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AP Physics
Week 5 Homework
Read Chapter 6
Key
Outcomes
Quiz 5 Preparation
Practice Problems
Item
#
Questions 1 - 2
1.
E
Question
A 2-kilogram block slides down a 30° incline as shown above with an acceleration of 2 meters per second squared.
Which of the following diagrams best represents the gravitational force W. the frictional force f, and the normal force N that act on
the block?
AP
2.
C
The magnitude of the frictional force along the plane is most nearly
(A) 2.5 N
(B) 5 N
(C) 6 N
( D) 10 N
( E) 16 N
FII-Fnet=Ff Fnet=m•a FII=mgsin (2kg)(10m/s2)(sin30°) – (2kg)(2m/s2) = 6N
AP
When the frictionless system shown above is accelerated by an
applied force of magnitude F, the tension in the string between
the blocks is
T between = Fnet on 1kg block
3.
E
(A) 2F
(B) F
(C)
2
3
F
(D)
1
F
2
(E)
1
F
3
AP
The acceleration of this object must be
(A) zero
(B) constant but not zero
(C) increasing
(D) decreasing
(E) equal to g
4.
A
The displacement x of an object moving along the x-axis is
shown above as a function of time t.
A uniform rope of weight 50 newtons hangs from a hook as
shown above. A box of weight 100 newtons hangs from the
rope.
5.
E
AP
What is the tension in the rope?
(A) 50 N throughout the rope
(B) 75 N throughout the rope
(C) 100 N throughout the rope
(D) 150 N throughout the rope
(E) It varies from 100 N at the bottom of the rope to 150 N at
the top.
AP
6.
Drop.
A ball falls straight down through the air under the influence of gravity. There is a retarding force F on the ball with
magnitude given by F = bv, where t is the speed of the ball and b is a positive constant. The magnitude of the
acceleration a of the ball at any time is equal to which of the following?
(A) (B) (C) (D) (E)
AP
7.
B
A block of mass 3m can move without friction on a horizontal table. This block is attached to another block of mass m
by a cord that passes over a frictionless pulley, as shown above. If the masses of the cord and the pulley are
negligible, what is the magnitude of the acceleration of the descending block?
(A) Zero
(B) g/4
(C) g/3
(D) 2g/3
(E) g
AP
Questions 8-9
The mass of the block is most nearly
(A) 1.0 kg
(B) 1.2 kg
(C) 1.6 kg
(D) 2.0 kg
(E) 2.5 kg
8.
D
A plane 5 meters in length is inclined at an angle of 37, as
shown above. A block of weight 20 newtons is placed at the
top of the plane and allowed to slide down.
The magnitude of the normal force exerted on the block by the plane is most nearly
9.
C
AP
(A) l0 N (B) 12N (C) 16 N (D) 20 N (E) 33 N
AP
The largest acceleration that can be given to the block by
pulling up on it with the rope without breaking the rope is most
nearly
(A)
(B)
(C)
(D)
(E)
10.
B
A rope of negligible mass supports a block that weighs 30 N,
as shown above. The breaking strength of the rope is 50 N.
6 m/s2
6.7 m/s2
10 m/s2
15 m/s2
16.7 m/s2
Fnet = ma Fnet/m =a =20N/(30N/10m/s2)= 6.7 m/s2
AP
If the reading on the scale is 3 N, then the buoyant force that
the fluid exerts on the object is most nearly
(A)
(B)
(C)
(D)
(E)
11.
B
The figure shows an object of mass 0.4 kg that is suspended
from a scale and submerged in a liquid.
1.3 N
1.0 N
0.75 N
0.33 N
0.25 N
w – Fb = Fscale w-Fscale= Fb (0.4kg)(10m/s2) -3N =1N
AP
What is the magnitude of the horizontal acceleration of the
cart?
(A)
0.5 m/s2
(B)
1.6 m/s2
(C)
2.0 m/s2
(D)
2.5 m/s2
(E)
2.6 m/s2
12.
A
The cart of mass 10 kg shown above moves without frictional
loss on a level table. A 10 N force pulls on the cart horizontally
to the right. At the same time, a 30 N force at an angle of 600
above the horizontal pulls on the cart to the left.
Fnet = ma Fnet/m=a 5N/10kg= 0.5m/s2
Fnet = Fx – Ff= 15N-10N=5N
Ff=10N
Fx=Fcos = (30N)(cos60°)=15N
AP
13.
C
An object of mass m is initially at rest and free to move without friction in any direction in the xy-plane. A constant net force of
magnitude F directed in the +x direction acts on the object for 1 s. Immediately thereafter a constant net force of the same
magnitude F directed in the +y direction acts on the object for 1 s. After this, no forces act on the object.
Which of the following vectors could represent the velocity of the object at the end of 3 s, assuming the scales on the x and y
axes are equal?
AP
Which two graphs represent the motion of an object on which the net force is zero?
14.
A
A force F1, and a 200.-newton force, F2, are applied simultaneously to the same point on a large crate resting
on a frictionless, horizontal surface. Which diagram shows the forces positioned to give the crate the greatest acceleration?
15.
A
A)
B)
) A bird feeder with two birds has a total mass of 2.0 kilograms
and is supported by wire as shown in the diagram below.
C)
D)
The force in the top wire is approximately
A) 20. N
B) 10. N
C) 14 N
D) 39 N
16.
A
The diagram below represents a box shown sliding down an
incline plane.
Toward which point will the force of friction on the
box be directed?
17.
D
A) 1
B) 2
C) 3
A box decelerates as it moves to the right along a horizontal surface, as shown in the diagram below.
18.
A
Which vector best represents the force of friction on the box?
D) 4
Block A is pulled with constant velocity up an incline as shown
in the diagram below.
19.
A
20.
B
Which arrow best represents the direction of the force of friction
acting on block A?
A constant unbalanced force of friction acts on a 15.0-kilogram mass moving along a horizontal surface at 10.0 meters per
second. If the mass is brought to rest in 1.50 seconds, what is the magnitude of the force of friction?
F=ma (m)(∆v/∆t) = (15kg) (-10m/s) / (1.5 s) = 100 N
A) 147 N
B) 100. N
In the graph below, the acceleration of an object is plotted
against the unbalanced force on the object.
C) 150. N
D) 10.0 N
What is the object's mass?
A) 2 kg
B) 1 kg
C) 0.5 kg
D) 0.2 kg
21.
C
F= ma F/a = m (3N)/(6m/s2 )= 0.5 kg
An 800-newton person is standing in an elevator. If the upward force of the elevator on the person is 600 newtons, the
person is
22.
A
A) accelerating downward
B) accelerating upward
C) moving downward at constant speed
D) at rest
Two forces are applied to a 2.0-kilogram block on a frictionless,
horizontal surface, as shown in the diagram below.
23.
D
The acceleration of the block is
A) 3.0 m/s2 to the left
B) 5.0 m/s2 to the left
C) 5.0 m/s2 to the right
D) 3.0 m/s2 to the right
Fnet = ma Fnet/m = a (8N-2N)/(2kg) = 3m/s2
The graph below shows speed as a function of time for four
cars, A, B, C, and D, in straight-line motion.
24.
D
Which car experienced the greatest average acceleration
during this 6.0-second interval?
A) car A
B) car B
C) car C
D) car D
The graph below represents the velocity-time relationship for a
2.0-kilogram mass moving along a horizontal frictionless
surface.
A) 2.0 N
B) 1.0 N
C) 0 N
D) 4.0 N
25.
C
A copper coin resting on a piece of cardboard is placed on a
beaker as shown in the diagram below. When the cardboard is
rapidly removed, the coin drops into the beaker.
26.
B
The net force on the mass during interval DE is
The two properties of the coin which best explain its fall are its
weight and its
A) electrical resistance
B) inertia
C) volume
D) temperature
I. An aluminum block weighing 20. newtons, sliding from left to right in a straight line on a horizontal steel surface, is acted on
by a 2.4-newton friction force. The block will be brought to rest by the friction force in a distance of 10. meters.
(a) On the diagram of the block, draw an arrow to identify the direction of each force acting on the block while it is still
moving, but is being slowed by the friction force. Identify each force by appropriately labeling the arrow that represents
its line of direction.
(b) Determine the magnitude of the acceleration of the block as it is brought to rest by friction force. [Show all work.]
a=Fnet/m (2.4 N) / ((20N/10m/s2) = 1.2 m/s2
II. In a laboratory exercise, a student collected the following data as the unbalanced force applied to a body of mass M was
changed.
(a) Label the axes of the graph with the appropriate values for force and acceleration.
(b) Plot an acceleration versus force graph for the laboratory data provided.
(c) Using the data or your graph, determine the mass, M, of the body. [Show all calculations.]
III. A ball of weight 5 newtons is suspended by two strings as shown above.
a.
In the space below, draw and clearly label all the forces that act on the ball.
b. Determine the magnitude of each of the forces indicated in part (a).
Since the ball is at rest, the net force = 0. T1+T2+w=0
T1y+T2y+w=0
T1sin53°+ T2sin37°+(-w)=0
T2x+T2x=0
T1cos53°+ (-T2cos37°) =0
T1sin53°+ T2sin37°+(-w)=0
T1sin53°+ T2sin37°= (w)
T1cos53°=T2cos37°
T1cos53°/cos37°= T2
0.75T1= T2
T1sin53°+ 0.75T1sin37°= (w)
T1•0.8+ 0.75T1•0.6= (w)
0.8T1+ 0.45T1= (w)
1.25 T1= (w)
T1= (w)/1.25 = 4N
4N•cos53°+ (-T2cos37°) =0
2.4N/cos37°= T2
3N= T2
III. A 10-kilogram block rests initially on a table as shown in cases I and II above. The coefficient of sliding friction between the block
and the table is 0.2. The block is connected to a cord of negligible mass, which hangs over a massless frictionless pulley. In case I a
force of 50 newtons is applied to the cord. In case II an object of mass 5 kilograms is hung on the bottom of the cord. Use g = 10
meters per second squared.
a. Calculate the acceleration of the 10-kilogram block in case I.
a= Fnet/msystem Fnet = T-Ff Ff=µFn Fn = -w w=mg a= T-µmg/msystem = ((50N)-(0.2)(10kg)(10m/s2))/(10kg) = 3m/s2
b. On the diagrams below, draw and label all the forces acting on each block in case II
c. Calculate the acceleration of the 10-kilogram block in case II.
a= Fnet/msystem Fnet = T-Ff Ff=µFn Fn = -w w=mg a= T-µmg/msystem = [(5kg)(10m/s2)-(0.2)(10kg)(10m/s2)]/(10kg+5kg) = 2m/s2
IV. A crane is used to hoist a load of mass m 1 = 500 kilograms. The load is suspended by a cable from a hook of mass m2 = 50
kilograms, as shown in the diagram above. The load is lifted upward at a constant acceleration of 2 m/s 2.
a. On the diagrams below draw and label the forces acting on the hook and the forces acting on the load as they accelerate upward
b.
Determine the tension T1 in the lower cable and the tension T2 in the upper cable as the hook and and load are accelerated upward
at 2 m/s2. Use g = 10 m/s².
Tupper = whook + wload + Fnet hook + Fnet load = m hookg + m loadg + m hooka + m loada
= (50kg)(10m/s2) + (500kg)(10m/s2) + (50kg)(2m/s2) + (500kg)(2m/s2) = 6600 N
Tlower = wload + Fnet load = (500kg)(10m/s2) + (500kg)(2m/s2) = 6000 N
V. A student whose normal weight is 500 newtons stands on a scale in an elevator and records the scale reading as a function of time.
The data are shown in the graph above. At time t = 0, the elevator is at displacement x = 0 with velocity v = 0. Assume that the positive
directions for displacement, velocity, and acceleration are upward.
a. On the diagram below, draw and label all of the forces on the student at t = 8 seconds.
b.
Rider mass = 50kg= w/g = 500N/10m/s2
Calculate the acceleration a of the elevator for each 5-second interval.
i. Indicate your results by completing the following table.
Time Interval (s)
a (m|s2)
0-5
0
5-10
200N/50kg= 4m/s2
10-15
0
15-20
-200N/50kg=-4m/s2
ii. Plot the acceleration as a function of time on the following graph.
c.
Determine the velocity v of the elevator at the end of each 5-second interval.
i. Indicate your results by completing the following table.
Time (s)
0-5
5-10
10-15
15-20
v (m/s)
0 v=at = (4m/s2)(5s)=20m/s
20m/s
v=vo+at = (20m/s)+(-4m/s2)(5s) = 0
ii. Plot the velocity as a function of time on the following graph.
d.
Determine the displacement x of the elevator above the starting point at the end of each 5-second interval.
i. Indicate your results by completing the following table.
Time (s) 0-5
5-10
10-15
15-20
x (m)
0 x=1/2at2=50m x=xo+vt=150m
x=xo+vt+1/2at2=200m
ii. Plot the displacement as a function of time on the following graph.
VI.
Two small blocks, each of mass m, are connected by a string of constant length 4h and negligible mass. Block A is placed on a smooth
tabletop as shown above, and block B hangs over the edge of the table. The tabletop is a distance 2h above the floor. Block A is then
released from rest at a distance h above the floor at time t = 0.
Express all algebraic answers in terms of h, m, and g.
a. Determine the acceleration of block A as it descends.
a= fnet/msystem mg/2m = g/2
b.
Block B strikes the floor and does not bounce. Determine the time t = t 1 at which block B strikes the floor.
t=
h=1/2at2
2h
h
=2
g
g
2
c. Describe the motion of block A from time t = 0 to the time when block B strikes the floor.
uniform acceleration
d. Describe the motion of block A from the time block B strikes the floor to the time block A leaves the table.
uniform velocity
e.
Determine the distance between the landing points of the two blocks.
x= vt
v 2 = v o2 +2ah
v= 2ah
v=
2gh
2
v= gh
x = vt =
(
æ
ö
h÷
gh çç2
÷ = 2h
è gø
)