Download Practice Exam 1

Physics 218 Midterm 1
Summer 2010
Dr. Edward Rhoads
______________________________________ Name
Multiple Choice (2 points each):
Please circle your answers.
Good luck!
1) The friction force on a car is 1000 Newtons (backwards). The car has a weight of
5000 Newtons. If you ignore air resistance what force does the engine need to
provide the car in order for the car to drive at a constant velocity?
a) 0 Newtons
b) 1000 Newtons
c) 6000 Newtons
d) It is impossible for the car to maintain a constant velocity because of the
frictional force.
2) You throw a rock straight up into the air with an initial velocity of 50 m/s up. 7
seconds later what will the velocity of the rock be?
a) 50 m/s up
b) 50 m/s down
c) 20 m/s up
d) 20 m/s down
3) A car has a weight of 4900 Newtons. If you ignore air resistance and friction then
about what force does the engine need to maintain in order to accelerate at a
constant rate of 0.4 m/s2?
a) 100 Newtons
b) 200 Newtons
c) 300 Newtons
d) 700 Newtons
4) A block slides across a frictionless surface at a velocity of 10 m/s. How far will
the block have slid after 5 seconds?
a) 10 m
b) 25 m.
c) 50 m
d) 100 m
5) A ball falls through the air at an acceleration of 8 m/s2. If the weight of the ball is
10 Newtons then what is the force of air resistance (hint, using the net force will
help)?
a) 0 Newtons
b) 2 Newtons upward
c) 5 Newtons upward
d) 8 Newtons upward
6) An untethered 65 kg astronaut finds himself or herself all alone in orbit around the
Earth with no spacecraft to call home. If the astronaut is in very low earth orbit
then which of the following is closest to the gravitational force exerted on the
astronaut by the Earth? SELECT THE ANSWER THAT IS THE CLOSEST
a) 0 N down
b) 600 N down
c) 65 N down
d) 65 N up
7) A 1000 kg car travels along a straight road at a constant speed. What is the net
force on the car?
a) Negative and non-zero
b) Zero
c) Greater than zero
d) The sign of the net force depends on the velocity.
8) A 1000 kg car travels around in a circle at constant speed. What is the
acceleration on the car?
a) Zero
b) Non-Zero
c) It depends on whether the car is traveling clockwise or counter clock wise
d) It is impossible to travel in a circle at a constant speed.
9) A string is tied to a ceiling (call this the top string). Hanging from the string is a 5
kg weight. Below the 5 kg weight another string is tied (call this the bottom
string). Hanging from that string is another 5 kg weight. Which of the following
is true about the tensions of the top string and the bottom string?
a) both strings have the same tensions
b) the top string has more tension than the bottom string
c) the bottom string has more tension than the top string
d) it is impossible to tell how much tension either string has
10) A bird of weight 5 Newtons is flying at a constant altitude and velocity. What is
the upward force on the wings due to air resistance?
a) 5 Newtons upward
b) 0 Newtons
c) 5 Newtons downward
d) 10 Newtons upward
SHORT ANSWER: (20 points each)
Be sure to answer every part of every question. Be sure to label the parts of questions
you are answering. Be sure to show work and include units and directions. The lowest
score of the 5 will be dropped but please still do all 5.
1) A sky diver jumps from a plane in a lying face down position. After a handful of
seconds he reaches "terminal velocity". At this point he is falling at a constant
velocity.
A) What is the net force on the sky diver?
B) If the weight of the sky diver is 800 N then what is the force of air resistance on
the sky diver.
C) The sky diver rotates so that he is upright. He now has less surface area to the air
as he falls. This means that he has less air resistance. Will the sky diver's "terminal
velocity" increase, decrease, or stay the same. Explain why.
D) The sky diver pulls his parachute. When the chute opens (lots of surface area
now) what initially happens to the air resistance force on the sky diver and how does
that affect his velocity?
E) Eventually the sky diver will reach "terminal velocity" again - although in this case
that velocity will be fairly small. Once the sky diver reaches that velocity how will
the force of air resistance on him compare to the force of air resistance that was on
the sky diver when he was at "terminal" velocity without the parachute? HINT:
Calculate the force on the diver with the parachute and compare that to your answer
from part B.
2) Flight of the Navigator
As part of an orienteering contest a woodsman is given a set of distances and
directions to go. However, the woodsman realized that he left his compass at home!
Oops. Now the only two directions he can figure out are West and North. The
instructions are as follows:
go 120 m in a direction 30 degrees West of North
go 80 m in a direction 15 degrees West of South’
go 50 m in a direction 20 degrees North of West
go 110 m in a direction 40 degrees South of East
Determine for the woodsman:
A) What is the total distance the woodsman should go North (note it might be
possible for this answer to be negative)?
B) What is the total distance the woodsman should go West (note it might be possible
for this answer to be negative)?
C) What is the magnitude of the distance the woodsman will travel?
D) For someone that uses the compass and does the full route it will take them 9 min.
In m/s find both the average SPEED and magnitude of the average VELOCITY that
they walk in that 9 min time (no they won’t be the same).
3) Tortoise vs. the Hare.
There are 2 rocket cars. The first is called the Hare. The Hare starts out every race at
a velocity of 190 m/s and travels at a constant velocity. The 2nd car is called the
Tortoise. It starts out at rest but accelerates at a rate of 12 m/s2 for the entire race.
A) After 13 seconds how far will the Hare have traveled?
B) It takes 50 seconds for the Tortoise to complete the race. After 50 seconds how far
will the Tortoise have traveled?
C) What speed is the Tortoise traveling when it crosses the finish line?
D) Seeing as your answer to 3B is the length of the race, how long does it take the
Hare to complete the race?
E) Which car wins the race?
4) Aim the mini cannon:
A mini cannon is taken to a tall building. Across the street from the mini cannon is
something that for some reason needs shot at. The two buildings are separated by a
horizontal distance of 71 meters. The mini cannon has to be set up so that it aims at an
angle of 35 degrees above the horizontal. The muzzle velocity of the mini cannon (i.e.
the SPEED the cannonball comes out at) of 36 m/s. Find:
A) The vertical AND the horizontal velocities of the cannonball when it leaves the mini
cannon (please specify both)
B) The amount of time that the cannon ball will take to hit the nearby building.
C) The magnitude of the velocity of the cannonball be the moment before it hits the
building (yes you will have to find the vertical and horizontal velocities to solve this)?
D) If the object trying to hit is on the 7th floor of the building you are shooting at then
what floor do you want to shoot your cannon from? You can assume that each floor is
about 10 meters high.
5) Challenge QUESTION:
Larry, Curly, and Mo are at it again. They are trying to push a sled which is full of
medical supplies to a ski resort. The fresh powder has a frictional coefficient of 0.04.
The sled with the supplies has a MASS of 150 kg. Larry has a mass of 80 kg. Curly
has a mass of 120 kg. Mo has a mass of 60 kg.
Mo pulls on the front end of the sled at an angle of 15 degrees above the horizontal
with a force of magnitude 150 N.
Curly sits on the sled (and the 150 kg sled does NOT include Curly).
Larry pushes on the back end of the sled at an angle of 30 degrees below the
horizontal with a force of magnitude 120 N.
A) What is the Net force for the sled in the vertical direction?
B) What is the normal force for the sled?
C) What is the Net force for the sled in the horizontal direction (and no it won’t be
zero N)?
D) What is the acceleration of the sled (here you will have to include the fact that
Curly is sitting on it)?
EQUATIONS:
X = Xo + Vo t + ½ at2
V = Vo + at
W = mg
F = ma
F = μN
F = G M1 M2 / r2 (towards object creating force)
G = 6.67 * 10-11 N m2 / kg2
g = 9.80 m/s2
Net Force = sum of all forces
Up = - Down
sin(theta) = opposite / hypotenuse
cos(theta) = adjacent / hypotenuse
tan(theta) = sin(theta) / cos(theta) = opposite / adjacent
c2 = a2 + b2
Quadratic Equation: x = [-b + (b2 – 4ac)1/2]/2a