highest speed?
(A) A.
(B) B.
(C) C.
(D) D.
4. A container of water is lifted vertically 3.00 m then returned to its original position. If the
total weight is 30 N, how much work was done?
(A) 45.0 J.
(B) 90.0 J.
(C) 180 J.
(D) No work was done.
(E) Negative work is done.
5. What is the SI unit of momentum?
(A) N∙m.
(B) N/s.
(C) N∙s.
(D) N/m.
(E) W.
6. The area under the curve on a Force versus time (F vs. t) graph represents
(A) Impulse.
(B) Momentum.
(C) Work.
(D) Kinetic energy.
(E) Does not represent a physical quantity.
7. The force on a 10.0 kg-mass object, acting along the x axis, varies as shown below. Determine
the work done by this force to move the object
(a) from x  0.0 to x  4.00 m, and
W = ½ (4.00 m)(500 N) = 1000 J
(b) from x  0.0 to x  6.00 m.
W = ½ (4.00 m)(500 N) - ½ (1.00 m)(200 N) – (1.00 m)(200 N) = 700 J
(c) If the object is at rest initially, what is the final speed of the object?
W = K
Kf = ½ mVf 2 = W
Vf 2 = 2 W /m
Vf = (2 W /m)1/2 = 11.8 m/s
8. A sled of 20.0 kg mass is initially given a shove up a 28.0º incline. Its initial speed is 5.15
m/s.
(a) How far would it move along the slope before coming to rest if the slope were frictionless?
Ei = Ef
Ei = ½ m Vi 2
Ef = m g yf
½ m Vi 2 = m g yf
½ Vi 2 = g yf
yf = (½ Vi 2)/g = 1.35 m
d = yf /sin (28.0º) = 2.88 m
(b) You find that the sled stops after traveling through a distance of 2.30 m (measured along the
slope). How much work is done on the sled by friction?
Wf = Ef – Ei = m g yf - ½ m Vi 2
yf = d sin (28.0º) = 1.08 m
Wf = -53.3 J
(c) Find the average friction force that acts on the sled.
Wf = f d cos (180o) = -f d
f = - Wf / d = - (-53.3 J)/(2.30 m) = 23.2 N
9. A 15.0-kg object moving in the  x direction at 5.5 m s collides head-on with a 10.0-kg object
moving in the  x direction at 4.0 m s . Find the final velocity of each mass if:
(a) the objects stick together;
m1V1i + m2V2i = (m1 + m2 )Vf
Vf = (m1V1i + m2V2i) /(m1 + m2 ) = 1.7 0m/s
(b) the 15.0-kg object is at rest after the collision;
m1V1i + m2V2i = m2 Vf
Vf = (m1V1i + m2V2i) /(m2 ) = 4.25 m/s
(c) the 10.0-kg object is at rest after the collision.
m1V1i + m2V2i = m1 Vf
Vf = (m1V1i + m2V2i) /(m1 ) = 2.83 m/s
10. A 6.00-kg block is pushed 8.00 m up a rough 37º inclined plane by a horizontal force of 75
N. If the initial speed of the block is 2.2 m s up the plane and a constant kinetic friction force of
25 N opposes the motion, calculate
(a) the initial kinetic energy of the block;
Ki = ½ mVi2 = ½ (6.00 kg)(2.20 m/s)2 =14.5 J
(b) the work done by the 75-N force;
WA = F d cos (37.0o) = (75.0 N) (8.00 m) cos (37.0o) = 479 J
(c) the work done by the friction force;
Wf = f d cos (180o) = (25.0 N) (8.00 m) cos (180o) = -200 J
(d) the work done by gravity;
Wg = w d cos (180o) = (58.9 N) (8.00 m) cos (127o) = -283 J
(e) the work done by the normal force;
Wn = 0
(f) the final kinetic energy of the block.
Wnet= K = Kf – Ki
Kf = Ki + Wnet
Kf = 14.5 J + (-4.00 J) = 10.5 J
TIME
OF
NAME_________SOLUTION____________________
PHYS 1111, Exam 2
Version 1
Section 1
March 30, 2005
1.
problems on nine (9) pages.
2.
3.
4.
5.
There are six (6) multiple choice and four (4) calculation problems. Do all problems. Show
all work; partial credit will be given for correct work shown.
6.
be located in the room.
7.
Start:
Stop:
10:30 a.m.
11:45 a.m.
PROBLEM
POINTS
1-6
24
7
19
8
19
CREDIT
9
19
10
19
TOTAL
100
POINTS EARNED
PARTIAL CREDIT.
Wt.
No.
04
1.
04
04
2.
5.
A block of mass, m, is released from rest at point A and slides down the frictionless
surface in the figure below. The kinetic energy of the block at point B is equal to
a.
(0.300)mgh.
b.
(0.700)mgh.
c.
mgh.
d.
None of the above.
If the work done by all the external forces acting on an object is 5.80 J, and the object
is initially at rest, its final kinetic energy is
a.
zero.
b.
2.90 J
c.
5.80 J
d.
cannot be determined.
A wheel is initially rotating with an angular velocity of 20.0 radians/s. If the wheel
rotates through an angular displacement of 100 radians while slowing down to rest,
the constant angular acceleration was
a.
-2.00 radians/s2.
 = o2 + 2  ()
b.
0.200 radians/s2.
 - o2)/ (2()) = 
rad/s) )/(2 x 100 rad)
 (20.0
2
`
Wt.
No.
04
4.
04
5.
c.
5.00 radians/s2.
d.
-200 radians/s2.
A person riding a bicycle is traveling due North, and decreases her linear velocity
from 10.0 m/s to 3.00 m/s. The direction of the angular acceleration vector of the
front wheel points
a.
East.
b.
West.
c.
North.
d.
downward.
The curve shown below represents the magnitude of the force exerted on a ball during
a collision with a wall plotted against time. The area under the curve represents:
a.
the force on the ball.
b.
the work done on the ball.
c.
the change in linear momentum of the ball.
d.
the power delivered by the wall to the ball.
PHYS 1111 Exam 2, Version 1
Spring 2005
2
Wt.
No.
19
7.
An applied force of magnitude A = 50.0 N, oriented at an angle of 35.0 with
respect to the vertical as shown in the diagram below, pushes a block of mass,
m = 3.00 kg, up a wall a distance of 1.80 m. The coefficient of kinetic friction
between the block and the wall is 0.230.
(04)
a.
Draw a free body diagram for the block.
(04)
b.
Calculate the work done by the applied force.
Spring 2005
3
WA = A d cos() = (50.0 N)(1.80 m)cos (35.0o)
=73.7J
(04)
c.
Calculate the work done by the weight of the block.
Ww = w d cos() = (3.00 kg)(9.80 m/s2)(1.80 m)cos (180o) = -52.9 J
(07)
d.
Calculate the work done by the kinetic friction force.
Wf = fk d cos() = k n d cos()
n = A sin (35.0o) = 28.7 N
Wf = (0.230)(28.7 N)(1.80 m)cos (180o) = -11.9 J
Wt.
No.
19
9.
(05)
A bullet of mass, m1 = 0.0200 kg, is fired horizontally into a block of wood of
mass, m2 = 8.00 kg. The initial speed of the bullet is 300 m/s. The block rests on
a horizontal surface.
a.
If the collision between the bullet and the block is perfectly inelastic, calculate
the final speed of the masses immediately after the collision.
m1V1i = (m1 + m2)Vf
Vf = m1V1i /(m1 + m2) = 0.748 m/s
Spring 2005
4
(14)
b.
A wooden block, m2 = 8.00 kg, as in part a above is now attached to an
unextended horizontal spring of spring constant of 300 N/m. A bullet,
m1 = 0.0200 kg, is again fired horizontally into the block and again undergoes
a perfectly inelastic collision. Assume the speed of the masses immediately
after the collision is 0.748 m/s.
If the maximum compression of the spring is 0.0668 m, calculate the
coefficient of kinetic friction between the block with the bullet and the
horizontal surface.
WNC = TEf - TEi = ½ k x2 – ½ (m1 + m2) Vf2
WNC = fk x cos (180o) = - (m1 + m2) g k x
- (m1 + m2) g k x = ½ k x2 – ½ (m1 + m2) Vf2
k = - (½ k x2 – ½ (m1 + m2) Vf2)/[(m1 + m2) g x]
k = 0.300
Wt.
No.
19
10.
A coin is located 0.250 m from the center of a rotating turntable. The coefficient
of static friction between the coin and the turntable is 0.500.
Spring 2005
5
(06)
a.
Draw a complete free body diagram for the coin.
(13)
b.
Calculate the magnitude of the maximum tangential velocity of the coin so
that it does not slip off of the turntable. Show your work.
FC = m aC = mV2/R
FC = fS = S n = S mg
mV2/R = S mg
Spring 2005
6
V2 = S g R
V = ( S g R)½
V = 1.11 m/s
TIME
OF
NAME____SOLUTION_________________________
PHYS 1111, Exam 2
Version 1
Section 2
March 29, 2006
1.
Check your examination for completeness prior to starting. There are a total of nine (9)
2.
3.
4.
5.
There are five (5) multiple choice and four (4) calculation problems. Work all multiple choice
problems and 4 calculation problems. Show all work; partial credit will be given for correct work
shown.
6.
7.
Start:
Stop:
6:00 p.m.
7:15 p.m
PROBLEM
POINTS
1-5
20
6
20
Spring 2005
7
CREDIT
7
15
8
20
9
15
TOTAL
100
PERCENTAGE
PARTIAL CREDIT.
1. When is kinetic energy conserved?
a. In elastic collisions
b. In inelastic collisions
(4)
c. In any collision in which the objects do not stick together
d. In all collisions
2. Angular acceleration is expressed in units of
a. Meters per second squared.
b. Radians per second squared.
(4)
c. Alphas per second squared.
d. Arcs per second squared.
3. A container of water is lifted vertically 3.0 m then returned to its original position. If the total
weight is 30 N, how much work was done?
Spring 2005
8
a. 45 J
b. 90 J
(4)
c. 180 J
d. No work was done.
4. The quantity 1/2 mv2 is
a. The kinetic energy of the object.
b. The potential energy of the object.
(4)
c. The work done on the object by the force.
d. The power supplied to the object by the force.
5. Describe the energy of a car driving up a hill.
a. Entirely kinetic.
b. Entirely potential.
(4)
c. Both kinetic and potential.
d. Gravitational.
6. A pitcher throws a baseball with mass 0.145 kg straight upward with an initial speed of
25.0 m/s.
a. How much work has gravity done on the baseball when it reaches a height of 20.0 m above the
pitcher’s hand?
Wg = m g yi - m g yf = 0 – (0.145 kg)(9.81 m/s2)(20.0 m) = -28.4 J
Spring 2005
9
b. Use the energy approach to calculate the speed of the baseball at a height of 20.0 m above
the pitcher’s hand. You can ignore air resistance.
m g yi + ½ m Vi 2 = m g yf + ½ m Vf 2
½ Vi 2 = g yf + ½ Vf 2
Vi 2 = 2 g yf + Vf 2
Vf = (Vi 2 - 2 g yf) 1/2
Vf = 15.3 m/s
c. Does the answer depend on whether the baseball is moving upward or downward at a
height of 20.0 m? Explain.
No, the speed of the baseball will not depend on whether the ball is moving up or down, the
velocity though will.
d. Bonus (5 points): Assuming that the ball reaches the height of 20.0 m with the speed of 10.0 m/s,
how much work was done on the baseball by air resistance?
WNC = Ef – Ei = (m g yf + ½ m Vf 2) – (m g yi+ ½ m Vi 2) = -9.61 J
7.
A bicycle wheel has an initial angular velocity of 1.50 rad/s.
a. If its angular acceleration is constant and equal to 0.300 rad/s2, what is its angular velocity
at t = 2.50 s?
o = 1.50 rad/s
 = 0.300 rad/s2
t = 2.50 s

 =o +t = (1.50 rad/s) + (0.300 rad/s2)(2.50 s) = 2.25 rad/s
Spring 2005
10
b. Through what angle has the wheel turned between t = 0 and t = 2.50 s?
 =0 +0 t +t2 = (1.50 rad/s)(2.50 s) + ½ (0.300 rad/s2)(2.50 s)2 = 4.69 rad
c. Draw the wheel and indicate the direction of its angular velocity and its angular
acceleration.
Assuming that the wheel rotates in counterclockwise direction, both angular velocity and angular
acceleration point out of the page.
8.
A 2.00 kg piece of wood slides on the surface shown below. The curved sides are perfectly
smooth, but the rough horizontal bottom is 10.0 m long and has a coefficient of kinetic
friction with the wood of 0.200. The piece of wood starts from rest 4.00 m above the rough
bottom.
a. Find the magnitude of the kinetic friction that acts on the wood as it moves over the rough
bottom.
fk = k n = k m g = (0.200)(2.00 kg)(9.81 m/s2) = 3.92 N
b. How much work is done by friction on the wood as the wood crosses the rough bottom?
Spring 2005
11
Wf = F d cos () = (3.92 N)(10.0 m)cos (180o) = -39.2 J
c. Where will this wood eventually come to rest?
Wf = Ef – Ei = (m g yf) – (m g yi)
Wf /(m g) = yf – yi
yf = Wf /(m g) + yi = 2.00 m

9. Your 1050-kg sports car, parked on a hill without the parking brake set, has rolled to the bottom of
the hill and is moving westward on a level road at a speed of 15.0 m/s. A truck driver driving
east on the same road sees the sports car coming, and decides to stop the car by running his
6320-kg truck into it. The two vehicles remain locked together after the collision.
a. At what speed should the truck be moving so that the truck and car are both stopped in the
collision?
m1 V1i + m2 V2i = (m1 + m2)Vf
Vf = 0
m1 V1i + m2 V2i = 0
m1 V1i = - m2 V2i
V1i = - (m2 V2i )/m1
V1i = 2.49 m/s
b. How much energy will be lost in the collision?
Ei = ½ m1 V1i 2 + ½ m2 V2i 2 = 137, 717 J
Ef = 0
E = -137, 717 J
Spring 2005
12
TIME
OF
NAME___SOLUTION__________________________
PHYS 1111, Exam 2
Version 1
Section 1
March 26, 2007
1.
2.
3.
4.
5.
shown.
6.
7.
Start:
Stop:
6:00 p.m.
7:15 p.m
PROBLEM
POINTS
1-6
20
7
20
8
20
9
20
10
20
Spring 2005
13
CREDIT
TOTAL
100
PERCENTAGE
PARTIAL CREDIT.
1. In a two-body collision, if the momentum of the system is conserved, then which of the
following best describes the kinetic energy after collision?
a. Must be less.
b. Must also be conserved.
(4)
c. May also be conserved.
d. Is doubled in value.
2. The impulse experienced by a body is equivalent to its change in:
a. Velocity.
b. Kinetic energy.
(4)
c. Momentum.
d. None of the above choices are valid.
3. The unit of power, Watt, is dimensionally the same as:
a. Joule-second.
b. Joule/second.
(4)
c. Joule-meter.
Spring 2005
14
d. Joule/meter.
4. As an object is lowered into a deep hole in the surface of the earth, which of the following is
true in regard to its gravitational potential energy?
a. Potential energy increases.
b. Potential energy decreases.
(4)
c. Potential energy stays constant.
d. Not enough information given.
5. A very light cart holding a 300 N box is moved at a constant velocity across a 15 m level
surface. What is the net work done in the process?
a. Zero.
b. 1/20 J.
(4)
c. 20 J.
d. 2000 J.
6. You drop a 60.0 g golf ball from 2.00 m high. It rebounds to 1.50 m. How much energy is
lost?
a. 0.294 J.
b. 0.500 J.
Spring 2005
15
(4)
c. 0.883 J.
d. 1.00 J.
7. A 55-kg hiker starts at an elevation of 1600 m and climbs to the top of a 3300-m peak.
a. What is the hiker’s change in potential energy?
Ui = m g yi
Uf = m g yf
U = m g yf - m g yi = m g (yf - yi) = (55.0 kg)(9.81 m/s2)(3300 m – 1600 m) =
917,235 J
b. What is the minimum work required of the hiker?
Wmin = 917,235 J, in the perfect world without friction, air resistance, etc.
c. Can the actual work done be more than this? Explain why.
Yes, because he will work against friction, air resistance, as well as possibly increase his kinetic
energy.
8. A 65-kg trampoline artist jumps vertically upward from the top of a platform with a speed of
5.0 m s .
a. How fast is he going as he lands on the trampoline, 3.0 m below?
yi = 3.00 m
yf = 0
Spring 2005
16
Vi = 5.00 m/s
Vf = ?
Ei = Ef
m g yi + ½ m Vi 2 = m g yf + ½ m Vf 2
g yi + ½ Vi 2 = ½ Vf 2
2 g yi + Vi 2 = Vf 2
Vf = (Vi 2 - 2 g yf) 1/2
Vf = 9.16 m/s
4
b. If the trampoline behaves like a spring with spring stiffness constant 6.2 10 N m , how far does
he depress it?
m g yi + ½ m Vi 2 = m g yf + ½ m Vf 2 + ½ k yf 2
now Vf = 0,
m g yi + ½ m Vi 2 = m g yf + ½ k yf 2
This is a quadratic equation which you need to solve for yf
yf = - 0.286 m
8. A skier traveling 12.0 m s reaches the foot of a steady upward 18.0º incline and glides
12.2 m up along this slope before coming to rest.
a. How much work was done by friction?
Wf = Ef – Ei = m g yf – ½ m Vi 2
Spring 2005
17
yf = (12.2 m)sin (18.0º) = 3.77 m, Vi = 12.0 m/s
Wf = -2,100,978 J
b. What was the magnitude of the average friction force acting on the skier?
Wf = f d cos (180o) = - f d
f = - Wf /d = - (- 2,100,978 J)/(12.2 m) = 172 N
10. A 9300-kg boxcar traveling at 15.0 m s strikes a second boxcar at rest. The two stick
together and move off with a speed of 6.0 m s .
a. What is the mass of the second car?
m1 V1i + m2 V2i = (m1 + m2)Vf
Vf = 6.00 m/s, V1i = 15.0 m/s, V2i = 0
m1 V1i = (m1 + m2)Vf
m1 V1i = m1 Vf + m2 Vf
m2 = (m1 V1i - m1 Vf )/Vf
m2 = (9300 kg)(15.0 m/s) – 6.00 m/s)/(6.00 m/s) = 13, 950 kg
b. How much kinetic energy was lost during collision?
Ei = ½ m1 V1i 2 = 1,046, 250 J
Spring 2005
18
Ef = ½ m1 Vf 2 + ½ m2 Vf 2 = 418,500 J
E = -627, 750 J
TIME
OF
NAME__SOLUTION___________________________
PHYS 1111, Exam 2
Version 1
Section 1
April 2, 2008
1.
2.
3.
4.
5.
shown.
6.
7.
Start:
Stop:
6:00 p.m.
7:15 p.m
PROBLEM
POINTS
1-6
20
7
20
8
20
Spring 2005
19
CREDIT
9
20
10
20
TOTAL
100
PERCENTAGE
PARTIAL CREDIT.
A mass is attached to one end of a string. The other
end is attached to a rigid support. The mass is
released from rest at A and swings in a vertical arc to
points B, C, and D, as shown below. Neglect air
resistance.
1. At what point does the mass have the least gravitational potential energy?
(A) A.
(B) B.
(C) C.
(D) D.
2. At what point does the mass have the least kinetic energy?
(A) A.
(B) B.
(C) C.
(D) D.
Spring 2005
20
3. At what point does the mass have negative gravitational potential energy?
(A) A.
(B) B.
(C) C.
(D) D.
(E) Answer depends on the choice of zero level for gravitational potential energy.
4. In an inelastic collision, if the momentum is conserved, then which of the following
statements is true about kinetic energy?
(A) Kinetic energy is also conserved.
(B) Kinetic energy is gained.
(C) Kinetic energy is lost.
(D) Kinetic energy can either be gained or lost.
(E) None of the above is correct.
5. A person applies a constant force on an object of mass 20.0 kg that causes the object to move
horizontally at a constant speed of 0.200 m/s through a distance of 0.800 m along a rough
surface. What is the net work done on the object?
(A) 160 J.
(B) 10.0 J.
(C) 16.0 J.
(D) 0 J.
(E) Cannot be determined without knowing the magnitude of the applied force.
6. A ball drops some distance and loses 30.0 J of gravitational potential energy. Do NOT ignore
Spring 2005
21
air resistance. How much kinetic energy did the ball gain?
(A) More than 30.0 J.
(B) Exactly 30.0 J.
(C) Less than 30.0 J.
(D) Exactly 0.0 J.
(E) Exactly 60.0 J.
7. A 100 kg warthog (Phacochoerus aethiopics) squeals with delight as it slides from rest down
the greased (frictionless) slope and plane as pictured below.
a. How fast does the hog move as it just reaches the bottom of the slope?
Ei = m g yi + ½ m Vi 2 = m g yi
Ef = m g yf + ½ m Vf 2 = ½ m Vf 2
Ei = Ef
m g yi = ½ m Vf 2
g yi = ½ Vf 2
Vf 2 = 2 g yi
Vf = 9.90 m/s
b. As the hog continues along the plane, it encounters another hog of 150 kg mass, initially at
Spring 2005
22
rest. The hogs collide and as a result of the collision the 150-kg hog acquires 5.20 m/s velocity to
the right. How fast does the 100-kg hog move after the collision and in what direction?
m1 V1i + m2 V2i = m1 V1f + m2 V2f
m1 V1f = m1 V1i + m2 V2i - m2 V2f
V1f = (m1 V1i + m2 V2i - m2 V2f )/m1
V1f = ((100 kg)(9.90 m/s) - (150 kg)(5.20 m/s))/(100 kg) = 2.10 m/s, to the right
10. You are pushing a 7.00 kg box along a rough ceiling with a force of 180 N directed at 35.0o
to the horizontal. The magnitude of the friction force acting on the box is 15.0 N.
a. Calculate the work done by each force on the box after you pushed it through 0.750
m.
Spring 2005
23
W = F x cos ()
WA = A x cos() = (180 N) (0.750 m) cos(35.0o) = 147 J
Wn = n x cos() = n (0.750) cos(90.0o) = 0 J
Ww = wx cos() = (7.00 kg)(9.81 m/s2) (0.750 m) cos(90.0o)= 0 J
Wf = fx cos() = (15.0 N)(0.750 m) cos(180o)= - 11.3 J
b. Calculate the total work done on the box.
WNET = WA + Wn + Ww + Wf = 147 J + 0 J + 0 J - 11.3 J = 136 J
c. If initially the box was at rest, what is the final speed of the box after you pushed it
through 0.750 m?
WNET = ½ m Vf 2 – ½ mVi 2
Spring 2005
24
Vi = 0 m/s
½ mVf 2 = WNET
Vf = √( 2WNET/m) = 6.23 m/s
11. The force shown in the force-time diagram below acts on a 1.50 kg object. The force is
directed along the x-axis. Find
a. The impulse of the force.
I = (2.00 N)(3.00 s) + ½ (2.00 N)(2.00 s) = 8.00 N-s
b. The final velocity of the object if it is initially at rest.
I = m Vf – mVi
Vi = 0 m/s
Vf = I/m = 5.33 m/s
c. The final velocity of the object if it is initially moving along the x axis with a
velocity of -2.00 m/s.
I = m Vf – mVi
Spring 2005
25
Vi = -2.00 m/s
Vf = (I + mVi )/m = 3.33 m/s
12. The launching mechanism of a toy gun consists of a spring of unknown spring constant,
as shown below. If the spring is compressed a distance of 0.120 m and the gun is fired
vertically as shown, the gun launches a 20.0-g projectile from rest to a maximum height
of 20.0 m above the starting point. Neglecting all resistive forces, determine
a. The spring constant,
b. The speed of the projectile as it moves through the equilibrium position of the
Spring 2005
26
spring.
TIME
OF
NAME____SOLUTION_________________________
PHYS 1111, Exam 2
Version 1
Section 1
July 18, 2005
1.
problems on eight (8) pages.
2.
3.
4.
5.
There are six (6) multiple choice and four (4) calculation problems. Work all problems.
Show all work; partial credit will be given for correct work shown.
6.
7.
Start:
Stop:
11:30 a.m.
12:50 p.m
PROBLEM
POINTS
1-6
30
7
20
8
20
9
15
Spring 2005
27
CREDIT
10
15
TOTAL
100
PERCENTAGE
PARTIAL CREDIT.
1. A work done by static friction is
e. Always positive.
(5)
f. Always negative.
g. Always zero.
h. Can be either depending on circumstances.
2. For an object moving along a circular trajectory at a constant speed:
a. Its velocity is constant.
b. Its acceleration is zero.
(5)
c. Its acceleration is perpendicular to its instantaneous velocity.
d. None of these.
3. When you sit on a chair, the resultant force on you is
e. Zero.
f. Up.
(5)
g. Down.
Spring 2005
28
h. Depends on your weight.
4. An object in motion must have:
a. Acceleration.
b. Momentum.
(5)
c. Potential energy.
d. All of the above.
5. A garbage truck crashes head-on into a Volkswagen and the two come to rest in a cloud of flies.
Which experiences the greater impact force?
i. The truck.
j. The Volkswagen.
(5)
k. Both experience the same force.
l. Not enough information.
6.
Angular displacement is usually express in units of
e. Meters.
(5)
f. Radians.
g. Revolutions.
Spring 2005
29
h. Arcs.
.
7. A 1.00-kg object moving at 10.0 m/s to the right collides with a stationary 2.00-kg object. The
collision is perfectly inelastic. Surface is frictionless.
(20)
a. What are the magnitude and direction of the combined object’s velocity
immediately after collision?
m1V1i + m2V2i = (m1 + m2 )Vf
Vf = (m1V1i) / (m1 + m2 ) = (1.00 kg)(10.0 m/s) / (1.00 kg + 2.00 kg) = 3.33 m/s
b. What is the kinetic energy of the combined object immediately after collision?
KE = ½ m Vf2 = ½ (3.00 kg) (3.33 m/s)2 = 16.6 J
c. If we neglect friction, how far along the inclined plane will the combined object
travel?
KEi = PEf = (m1 + m2 ) g yf
yf = KEi /((m1 + m2 ) g) = 0.566 m or 0.940 m along the incline.
Spring 2005
30
8. A skier with a mass of 80.0 kg starts from rest at the top of a slope and skis down from an
elevation of 110 m. The speed of the skier at the bottom of the slope is 20.0 m/s.
a. Find the change in the skier’s potential energy.
PE = m g yf - m g yf = - (80.0 kg)(9.80 m/s2)(110 m) = - 86,240 J
b.
Find the change in the skier’s kinetic energy.
KE = ½ m Vf2 - ½ m Vi2 = ½ (80.0 kg) (20.0 m/s)2 = 16,000 J
c. How much work was done by friction force?
Wnc = KE +PE = - 84,240 J + 16,000 J = -70,240 J
d. What average friction force acted on the skier during his descent, if the total path
length along the slope was 250 m?
Wnc = -fk d
fk = - Wnc /d= 281 N
9. An
2
18.0-kg box is released on a 37.0º incline and accelerates down the incline at 0.270 m s .
a. Draw a free-body diagram.
Spring 2005
31
b. Find the friction force impeding its motion.
nx = n cos (90o) = 0
ny = n sin (90o) = N
fkx = fk cos (180o) = - fk
fky = fk sin (180o) = 0
wx = m g sin (37o)
wy = - m g cos (37o)
Fx = ax
- fk + m g sin (37o) = m ax
fk = m g sin (37o) - m ax = (18.0 kg) (5.90 m/s2 – 0.270 m/s2) = 101 N
c. Bonus: How large is the coefficient of kinetic friction?
Fy = 0 => n - m g cos (37o) = 0
n = m g cos (37o) = 140 N
k = fk / n = 0.721

Spring 2005
32
10. A merry-go-round makes 24.0 revolutions in 3.00-minute interval.
a. What is its angular velocity in rad/s?
 = (24.0 rev)(2 rad /1 rev)/(3.00 min)(1 min/60 sec) = 0.838 rad/s
b. The merry-go–round starts to slow down uniformly and comes to a rest after
completing 6.00 revolutions. What was its angular acceleration during that time?
)
))
.838 rad/s2 (6 x rad)) = -0.00931 rad/s2
c. How long did it take to come to rest?
t
t= 90.0 s
TIME
OF
NAME___SOLUTION__________________________
PHYS 1111, Exam 2
Version 1
Section 1
July 16, 2007
1.
2.
Spring 2005
33
3.
4.
5.
shown.
6.
7.
Start:
Stop:
1:30 p.m.
2:45 p.m
PROBLEM
POINTS
1-6
30
7
30
8
30
9
30
10
30
TOTAL
150
CREDIT
PERCENTAGE
NOTE: ALL MULTIPLE CHOICE PROBLEMS BELOW ARE WORTH 6 POINTS EACH.
PARTIAL CREDIT.
1. The quantity mgy is
(A) The kinetic energy of the object.
(B) The elastic potential energy of the object.
(C) The gravitational potential energy of the object.
Spring 2005
34
(D) The work done on the object by a force.
(E) The power supplied to the object by the force.
2. An object is held at a height h above the ground. A second object with four times the mass of
the first is held at the same height. The potential energy of the second object compared to the first
is
(A) One-forth as much.
(B) One-half as much.
(C) Twice as much.
(D) Four times as much.
(E) Eight times as much.
3. An acorn falls from a tree.
(A) Its kinetic energy increases and its gravitational potential energy decreases.
(B) Its kinetic energy increases and its gravitational potential energy increases.
(C) Its kinetic energy decreases and its gravitational potential energy increases.
(D) Its kinetic energy decreases and its gravitational potential energy decreases.
(E) Both kinetic and gravitational potential energies remain constant.
4. Two children are riding on a merry-go-round. Child A is at a greater distance from the axis of
rotation than child B. Which child has the larger angular displacement?
(A) Child A.
(B) Child B.
(C) They have the same zero angular displacement.
(D) They have the same non-zero angular displacement.
(E) There is not enough information given to answer the question.
Fall 2006
35
5. Identical forces act for the same time interval on two different masses. The change in
momentum of the smaller mass is
(A) Smaller than the change in momentum of the larger mass, but not zero.
(B) Larger than the change in momentum of the larger mass.
(C) Equal to the change in momentum of the larger mass.
(D) Zero.
(E) There is not enough information to answer the question.
6. It is more difficult to start moving a heavy box from rest than it is to keep pushing it with
constant velocity, because
(A) The normal force is greater when the box is at rest.
s < k
(C) Initially, the normal force is not perpendicular to the normal force.
Ds > k
s = k
7. A 6.00-kg block slides down a frictionless incline making an angle of 60.0o with the horizontal.
a. List all the forces acting on the block, and find the work done by each force when the block
slides 2.00 m (measured along the incline).
Gravitational force and normal force
Ww = (6.00 kg)(9.81 m/s2)(2.00 m)cos(30.0o) = 102 J
Wn = 0 J
Fall 2006
36
b. What is the total work done on a block?
Wnet = Ww + Wn = 102 J
`
c. What is the speed of the block after it slid 2.00 m if it starts with an initial speed of 1.50
m/s?
Wnet = Kf – Ki = ½ mVf2 - ½ mVi2
½ mVf2 = Wnet + ½ mVi2
Vf2 = 2 Wnet /m + Vi2
Vf = (2 Wnet /m + Vi2) 1/2
V f = 6.02 m/s
8. A 5.00 –kg object moving to the right with a speed of 4.00 m/s collides head-on with a 10.0kg object moving toward it with a speed of 3.00 m/s. The 10.0-kg object stops dead after the
collision.
a. What is the final velocity of the 5.00-kg object? Please specify both magniude and
direction.
m1V1i + m2V2i = m1V1f + m2 V2f
V2f = 0
m1V1i + m2V2i = m1 V1f
V1f = (m1V1i + m2V2i)/ m1
V1f = ((5.00 kg)(4.00 m/s) + (10.0 kg)(-3.00 m/s))/(5.00 kg) = -2.00 m/s (after collision the
5.00-kg object is moving with a speed of 2.00 m/s to the left)
b. Is this collision elastic or inelastic?
Ki = ½ m1V1i2 + ½ m2V2i2 = ½ (5.00 kg) (4.00 m/s)2 + ½ (10.0 kg) (3.00 m/s)2 = 85.0 J
Fall 2006
37
Kf = ½ m1V1f2 = ½ (5.00 kg) (2.00 m/s)2 = 10.0 J
Kinetic energy is not conserved, collision is inelastic.
9. A compact disc rotates from rest to 500 rev/min in 5.50 s.
a. What is the angular acceleration of the disc assuming it is constant?
(500 rev/min)(2 rad/rev)(1 min/60s) = 52.4 rad/s
t
(t = ((52.4 rad/s) – (0 rad/s))/(5.50 s) = 9.52 rad/s2
b. How many revolutions does it make in 5.50 s?
)
) = ()
) = (52.4 rad/s0 rad/s rad/s2) = 144 rad = 23.0 rev
c. What is the centripetal acceleration of the point on the rim of the disc (6.00 cm
from the center) at the moment the disc reaches angular velocity of 500 rev/min?
acp = R = (52.4 rad/s)2(0.0600 m) = 165 m/s2
10. You drop your physics textbook (of 4.00 kg mass) from the second floor window 7.00 m
above the ground.
a. Neglecting the air resistance, how fast do you expect it to move as it reaches the
ground?
Ei = m g yi + ½ mVi2 = m g yi
Fall 2006
38
Ef = m g yf + ½ mVf2 = ½ mVf2
Ei = Ef
m g yi = ½ mVf2
g yi = ½ Vf2
Vf = (2 g yi)1/2
Vf = (2 (9.81 m/s2) (7.00 m))1/2
Vf = 11.7 m/s
b. You measure the speed of the book just before it hits the ground and find it to be 9.00
m/s. You correctly assume that the air resistance slows the book down. How much
work is done on the book by air resistance?
Wf = Ef – Ei = ½ m Vf 2 - m g yi
Wf = Ef – Ei = ½ (4.00 kg) (9.00 m/s) 2 - (4.00 kg) (9.81 m/s2)(7.00 m) = -113 J
c. What is the average air resistance force that acts on the book during the fall?
Wf = f d cos (180o) = - f d
f = - Wf /d = - (- 113 J)/(7.00 m) = 16.1 N
Fall 2006
39
Fall 2006
40
Fall 2006
41

TIME OF COMPLETION_______________

Transcription

Similar documents

SHUTTLEBUS - ZOOM - City Of Saco, Maine

AL Bump, Florist • la L. Burton Orr

Belzona 1111 Super Metal

Cos d`Estournel France - Bordeaux Cabernet Sauvignon

file!

ALL-NEW! - Previews

file!

ASBOG EXAMINATION SCORE REQUEST

wdgl eagle 98.1