Name
Period
Date
Quiz: Combined Forces [Concepts]
AP Physics
1.
C _____ A force F (larger than the largest possible force of static friction) is applied to the left to an object moving to the
right on a horizontal surface. Then:
A. the object must be
B. F and the friction
C. the object must be
D. the object must be
E. the object must come
moving at constant
force act in opposite
slowing down
speeding up
to rest and remain at
speed
directions
rest
2.
A _____ A block of mass m is pulled at constant velocity along a rough horizontal floor by an
applied force T as shown. The magnitude of the frictional force is:
C. zero
D. mg
A. T cos θ
B. T sin θ
E. mg cos θ
3.
A _____ Equal forces F act on isolated bodies A and B. The mass of B is three times that of A.
The magnitude of the acceleration of A is:
A. three times that of B
B. 1/3 that of B
C. the same as B
D. nine times that of B
4. B _____ An example of an inertial reference frame is:
A. any reference frame B. a frame attached to
C. any reference frame
that is not accelerating
a particle on which
that is at rest
there are no forces
D. a reference frame
attached to the center
of the universe
E. 1/9 that of B
E. a reference frame
attached to Earth
5.
C _____ A coin is placed on a horizontal phonograph turntable. Let N be the magnitude of the normal force exerted by the
turntable on the coin, ƒ be the magnitude of the frictional force exerted by the turntable on the coin, and ƒs, max be the
maximum possible force of static friction. The speed of the turntable is increased in small steps. If the coin does not slide, then
A. N increases, ƒ
B. N increases, ƒ
C. ƒ increases and both D. N, ƒ, and ƒs, max all
E. N, ƒ, and ƒs, max all
increases, and ƒs, max
increases, and ƒs, max
N and ƒs, max stay the
stay the same
increase
stays the same
increases
same
6. A _____ If we know an object is moving at constant velocity, we may assume:
A. the net force acting
B. there are no forces
C. the object is
D. the object is losing
on the object is zero.
acting on the object.
accelerating.
mass.
7. B _____ Why do raindrops fall with constant speed during the later stages of their descent?
A. The gravitational
B. Air resistance just
C. The drops all fall
D. The force of gravity
force is the same for all balances the force of
from the same height
is negligible for objects
drops
gravity
as small as raindrops
8. D _____ An object moving in a circle at constant speed:
A. must have only one
B. is not accelerating
C. is held to its path by
force acting on it
centrifugal force
D. has an acceleration
of constant magnitude
E. the net force acting
on the object isn't zero.
E. Gravity cannot
increase the speed of a
falling object to more
than 9.8m/s
E. has an acceleration
that is tangent to the
circle
9.
D _____ A giant wheel, having a diameter of 40 m, is fitted with a cage and platform on which a man of mass m stands. The
wheel is rotated in a vertical plane at such a speed that the force exerted by the man on the platform is equal to his weight
when the cage is at X, as shown. The net force on the man at point X is:
A. zero
B. mg, down
C. mg, up
D. 2mg, down
E. 2mg, up
10. A _____ A heavy steel ball B is suspended by a cord from a block of wood W. The entire
system is dropped through the air. Neglecting air resistance, the tension in the cord is:
A. zero
B. the difference in the
C. the difference in the
D. the weight of B
masses of B and W
weights of B and W
Quiz-Combined Forces
4
E. none of these
11/10/14
Name
Period
Date
AP Physics
Quiz: Combined Forces [Graphs]
1. A groups of washers of mass m2=0.025 kg is swung around in a level circular path on the end of a
string. The radius of the path, r, is increased over several trials and the speed is measured for each trial.
The supported mass, m1, and the washer mass, m2 is kept the same for all trials.
a.
3.5 m/s ___________ Estimate the speed of m2 for a radius of 0.8 meter.
b.
0.4 N _____________ Calculate the tension force acting on m2.
6
Fnet=mv2/r = Tà(0.025)(3.52)/0.8 = T = 0.38 N
c.
Sketch what velocity squared vs. radius would look like for this
data.
Speed, m/s
5
4
3
2
1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Radius, m
2. The vertical and horizontal position of a toy helicopter is graphed, relative to a stationary observer.
b.
5 m ______________ Calculate the magnitude of the initial
position vector of the helicopter.
1.2 m/s ___________ Calculate the magnitude of the initial
velocity of the helicopter.
Use both slopes: Vy≈2/3 m/s= 0.67 m/s
Vx≈1; 12+0.672=V2 à V=1.204
c.
D _____ At which moment (from these choices) is the helicopter
moving with the largest speed?
A. 0 sec
B. 1 sec
C. 7.5 sec D. 9 sec
E. The speed is
constant
throughout the
graph.
d.
10
9
8
Vertical
7
Position, m
a.
6
5
riz
Ho
4
3
2
1
0
0
1
2
3
4
5
6
7
8
Times, seconds
Sketch graphs of horizontal velocity and horizontal acceleration for the helicopter based on the original graph.
e. C _____ At which moment is the helicopter affected by the largest net force?
A. 0 s
B. 6 s
C. 7.5 s
D. 9 s
f.
tal
on
E. Fnet is the same at all times
0 ________________ Estimate the net force acting on the helicopter at t=4 seconds.
Velocities are both constant
Quiz-Combined Forces
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10
Name
Period
Date
AP Physics
Quiz: Combined Forces [Calculations]
Give answers with correct precision and units. Show work for credit.
1. The blocks are connected by a light string which passes over two pulleys. The blocks have relative masses M, 3M and 4M, as
shown. Assume all surfaces and the pulleys are frictionless and that they are initially at rest.
a.
a=³⁄₈ g _____________ Develop an expression for the acceleration
of the 3M block in terms of M and g(the acceleration of gravity).
Forces on connected system:
Fnet=4Mg–T2+T2–T1+T1–Mg = 3Mg & Fnet=8Ma à 3Mg=8Ma àa=³⁄₈ g
Since the tension forces act in both directions on different parts of the system, they
cancel out.
b. ¹¹⁄₈ g _______________ Find an expression for T1 in terms of M and g.
On “M”:
Fnet=T1–Mg and Fnet= Ma à T–Mg=M(³⁄₈g) àT=³⁄₈g + ⁸⁄₈ g = ¹¹⁄₈ g
Can also do this with multiple expressions
(1) Fnet=Ma=T1–Mg à
(2) Fnet=3Ma=–T1+T2àT1=T2–3Ma
(3) Fnet=4Ma=–T2+4Mg
Ma=T1–Mg = (T2–3Ma)–Mgà4Ma+Mg=T2
4Ma=–T2+4Mg=–(4Ma+Mg)+4Mgà
4Ma=–4Ma–Mg+4Mg
8Ma = 3Mg
a=³⁄₈ g
¾gy________________ Calculate the instantaneous velocity of the 4M block after it has fallen
through a vertical displacement of y. Your answer should be in terms of the given variables.
vo=0, a=³⁄₈g à v2=vo2=2a∆y àv2=2(³⁄₈g)y = ¾gy
c.
2. A ball of mass m1=130 grams it attached to a string which passes through a hole in a frictionless surface. The other end of the
string supports block of mass m2=450 grams. The ball is spun around in a circular path so that it maintains contact with the flat
surface at a constant radius of r=23 cm.
a. 4.4 N _______________ Calculate the tension of the string acting on m1.
on m1: Fnet=T=m2g = 0.45 • 9.8 = 4.41 N
b. 2.8 m/s _____________ Calculate the instantaneous speed of the ball.
Fnet=T=m2g and Fnet=m1v2/r à m2g=m1v2/ràv=m2gr/m1=v2 à v= m2gr = (0.450)(9.8)(0.23) =2.79 m/s
m1
0.130
c. 0.52 sec ____________ Calculate the time required for the ball to complete one complete trip around its circular path.
v=2πr/t à t=2πr/v = 0.518 sec
3. An adult exerts a horizontal force on a swing that is suspended by a rope of length , L, holding it at an angle θ with the
vertical. The child in the swing has a weight W and dimensions that are negligible compared to L. The weights of the rope and of
the seat are negligible. In terms of W and θ determine:
a. T=W/cosθ ___________ The tension in the rope.
y:Fnet=0 & Fnet=Ty–W; Ty=Tcosθ ;
0= Tcosθ–W à Tcosθ=W àT=W/cosθ
Ty
b. Wtanθ ______________ The horizontal force exerted by the adult.
x:Fnet=0 & Fnet= Tx–F; Tx=Tsinθ; T=W/cosθ; 0=Tsinθ – Fà (W/cosθ)sinθ=FàF=Wtanθ
The adult releases the swing from rest. In terms of W and θ, determine:
T=Wcosθ ___________ The tension in the rope just after the release (the swing is instantaneously at
rest).
Calculated for forces along rope since centripetal force acts directly down length of rope:
Fnet=mv2/L = 0 & Fnet=T–WropeàT=Wrope
T=Wcosθ
d. 3W–2Wcosθ _________ The tension in the rope as the swing passes through the lowest point.
Fnet=mv2/L =T–W [W is directly down here so Wrope=W]
Need to find v2, m and r in terms of W and θ
L-Y
• m=W/g
•cosθ=(L-Y)/Là Lcosθ=(L-Y)àY=L–Lcosθ=L(1–cosθ)
• v2=vo2+2ay = 2gy=2gL(1–cosθ) à v2=L(1–cosθ)
àmv2/L=(W/g)2gL(1–cosθ)/L = 2W(1–cosθ)
Y
2W(1–cosθ)=T–W
2W–2Wcosθ=T–Wà3W–2Wcosθ
Quiz-Combined Forces
6
T
Tx
F
c.
W
T
Wrope
θ
W
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