Download Actual exam 2 - Sarah Spolaor

Your Name: ___________________________ TEST #2
Print clearly.
On the Scantron, fill out your student ID, leaving the first column empty and starting in the
second column. Also write your name, class time (11:30 or 12:30), and “Test 1” at the bottom.
There are 20 equally-weighted questions on this test. There is only one correct answer per
question. Mark your answer on the Scantron. The second to last page is blank for extra space
if needed. The formulas are on the last page so you can separate it for easy access. The key will
be posted online after all make-up tests are completed.
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1. A 10 kg box is being pulled across a level surface at a constant velocity by a rope with a
magnitude of 5.0 N at a 15 degree angle above the horizontal. What is the magnitude of the
force of kinetic friction opposing the motion?
a) 1.3 N
b) 3.7 N
c) 4.8 N
d) 5.2 N
e) 8.2 N
2. If an object of mass m moves with constant velocity v, the net force on the object is
a) mv
b) a
c) mg
d) 0
e) None of these answers are correct.
3. What is the magnitude of the acceleration of a block sliding down a frictionless 25 degree
incline?
a) 2.1 m/s2
4.
b) 3.1 m/s2
c) 4.1 m/s2
d) 8.9 m/s2
A spaceship drifts sideways in outer space from point P to point Q as shown below. The
spaceship is subject to no outside forces. When the spaceship reaches position Q, the
spaceship’s engine turns on and produces a constant force on the spaceship at a right angle to
the line PQ. The constant force is maintained until the spaceship reaches a point R in space
(also shown below).
As the spaceship moves from point Q to point R, its speed is
a) constant.
b) continuously increasing.
c) continuously decreasing.
d) increasing for a while and constant thereafter.
e) constant for a while and decreasing thereafter.
5. Consider a 25 kg dresser at rest on an inclined ramp that makes an angle of 20 degrees with
the ground. The frictional force acting on the ramp is 30 N. What is the normal force on the
dresser? (I purposely have not listed the coefficient of friction, as it is not required to solve
the problem.)
a) 10 N
b) 30 N
c) 200 N
d) 220 N
e) 230 N
6. A large crate is placed in the back of a pickup truck. It is not tied down. As the truck
accelerates forward, the crate remains at rest relative to the truck (i.e., the crate does not
slide). What force causes the crate to accelerate forward?
a) The normal force
b) The gravitational force
c) The static friction between the crate and the floor of the truck
d) The difference between the normal force and the weight
e) The kinetic friction between the crate and the floor of the truck
7. You’re helping a friend move a heavy couch (mass 200 kg) across a carpet, and find you
must exert a horizontal force of 700 N to get the couch to barely move. Once the couch starts
moving, you continue to push with 700 N. Your friend, a physics major, estimates that it then
accelerates at 1.10 m/s2. What is the coefficient of static friction between the couch and the
carpet?
a) 0.12
b) 0.31
c) 0.36
d) 0.55
e) 0.80
8. For the couch and carpet in the previous problem, what is the force of kinetic friction?
a) 220 N
b) 360 N
c) 480 N
d) 630 N
e) 700 N
9. A mass, M1 = 30kg, is at rest on an incline of θ = 35o, connected by a string to a wall as
shown in the below diagram. What is the tension in the string connected to the wall? Assume
the incline is frictionless.
a) 1 N
b) 66 N
c) 100 N
d) 169 N
e) 1050 N
10. A firefly is flying peacefully through the forest and sees beautiful, swiftly-moving lights just
ahead. It flies toward the lights and comes upon a road, then gets hit by an oncoming bus.
Which statement is true?
a) The magnitude of the force of the bus’s windshield on the fly is equal to that of the fly on the
windshield.
b) The force of the bus’s windshield on the fly has a greater magnitude than the force of the fly
on the bus’s windshield.
c) The force of the fly on the bus’s windshield has a greater magnitude than the force of the bus’s
windshield on the fly.
d) Not enough information to determine.
11. A 10 kg box is being pulled a distance of 3.0 m across a level surface at a constant velocity
by a rope with a magnitude of 5.0 N at a 15 degree angle above the horizontal. How much
work is done by the rope on the box?
a) 10 J
b) 12 J
c) 15 J
d) 30 J
e) 42 J
12. A truck weighs twice as much as a car, and is moving at twice the speed of the car. Which
statement is true about the truck’s kinetic energy compared to that of the car?
a)
b)
c)
d)
e)
They have the same kinetic energy.
The truck has 2 times the kinetic energy of the car.
The truck has 4 times the kinetic energy of the car.
The truck has 8 times the kinetic energy of the car.
The truck has 16 times the kinetic energy of the car.
13. A 50 N grocery bag is moved from a height of 1 m above the floor to a height of 3 m above
the floor. What is the increase in its gravitational potential energy?
a) 50 J
b) 100 J
c) 150 J
d) 980 J
e) 1470 J
14. A house cat with a mass of 5.0 kg jumps with an initial speed of 4.0 m/s from a 1.2 meter
high countertop onto the floor. Neglecting air resistance, what is the cat’s speed immediately
before it hits the surface of the floor?
a) 1.2 m/s
b) 3.4 m/s
c) 4.0 m/s
e) 5.3 m/s
3) 6.3 m/s
15. A car of mass m traveling at speed v crashes into the rear of a truck of mass 2m that was at
rest and in neutral at an intersection. If the collision is perfectly inelastic, what is the speed of
the combined car and truck after the collision?
a) v/3
b) v/2
c) v
d) 2v
e) none of these
16. A 30,000 kg semi-truck is rolling through an intersection at 5 km/h. A 1000 kg sports car is
driving on the highway with a speed limit of 120 km/h. In order for the car’s momentum to
be greater than the truck’s momentum what must be true?
a)
b)
c)
d)
The car must drive the same speed as the truck.
The car must drive 25 km/h.
The car must exceed the speed limit.
It is not possible for the car’s momentum to be greater than the truck’s momentum.
17. Two balls collide, as shown in the figure below. The white ball has a mass 1.0 kg, and the
grey ball has a mass 5.0 kg. Before the collision, the white ball is travelling with a velocity
5.0 m/s and the grey ball is travelling towards the white one with a velocity of -10 m/s. After
the collision, the white ball is travelling at a velocity of -5.0 m/s. What is the velocity of the
grey ball after the collision?
a) -8.0 m/s
b) -5.3 m/s
c) -5.0 m/s
d) +8.0 m/s
e) +10 m/s
18. The previous system was an example of an inelastic collision. What is the change in energy
of the system from before to after the collision?
a) 0 J
b) 45 J
c) 64 J
d) 90 J
e) 263 J
19. You’re stranded near the edge of a frozen, frictionless lake holding a rifle. To get to the edge
of the pond, you brace the rifle against your shoulder and fire horizontally, directly away
from the nearest lake edge. You know the bullet weighs 3.0 g and the bullet exits the rifle at a
speed of 800.0 m/s, and you know you and the rifle together weigh 100.0 kg, so you can
calculate, based on your recoil, how long it should take you to reach the edge of the pond
after firing the gun. The edge of the pond is only 2.0 meters away. How long will it take you
to reach the edge of the pond after firing the gun?
a) 3.2 s
b) 5.5 s
c) 1.4 min
d) 3.5 min
e) 1.6 hours
20. Let’s say a baseball player standing on the pitcher’s mount throws a pitch at 90 mph (40.2
m/s). Determine the impulse required for the pitcher to give a 0.145-kg baseball such a
momentum.
a) 4.2 kg m/s
b) 5.8 kg m/s
c) 6.4 kg m/s
d) 11.4 kg m/s
e) 1.3 x 104 kg m/s
Possibly Useful Information (bold indicates a vector)
v = vo + a t
2
2
Dx = vot + ½ a t
2
v = vo + 2 a Dx
2
Weight = W = m g
g = 9.8 m/s
sin θ = opposite/hyp
cos θ = adjacent/hyp
1 kg = 2.2 pounds
1 m = 3.28 ft
fk = µ k n
fs £ µ s n
tan θ = opposite/adj
quadratic: ax2+bx+c=0
1 in = 2.54 cm
1 mile = 1609.34 m
a2 + b2 = c2
x=(-b±(b2-4ac)^½)/2a
Newton’s 1st Law: Every object continues in its state of rest, or of constant speed in a straight
line, until a nonzero net force acts on it.
Newton’s 2nd Law: net F = m a
Newton’s 3rd Law: When one object exerts a force on a second object, the second object exerts
an equal and opposite force on the first object.
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Work = !|| ∆$ = (component of force in the direction of displacement) (displacement)
Kinetic Energy:
2
KE = ½ m v
Work-Energy Theorem
Power = Work/time
%&'( = %&* + %* = ∆,-
Gravitational Potential Energy:
Work done by gravity:
GPE = mgy
where y is vertical position
%. = −∆01-
Wnc + KEi + PEi = KEf + PEf
where Wnc is work done by frictional forces
KEi + PEi = KEf + PEf
if there is no friction
Momentum: p = m v
I (impulse) = change in momentum = Δp = F Δt
Conservation of momentum: m1v1i + m2v2i = m1v1f + m2v2f
Elastic collision: v1i-v2i = -(v1f-v2f) Perfectly inelastic collision: m1v1i + m2v2i = (m1+ m2)vf
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