AP Physics Final Exam Test Bank | More
than 300 Questions and Verified Answers |
100% Correct (2023/ 2024 Latest Update)
QUESTION
A 50. kg girl on... [Show More] ice skates, initially at rest, pushes against a wall, extending her arms 0.25m for
0.10s causing her to accelerate. Find the magnitude of the force applied by the girl to the wall.
Answer:
2500 N
Fg=ma
x= (1/2) a t2 + v0t
a = 2 x / t2
a = 2*0.25 / 0.12= 50 m/s2
Fg=ma = 50 kg * 50 m/s2 = 2500N
Third Law
Fg=Fw=2500N
QUESTION
Assume an astronaut, including his space suit, has a mass of 150kg. In space, if he experiences
an acceleration of 0.45 m/s2 as he pushes on a 500. kg satellite, what is the magnitude of the
acceleration of the satellite?
Answer:
0.14 m/s2
F = ma = 150 kg * 0.45 m/s2 = 67.5 N
By the third law, the reaction force has the same magnitude:
a = F/m = 67.5N / 500 kg = 0.135 m/s2
QUESTION
A 1500. kg truck is traveling along a straight, level road at a constant speed of 15.0 m/s when the
driver removes his foot from the accelerator. After 18.0 s, the truck's speed is 10.0 m/s.
A. Calculate the acceleration of the truck.
B. Assuming a constant net force was applied opposite the direction of motion, what was the
magnitude of the net force acting on the truck during the 18.0 s?
C. If the same force had been applied to a truck with a mass of 2500. kg, also traveling at 15.0
m/s, how far it would it travel before coming to a complete stop?
Answer:
A. v=at+v0 1 points
a=(v-v0)/t
a=(10-15)/18 1 points
a=0.278 m/s2 1 points
B. F=ma 1 points
F=(1500)(0.278)
F=417 N 1 points
C. a=F/m 1 points
a=417/2500
a=0.167 1 points
v2=v02+2ax 1 points
x=(v2-v02)/2a
x=(02-152)/2(0.167) 1 points
x=1350 m 1 points
QUESTION
A net force is exerted on an object toward the south. The object
Answer:
is accelerating toward the south.
QUESTION
A plane is flying horizontally with its engines applying a force. A strong wind blows upward at
an angle pushing the plane to go faster. Which of the following could be the FBD for this
situation?
Answer:
west,north/west/south
QUESTION
A box is lifted by two ropes. The first rope exerts a force of 600 N at an angle of 35° to the right
of vertical. The second cable exerts a force of 1300 N. If the box is accelerating directly upward,
at what angle to the left of the vertical is the second rope pulling?
Answer:
15°
For the box to accelerate directly upward, the forces in the x direction must sum to zero. Since
the angle is measured from vertical, the x component are the force times the sin of the angle. 600
sin 35°-1300 sinθ=0. 344=1300 sinθ. θ = sin-1(344/1300) = 15°
QUESTION
A ball is thrown directly upward. Neglecting air resistance, which one of the following
statements is true about the net force acting on the ball at the top of the path?
Answer:
The net force is always directed downward throughout the entire freefall problem, on the way up,
on the way down AND at the top; and the magnitude of the net force is mg, the weight of the
ball.
QUESTION
A person with a mass of 45 kg weighs how much on Mars, where g = 3.78 m/2 ?
Answer:
170 N
W=mg=(45)(3.78)= 170 N
QUESTION
What is the mass of a child with a weight of 350 N at the surface of the Earth?
Answer:
36 kg
W=mg
m=W/g=(350)/(9.8)= 36 N
QUESTION
A box resting on a table experiences the force of gravity pulling downward and the normal force
from the table pushing upward. These two forces are opposite in direction and
Answer:
According to Newton's first law, if the net force on an object is zero (gravity pulling down on the
box, while the table is pushing up on the box with a force of the same magnitude), the object is in
equilibrium, stationary in this case.
QUESTION
If you are applying an upward force of 3.0 N to a book weighing 2.0N, the normal force on your
hand due to the book is:
Answer:
3.0 N downward
By Newton's third law, the force of the hand on the book (3.0 N upward) is equal and opposite to
the force of the book on the hand (3.0 N downward).
QUESTION
Apparent weight is equal to your weight when the surface you are standing on
Answer:
Moves at a constant velocity or is at rest.
From the FBD, N-mg=ma, so N=mg+ma = mg + 0 for an object at rest or travelling at a constant
velocity.
QUESTION
An elevator is accelerating with a 57.0 kg girl standing on a scale in an elevator. If the scale
reads 450. N, what is the magnitude of the acceleration of the elevator?
Answer:
1.91 m/s2
From the FBD, N-mg = ma, a=(N-mg)/m = (450-57*9.8)/57 = 1.91 m/s2 [Show Less]
