A heavy block, labeled “A”, is sitting on a table. On top of that block is a lig
ID: 1339342 • Letter: A
Question
A heavy block, labeled “A”, is sitting on a table. On top of that block is a lighter block, labeled “B” as shown in the figure at the right. Consider three cases.
1) The finger is pushing but not hard enough. Neither block moves.
2) The finger is pushing hard enough that the two blocks are speeding up.
3) The blocks have sped up, and are now moving at a constant speed. The finger still has to push to keep them going at a constant velocity.
In all cases where the blocks are moving, they are moving together. (Block B is not sliding on block A.) Block A has a mass of 0.8 kg; block B has a mass of 0.1 kg. the coefficient of friction between the block and the table is 0.29, and the coefficient of friction between the two blocks is 0.18. (You may use g = 10 N/kg, and you may treat kinetic and static friction as the same.)
A) For case (1), what is the maximum force the finger can exert without the blocks beginning to move?
Answer:
B) For case (2)...
...while the blocks are speeding up, the finger is pushing with a force of 4.1 N. What is the acceleration of block A?
Answer:
...find the time t needed for this force to bring the blocks from rest to a speed of 60 cm/s.
Answer:
...we are assuming that the blocks are moving together. This means that block B is speeding up.
Which of the following forces is responsible for the acceleration of block B?
i.Force applied by the finger, to the right
ii.Force applied by the finger, to the left
iii.Friction force exerted by the floor, to the right
iv.Friction force exerted by the floor, to the left
v.Friction force exerted by block A, to the right
vi.Friction force exerted by block A, to the left
vii.Some other force, to the right
viii.Some other force, to the left
Answer:
...find the magnitude of this force.
Answer:
C) For case (3), what force do you need to exert to keep the blocks moving at a constant speed of 60 cm/s?
Answer:
Explanation / Answer
maximum static friction( taking both the blocks as a single system so that the friction between two blocks wont come into picture)
=0.29*(0.8+0.1)*10=2.61 N
hence in case 1, maximum force that can be applied by the finger is 2.61 N
b)net force exerted on both the blocks=4.1-2.61=1.49 N
hence acceleration=force applied/total mass=1.49/0.9=1.656 N/m^2
c)
as we know,
final veloicty=initial veloicty+acceleration*time
==>0.6 m/s=0 + 1.656*t^2
==>t=0.602 seconds
d)the forces responsible for acceleration of block B:
v.Friction force exerted by block A, to the right
C)
in order to keep the block moving at constant speed, net force on the blocks=0
hence force applied=friction force=2.61 N
hence force of 2.61 N is required to be applied so that blocks wil move at constant speed of 60 cm/s.
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