Review- PHYS114 Final
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Review- PHYS114 Final
Due: 1:43pm on Monday, March 24, 2014
To understand how points are awarded, read the Grading Policy for this assignment.
Part Problem I.5-9
Description: Animal Athletes Different animals have very different capacities for running. A horse can maintain a top
speed of 20 m/s for a long distance but has a maximum acceleration of only 6.0 m/s^2, half what a good human
sprinter can achieve with a block to ...
Animal Athletes Different animals have very different capacities for running. A horse can maintain a top speed of 20 m/s
for a long distance but has a maximum acceleration of only 6.0 m/s 2 , half what a good human sprinter can achieve with a
block to push against. Greyhounds, dogs especially bred for feats of running, have a top speed of 17 m/s , but their
acceleration is much greater than that of the horse. Greyhounds are particularly adept at turning corners at a run.
Part A
If a horse starts from rest and accelerates at the maximum value until reaching its top speed, how much time elapses,
to the nearest second?
ANSWER:
4s
1s
3s
2s
Part B
If a horse starts from rest and accelerates at the maximum value until reaching its top speed, how far does it run, to
the nearest 10 m ?
ANSWER:
40 m
10 m
20 m
30 m
Part C
A greyhound on a racetrack turns a corner at a constant speed of 16m/s with an acceleration of 7.7m/s 2 . What is
the radius of the turn?
ANSWER:
33m
22m
44m
11m
Part D
A human sprinter of mass 79kg starts a run at the maximum possible acceleration, pushing backward against a block
set in the track. What is the force of his foot on the block?
ANSWER:
470N
1700N
780N
950N
Part E
In the photograph of the greyhounds in the figure, what is the direction of the net force on each dog?
ANSWER:
right, toward the inside of the turn
left, toward the outside of the turn
down
up
Part Problem I.14-16
Description: Pulling Out of a Dive Falcons are excellent fliers that can reach very high speeds by diving nearly
straight down. To pull out of such a dive, a falcon extends its wings and flies through a circular arc that redirects its
motion. The forces on the...
Pulling Out of a Dive Falcons are excellent fliers that can reach very high speeds by diving nearly straight down. To pull
out of such a dive, a falcon extends its wings and flies through a circular arc that redirects its motion. The forces on the
falcon that control its motion are its weight and an upward lift force-like an airplane-due to the air flowing over its wings. At
the bottom of the arc, as in the figure, a falcon can easily achieve an acceleration of 15 m/s 2 .
Part A
At the bottom of the arc, as in the figure, what is the direction of the net force on the falcon?
ANSWER:
to the left, opposite the motion
down
to the right, in the direction of the motion
the net force is zero.
up
Part B
Suppose the falcon weighs 8.5N and is turning with an acceleration of 15m/s 2 at the lowest point of the arc. What is
the magnitude of the upward lift force at this instant?
ANSWER:
8.5N
13N
21N
17N
Part C
A falcon starts from rest, does a free-fall dive from a height of 30 m, and then pulls out by flying in a circular arc of
radius 50 m . Which segment of the motion has a higher acceleration?
ANSWER:
the circular arc
the free-fall dive
the two accelerations are equal.
Part Problem I.17-19
Description: Bending Beams If you bend a rod down, it compresses the lower side of the rod and stretches the top,
resulting in a restoring force. The figure shows a beam of length L, width w, and thickness t fixed at one end and free
to move at the other...
Bending Beams If you bend a rod down, it compresses the lower side of the rod and stretches the top, resulting in a
restoring force. The figure shows a beam of length L , width w, and thickness t fixed at one end and free to move at the
other. Deflecting the end of the beam causes a restoring force F at the end of the beam. The magnitude of the restoring
force F depends on the dimensions of the beam, the Young's modulus Y for the material, and the deflection d . For small
values of the deflection, the restoring force is
F = [ Y wt3 ]d. This is similar to the formula for the restoring force of a
3
4L
spring, with the quantity in brackets playing the role of the spring constant k .
When a 67kg man stands on the end of a springboard (a type of diving board), the board deflects by 7.0cm .
Part A
If a 39kg child stands at the end of the board, the deflection is
ANSWER:
8.1cm .
6.1cm .
4.1cm .
2.0cm .
Part B
A 67kg man jumps up and lands on the end of the board, deflecting it by 12cm . At this instant, what is the
approximate magnitude of the upward force the board exerts on his feet?
ANSWER:
750N
400N
1500N
1100N
Part C
If the board is replaced by one that is half the length but otherwise identical, how much will it deflect when a 67kg man
stands on the end?
ANSWER:
1.8cm
3.5cm
7.0cm
0.88cm
Part Problem I.20
Description: You go to the playground and slide down the slide, a L long ramp at an angle of theta with respect to
horizontal. The pants that you've worn aren't very slippery; the coefficient of kinetic friction between your pants and the
slide is mu. A friend...
You go to the playground and slide down the slide, a 3.2m long ramp at an angle of 34∘ with respect to horizontal. The
pants that you've worn aren't very slippery; the coefficient of kinetic friction between your pants and the slide is 0.39. A
friend gives you a very slight push to get you started.
Part A
How long does it take you to reach the bottom of the slide?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
t=
= 1.7
Part Problem I.21
Description: If you stand on a scale at the equator, the scale will read slightly less than your true weight due to your
circular motion with the rotation of the earth. (a) Draw a free-body diagram to show why this is so, taking the
downward direction to be...
If you stand on a scale at the equator, the scale will read slightly less than your true weight due to your circular motion with
the rotation of the earth.
Part A
Draw a free-body diagram to show why this is so, taking the downward direction to be towards the center of the earth.
Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The
length of the vectors will not be graded.
ANSWER:
Part B
By how much is the scale reading reduced for a person with a true weight of 850N ?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
= 2.9
mg − n =
Part Problem I.22
Description: Dolphins and other sea creatures can leap to great heights by swimming straight up and exiting the
water at a high speed. A m dolphin leaps straight up to a height of h. When the dolphin reenters the water, drag from
the water brings it to a stop in...
Dolphins and other sea creatures can leap to great heights by swimming straight up and exiting the water at a high speed.
A 230kg dolphin leaps straight up to a height of 7.2m . When the dolphin reenters the water, drag from the water brings it
to a stop in 1.8m . Assuming that the force of the water on the dolphin stays constant as it slows down,
Part A
How much time does it take for the dolphin to come to rest?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
t=
= 0.30
Part B
What is the force of the water on the dolphin as it is coming to rest?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
= 1.1×104
F=
Part Problem II.5-9
Description: Big Air A new generation of pogo sticks lets a rider bounce more than 2 meters off the ground by using
elastic bands to store energy. When the pogo's plunger hits the ground, the elastic bands stretch as the pogo and rider
come to rest. At the low...
Big Air A new generation of pogo sticks lets a rider bounce more than 2 meters off the ground by using elastic bands to
store energy. When the pogo's plunger hits the ground, the elastic bands stretch as the pogo and rider come to rest. At the
low point of the bounce, the stretched bands start to contract, pushing out the plunger and launching the rider into the air.
kg
For a total mass of 80 kg (rider plus pogo), a stretch of 0.40
m launches a rider 2.0 m above the starting point.
Part A
If you were to jump to the ground from a height of 2 meters, you'd likely injure yourself. But a pogo rider can do this
repeatedly, bounce after bounce. How does the pogo stick make this possible?
ANSWER:
The elastic bands warm up as the rider bounces, absorbing dangerous thermal energy.
The elastic bands absorb the energy of the bounce, keeping it from hurting the rider.
The elastic bands let the rider come to rest over a longer time, meaning less force.
The elastic bands simply convert the rider's kinetic energy to potential energy.
Part B
Assuming that the elastic bands stretch and store energy like a spring, how high would the 80
kg pogo and rider go for
a stretch of 0.20 m ?
ANSWER:
m
1.5 m
0.50 m
1.0 m
2.0
Part C
Suppose a much smaller rider (total mass of rider plus pogo of 49kg ) mechanically stretched the elastic bands of the
pogo by 0.40m , then got on the pogo and released the bands. How high would this unwise rider go?
ANSWER:
4.9m
3.3m
2.4m
6.5m
Part D
A pogo and rider of 80 kg total mass at the high point of a 2.0 m jump will drop 1.5 m before the pogo plunger
touches the ground, slowing to a stop over an additional 0.40 m as the elastic bands stretch. What approximate
average force does the pogo stick exert on the ground during the landing?
ANSWER:
1500N
700N
4000N
3000N
Part E
Riders can use fewer elastic bands, reducing the effective spring constant of the pogo. The maximum stretch of the
bands is still 0.40 m . Reducing the number of bands will
ANSWER:
not change the force on the rider but give a lower jump height.
make no difference to the force on the rider or the jump height.
reduce the force on the rider but give the same jump height.
reduce the force on the rider and give a lower jump height.
Part Problem II.10-12
Description: Testing Tennis Balls Tennis balls are tested by being dropped from a height of 2.5 m onto a concrete
floor. The 57 g ball hits the ground, compresses, then rebounds. A ball will be accepted for play if it rebounds to a
height of about 1.4 m; it will...
Testing Tennis Balls Tennis balls are tested by being dropped from a height of 2.5 m onto a concrete floor. The 57 g ball
hits the ground, compresses, then rebounds. A ball will be accepted for play if it rebounds to a height of about 1.4 m ; it will
be rejected if the bounce height is much more or much less than this.
Part A
Consider the sequence of energy transformations in the bounce. When the dropped ball is motionless on the floor,
compressed, and ready to rebound, most of the energy is in the form of
ANSWER:
elastic potential energy.
gravitational potential energy.
kinetic energy.
thermal energy.
Part B
If a ball is "soft," it will spend more time in contact with the floor and won't rebound as high as it is supposed to. The
force on the floor of the "soft" ball is ________ the force on the floor of a "normal" ball.
ANSWER:
greater than
the same as
less than
Part C
Suppose a ball is dropped from 2.5m and rebounds to 1.7m . How fast is the ball moving just before it hits the floor?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
vf =
= 7.0
Part D
What is the ball's speed just after leaving the floor?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
vi =
= 5.8
Part E
What happens to the "lost" energy? The "lost" energy is transformed into
ANSWER:
elastic potential energy.
kinetic energy.
gravitational potential energy.
thermal energy.
Part F
If the time of the collision with the floor is 6.1ms , what is the average force on the ball during the impact?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
F=
= 120
Part Problem II.13-16
Description: Squid Propulsion Squid usually move by using their fins, but they can utilize a form of "jet propulsion,"
ejecting water at high speed to rocket them backward, as shown in the figure. A 4.0 kg squid can slowly draw in and
then quickly eject 0.30 kg...
Squid Propulsion Squid usually move by using their fins, but they can utilize a form of "jet propulsion," ejecting water at
high speed to rocket them backward, as shown in the figure. A 4.0 kg squid can slowly draw in and then quickly eject 0.30
kg of water. The water is ejected in 0.10 s at a speed of 10 m/s . This gives the squid a quick burst of speed to evade
predators or catch prey.
Part A
What is the speed of the squid immediately after the water is ejected?
ANSWER:
10 m/s
m/s
7.5 m/s
0.75 m/s
1.3
Part B
What is the squid's approximate acceleration in g?
ANSWER:
1.0g
0.75g
10g
7.5g
Part C
What is the average force on the water during the jet?
ANSWER:
100 N
N
10 N
30 N
3.0
Part D
This form of locomotion is speedy, but is it efficient? The energy that the squid expends goes two places: the kinetic
energy of the squid and the kinetic energy of the water. Think about how to define "what you get" and "what you had to
pay"; then calculate an efficiency for this particular form of locomotion. (You can ignore biomechanical efficiency for
this problem.)
Express your answer using two significant figures.
ANSWER:
e = 7.0 %
Part Problem II.17-20
Description: Teeing Off A golf club has a lightweight flexible shaft with a heavy block of wood or metal (called the
head of the club) at the end. A golfer making a long shot off the tee uses a driver, a club whose 300 g head is much
more massive than the 46 g...
Teeing Off A golf club has a lightweight flexible shaft with a heavy block of wood or metal (called the head of the club) at
the end. A golfer making a long shot off the tee uses a driver, a club whose 300 g head is much more massive than the 46
g ball it will hit. The golfer swings the driver so that the club head is moving at 41m/s just before it collides with the ball.
The collision is so rapid that it can be treated as the collision of a moving 300 g mass (the club head) with a stationary 46 g
mass (the ball); the shaft of the club and the golfer can be ignored. The collision takes 5.6 ms , and the ball leaves the tee
with a speed of 71 m/s .
Part A
What is the change in momentum of the ball during the collision?
ANSWER:
1.4kg ⋅ m/s
1.9kg ⋅ m/s
3.3kg ⋅ m/s
5.2kg ⋅ m/s
Part B
What is the speed of the club head immediately after the collision?
ANSWER:
30m/s
26m/s
20m/s
11m/s
Part C
Is this a perfectly elastic collision?
ANSWER:
yes
no
There is insufficient information to make this determination.
Part D
If we define the kinetic energy of the club head before the collision as "what you paid" and the kinetic energy of the ball
immediately after as "what you get," what is the efficiency of this energy transfer?
ANSWER:
0.54
0.46
0.37
0.27
Part Problem II.21
Description: Football players measure their acceleration by seeing how fast they can sprint 40 yards (37 m). A zippy
player can, from a standing start, run 40 yards in t, reaching a top speed of about v. (a) For an m player, what is the
average power output for...
Football players measure their acceleration by seeing how fast they can sprint 40 yards (37 m ). A zippy player can, from a
standing start, run 40 yards in 4.2s , reaching a top speed of about 10m/s .
Part A
For an 73kg player, what is the average power output for this sprint?
ANSWER:
200W
870W
400W
700W
Part Problem II.22
Description: The unit of horsepower was defined by considering the power output of a typical horse. Working-horse
guidelines in the 1900s called for them to pull with a force equal to 10% of their body weight at a speed of 3.0 mph. (a)
For a typical working horse ...
The unit of horsepower was defined by considering the power output of a typical horse. Working-horse guidelines in the
1900s called for them to pull with a force equal to 10% of their body weight at a speed of 3.0 mph .
Part A
For a typical working horse of 1600lb , what power does this represent in W?
Express your answer using two significant figures.
ANSWER:
P=
= 960
W
Part B
What power does this represent in hp?
Express your answer using two significant figures.
ANSWER:
P=
= 1.3
hp
Part Problem II.23
Description: A m_1 football player is moving at v_1 to the east; a m_2 player is moving at v_2 to the west. They
meet, each jumping into the air and grabbing the other player. (a) While they are still in the air, how fast is the pair
moving? (b) While they are...
A 100kg football player is moving at 5.9m/s to the east; a 140kg player is moving at 5.4m/s to the west. They meet,
each jumping into the air and grabbing the other player.
Part A
While they are still in the air, how fast is the pair moving?
Express your answer using two significant figures.
ANSWER:
= 0.69
vf =
Part B
While they are still in the air, which way is the pair moving?
ANSWER:
to the east
to the west
Part Problem II.24
Description: A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the
hand is transformed into elastic potential energy of the bending board; if the board bends far enough, it breaks.
Applying a force to the center...
A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is
transformed into elastic potential energy of the bending board; if the board bends far enough, it breaks. Applying a force to
the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the force.
Ultimately the board breaks at an applied force of 850 N and a deflection of 1.1cm .
Part A
To break the board with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the
hand.
Express your answer to two significant figures and include the appropriate units.
ANSWER:
v=
= 4.3
Part B
If the hand is moving this fast and comes to rest in a distance of 1.1cm , what is the average force on the hand?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
F=
= 430
Part Problem II.25
Description: A child's sled has rails that slide with little friction across the snow. Logan has an old wooden sled with
heavy iron rails that has a mass of m_1 quite a bit for a m_2 child! Logan runs at v and leaps onto the stationary sled
and holds on tight as...
A child's sled has rails that slide with little friction across the snow. Logan has an old wooden sled with heavy iron rails that
has a mass of 9.0 kg quite a bit for a 33kg child! Logan runs at 3.8m/s and leaps onto the stationary sled and holds on
tight as it slides forward. The impact time with the sled is 0.26s .
Part A
Immediately after Logan jumps on the sled, how fast is it moving?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
vf =
= 3.0
Part B
What was the force on the sled during the impact?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
= 100
F=
Part C
How much energy was "lost" in the impact?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
∆K =
Also accepted:
Part D
Where did this energy go?
= -51
= 51
ANSWER:
elastic potential energy.
This energy was dissipated as
thermal energy.
kinetic energy.
gravitational potential energy.
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