Download Kinematics Multiples

Systems
From Princeton Review Book
1. Assuming a frictionless, massless pulley, determine the acceleration of the blocks once
they are released from rest. Hint: Check limiting values.
a.
m
g
( M  m)
b.
M
g
( M  m)
c.
M
g
m
d.
( M  m)
g
( M  m)
e.
( M  m)
g
( M  m)
*E. The net force is the difference in the weights (the solutions have factored out the “g”).
The net inertia is the sum of the masses.
2. In the figure shown below, the coefficient of sliding friction between the small block and
the tabletop is 0.2. If the pulley is frictionless and massless, what will be the acceleration
of the blocks once they are released from rest?
a. (0.45)g
b. (0.6)g
c. (0.7)g
d. (0.8)g
e. (0.9)g
*C. Apply Newton’s second law system trick:
a system 
TotalForce (3mg )  friction 3mg  (.2)( mg ) 2.8mg



 .7 g
TotalMass
4m
4m
4m
Note : some people want to also subtract t he weight of the block on top of the table.
Don' t do this! !! The weight of this block is completely canceled by its normal force.
The mass of the block does resist motion - -but this factor does not show up in the numerator,
it appears in the denominato r as part of the total inertia of the system.
==================================================
From Old AP’s
3. 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
* B. Use the system trick:
a
Fexternal
(mg )
g


m
(4mg ) 4
Some folks want to subtract the 3mg from the mg in the numerator. Don’t do this. The
weight of the top block is totally cancelled by the normal force. In addition, the weight is
perpendicular to the motion of the block, so it can have no effect on it. The mass of the top
block does act to slow the system down—you account for this by including its mass in the
denominator.
4. (1974) Three blocks: A, B, and C, each have mass m and are connected by massless
strings as shown above. Block C is pulled to the right by a force F, causing the entire threeblock system to accelerate. Assuming frictionless contact between the blocks and the level
table beneath them, what is the net force experienced by block B?
a. zero.
b. F/3
c. F/2
d. 2F/3
e. F
* B.
First, find the acceleration of the whole system:
a
F
F

m 3m
Now we only care about the NET force on block B, so we don' t really need to look at tension s.
 F 
F  ma  (1m) 
  F /3
 3m 
5. (1984) When the frictionless system shown below is accelerated by an applied force of
magnitude F, the tension in the string between the blocks is:
a. 2 F
b. 1 F
c. 2/3 F
d. ½ F
e. 1/3 F
* E. First, find the acceleration of the system:
a
F
F

m 3kg
Now, analyze the 1 kg block.
The only horizontal force acting on it is the tension, which is what we are solving for :
F  ma
 F  F
 
T  (1kg) 
 3kg  3
You could also use logic to infer that since the rope is only pulling 1/3 the total mass, the
tension in the rope is 1/3 the total force on the system.
6. Two blocks are pushed along a horizontal frictionless surface by a force of 20 N to the
right as shown above. The force that the 2 kg block exerts on the 3 kg block is:
a. 8 Newtons to the left.
b. 8 Newtons to the right.
c. 10 Newtons to the left.
d. 12 Newtons to the right.
e. 20 Newtons to the left.
* A. First, find the acceleration of the system:
a
F 20 N

 4m / s 2
m 5kg
Even thoug h the problem asks for the force on the 3 kg block, by Newton' s third law this is the
same as the force on the 2 kg block.
This is the only force acting on the 2 kg block, so it is easier to analyze it :
F  ma
Normal  (2kg)(4 m/s 2 )  8 Newtons
You could also look at the 3 kg block if you wished :
F  ma
20  Normal  (3kg)( 4m / s 2 )
So the normal  8 Newtons
And the normal acts to the LEFT on the 3 kg block since it is caused by electrical repulsion
from the surface of the 2 kg block.
7. (1993) Two 0.60 kg objects are connected by a thread that passes over a light,
frictionless pulley, as shown below. The objects are initially held at rest. If a third object
with a mass of 0.30 kg is added on top of one of the 0.60 kg objects as shown and the
objects are released, the magnitude of the acceleration of the 0.30 kg object is most
nearly:
a. 10.0 m/s2.
b. 6.0 m/s2.
c. 3.0 m/s2.
d. 2.0 m/s2.
e. 1.0 m/s2.
* D. All three blocks have the same acceleration, so I’m not sure why they asked for the
acceleration of the one block, unless it was just to confuse you.
F (.9kg)(10m / s 2 )  (.6kg)(10m / s 2 )
a

 2m / s 2
m
(.9kg  .6kg)
Note : Here you DO want to subtract t he weight of the " losing" mass becuase its weight
is not being supported by a surface below it. It' s weight is parallel to the motion.
8. (2004, 68%)
Three blocks of masses 3m, 2m, and m are connected to strings A, B, and C as shown
above. The blocks are pulled along a rough surface by a force of magnitude F exerted by
string C. The coefficient of friction between each block and the surface is the same. Which
string must be the strongest in order not to break?
a. A.
b. B.
c. C.
d. They must all be the same strength.
e. It is impossible to determine without knowing the coefficient of friction.
* C Although it might be tempting to choose "e" because you don't know how much
friction each rope has to fight, it is not relevant. Remember, every object has the same
acceleration. Thus, rope A has to exert enough force to "fight" the friction on block A and
the inertia of block A (even though inertia is not a force). Rope B has to exert enough force
to fight friction on two blocks and accelerate the inertia of two blocks. Rope C must fight
friction on all three blocks as well as accelerate all three blocks. Clearly this takes the most
force. It would be a different story if friction were helping the ropes.
9.
(2004, 40%)
As shown above, two students sit at opposite ends of a boat that is initially at rest. The
student in the front throws a heavy ball to the student in the back. What is the motion of the
boat at the time immediately after the ball is thrown and later, after the ball is caught?
(Assume that air and water friction are negligible.)
a.
b.
c.
d.
e.
Immediately
After the throw
After the catch
Boat moves forward
Boat moves forward
Boat moves forward
Boat moves backward
Boat moves backward
Boat moves forward
Boat moves backward
Boat does not move
Boat does not move
Boat moves forward
* C.
When the girl throws the ball, she and the boat recoil forward by Newton's third law.
However, when the boy catches the ball, the ball exerts a force on the boy and the boat,
causing the boat to come to a stop. The ball is internal to the system, and so the only way to
make the boat move for an indefinite period of time is for one of the kids to throw the ball
OFF the boat so that it is no longer part of the system.
10.
(2004, 54 %)
Two blocks of masses M and m, with M > m, are connected by a light string. The string
passes over a frictionless pulley of negligible mass so that the blocks hand vertically. The
blocks are then released from rest. What is the acceleration of the block of mass M?
a. g.
b.
( M  m)
g
M
c.
( M  m)
g
M
d.
( M  m)
g
( M  m)
e.
( M  m)
g
( M  m)
*E. This is an Atwood's machine question, even though they didn't draw a diagram.
The acceleration is the difference in the forces (because each block wants to accelerate in a
different direction) divided by the total inertia.
This is solution E.