(hph268) huynh HW #4 clark (56485)
This print-out should have 19 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
001 10.0 points
Two identical objects are moving directly toward one another at the same speed v .
m
v
v
8.
1
5M g
6
003 10.0 points
A force F is exerted by a broom handle on the
head of the broom, which has a mass m. The
handle is at an angle to the horizontal, as
shown below.
m
F
What is the total kinetic energy of the system of the two objects?
1. Zero since (v v )2 = 0
2. None of these
3. 4 m v 2 since (2 v )2 = 4 v 2
1
4. Still m v 2 from the denition of kinetic
2
energy
5. m v 2 since energy is a scalar
002 10.0 points
An object with mass M is attached to the end
of a string and is being lowered vertically at a
g
constant acceleration of .
6
If it has been lowered a distance from rest,
how much work has been done by the tension
in the string?
M g
6
M g
2.
6
M g
3.
3
5M g
4.
6
M g
5.
3
1.
What is the work done by the force on the
head of the broom as it moves a distance d
across a horizontal oor?
1. W = F m tan
2. W = F d sin
3. W = F m cos
4. W = F m d sin
5. W = F d cos
004 10.0 points
A block of mass m is pushed a horizontal distance D from position A to position B, along
a horizontal plane with friction coecient .
Then m is pushed from B to A.
If the force pushing m from A to B is P ,
and the force pushing m from B to A is P ,
what is the total work done by friction?
1. 2 ( m g P ) D
2. 0
3. 2 m g D
6. M g
4. 2 (P m g ) D
7. M g
5. +2 m g D
huynh (hph268) HW #4 clark (56485)
2
4. W = m g x tan
005 (part 1 of 4) 10.0 points
A box of mass m with an initial velocity of v0
slides down a plane, inclined at with respect
to the horizontal. The coecient of kinetic
friction is . The box stops after sliding a
distance x.
m
v0
k
5. W = m g x sin
6. W = m g ( cos sin )
m v0
2 ( + tan )
2
m v0
8. W =
2 ( + tan )
7. W =
9. W = 0
10. W =
How far does the box slide?
1. x =
2. x =
3. x =
4. x =
5. x =
6. x =
7. x =
8. x =
9. x =
10. x =
2
v0
2 g ( sin + cos )
2
v0
2 g ( sin cos )
2
v0
2 g cos
2
v0
2 g (sin cos )
2
v0
2 g ( cos sin )
2
v0
g ( sin 2 cos )
2
v0
2 g ( cos + sin )
2
v0
g ( sin + cos )
2
v0
2 g sin
2
v0
g (sin cos )
006 (part 2 of 4) 10.0 points
How much work is done by friction?
1. W = m g ( cos + sin )
2. W =
m v0
2 ( tan )
3. W = m g x cos
2
m v0 cos
2 ( sin cos )
007 (part 3 of 4) 10.0 points
How much work is done by the normal force?
1. W = m g x sin
2. W = m g ( cos sin )
3. W =
2
m v0 cos
2 ( sin cos )
4. W = m g x cos
5. W = m g ( cos + sin )
6. W =
m v0
2 ( + tan )
7. W = m g x tan
8. W =
m v0
2 ( tan )
9. W = 0
10. W =
2
m v0
2 ( + tan )
008 (part 4 of 4) 10.0 points
How much work is done by gravity?
1. W = 0
2. W =
m v0
2 tan
3. W = m g x sin
4. W = m g (cos sin )
huynh (hph268) HW #4 clark (56485)
3
How much time does it take the rock to
travel from the edge of the building to the
ground?
m v0
5. W =
tan
6. W = m g x tan
1. t =
2h
g
2. t =
2
m v0
7. W =
tan
2
m v0
8. W =
2 tan
h v0
2h
g
h
4. t =
v0
h v0
5. t =
g
3. t =
9. W = m g x cos
10. W = m g (cos + sin )
009 10.0 points
A block sliding on a horizontal surface has an
initial speed of 0.5 m/s. The block travels a
distance of 1 m as it slows a to stop.
What distance would the block have traveled if its initial speed had been 1 m/s?
011 (part 2 of 2) 10.0 points
What is the kinetic energy of the rock just
before it hits the ground?
1. Kf = m g h
1. 0.5 m
1
2
m v0
2
1
2
m v0 + m g h
2
1
2
3. Kf = m v0
2
1
2
4. Kf = m v0 m g h
2
2. Kf =
2. 1 m
3. 2 m
4. more information is needed to answer the
question
5. 3 m
6. 4 m
010 (part 1 of 2) 10.0 points
A rock of mass m is thrown horizontally o
a building from a height h. The speed of the
rock as it leaves the throwers hand at the
edge of the building is v0 as shown.
5. Kf = m g h
012 10.0 points
A sports car accelerates from zero to 30 mph
in 1.5 s.
How long does it take for it to accelerate
from zero to 60 mph, assuming the power of
the engine to be independent of velocity and
neglecting friction.
1. 2 s
v0
2. 6 s
3. 9 s
h
4. 12 s
5. 3 s
x
6. 4.5 s
huynh (hph268) HW #4 clark (56485)
013 10.0 points
A student weighing 700 N climbs at constant
speed to the top of an 8 m vertical rope in
10 s. The average power expended by the
student to overcome gravity is most nearly
1. 5,600 W
4
016 10.0 points
When an object is moved from rest at point
A to rest at point B in a gravitational eld,
the net work done by the eld depends on the
mass of the object and
1. both the positions of A and B and the
path taken between them.
2. 87.5 W
2. the path taken between A and B only.
3. 560 W
3. the positions of A and B only.
4. 875 W
4. the nature of the external force moving
the object from A to B .
5. 1.1 W
014 10.0 points
The amount of potential energy an object has
depends
1. only on its mass.
2. only on its speed.
3. on its mass and speed.
4. only on its height from the ground.
5. the velocity of the object as it moves
between A andB .
017 (part 1 of 2) 10.0 points
A bead slides without friction around a loopthe-loop. The bead is released from position
A at a height yA (which may vary) from the
bottom of the loop-the-loop which has a radius r .
The acceleration of gravity is g .
A
C
5. on its mass and its height from the
ground.
r
yA
015 10.0 points
As a bug drops vertically from a point meters above a bush to a leaf on the bush, its
gravitational potential energy Ug (y )
1. increases by the amount of work gravity
does on the bug as it drops.
2. remains constant.
3. increases by the amount of work you
would have to do against gravity to carry
the bug from the leaf to its original position.
4. decreases by the magnitude of the amount
of work gravity does on the bug as it drops.
B
What is the instantaneous kinetic energy
KC at C so that the bead would press the
1
track with an upward force Fpress = m g ?
2
1. KC = m g r
3
mgr
2
3
3. KC = m g r
4
2. KC =
4. KC = 2 m g r
huynh (hph268) HW #4 clark (56485)
2
mgr
3
1
6. KC = m g r
4
1
7. KC = m g r
2
5. KC =
018 (part 2 of 2) 10.0 points
Hint: Use energy conservation.
Denote the instantaneous kinetic energy at
C by KC . Express the height at A, yA , in
terms of KC , m and r .
KC
mg
mg
2. yA = r +
KC
mg
3. yA = 2 r +
KC
1. yA = 2 r +
4. yA = 2 r
KC
mg
mg
6. yA =
KC
5. yA =
7. yA = r +
KC
mg
8. yA = 3 r
9. yA = 2 r
KC
mg
019 10.0 points
You drop a 150 g baseball from a window 15 m
above the ground.
What is the kinetic energy of the baseball
when it hits the ground? The acceleration
due to gravity is 10 m/s2 .
1. 22.5 J
2. 2.25 104 J
3. 0 J
4. 1.125 J
5. 1.125 105 J
5