qA_20

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course phy 201

10/25 8:00 pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

020. Forces (inclines, friction)

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Question: `q001. Note that this assignment contains 3 questions.

. A 5 kg block rests on a tabletop. A string runs horizontally from the block over a pulley of negligible mass and with negligible friction at the edge of the table. There is a 2 kg block hanging from the string. If there is no friction between the block in the tabletop, what should be the acceleration of the system after its release?

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Your solution:

M1: 5 kg; m2: 2 kg; acceleration on m2: 9.8 m/s^2; force applied: 19.6 newtons; total weight of system: 7 kg; acceleration of entire system: 2.8 m/s^2;

confidence rating #$&*:32; 3

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Given Solution:

Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero.

The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons.

The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2.

STUDENT QUESTION

since it is a frictionless system i thought the answer would the acceleration of gravity since that would be the force pulling the objects down. INSTRUCTOR RESPONSE

A mass resting by itself on a level tabletop does now accelerate downward in response to the gravitational force exerted on it. This is because the tabletop pushes up on it. This occurs whether friction is high, low or absent. Of course if friction is very low, you have to be sure the tabletop is very nearly level, and if friction is absent you'd best be sure it is completely level.
These forces are still present if you add the hanging mass to the system. 
The mass on the tabletop does experience the downward pull of gravity, but that force is balanced by the supporting force of the tabletop. The mass therefore does not move in the direction of the gravitational pull. It is nevertheless part of the system being accelerated by the gravitational force on the hanging mass.

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Self-critique (if necessary):OK

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Self-critique rating: OK

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Question: `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.

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Your solution:

M1: 5 kg; m2: 2 kg; acceleration on both objects: 9.8 m/s^2; force of friction: 0.1 force on m1; force m1: 49 newtons; force m2: 19.6 newtons; force of friction: 4.9 newtons; force on system: 19.6 - 4.9 = 14.7 newtons; acceleration of system: 2.1 m/s^2;

confidence rating #$&*:32; 3

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Given Solution:

Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object.

The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.

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Self-critique (if necessary):OK

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Self-critique rating: OK

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Question: `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.

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Your solution:

M1: 5 kg; m2: 2 kg; acceleration on both objects: 9.8 m/s^2; force of friction: 0.1 force on m1; force m1: 49 newtons; force m2: 19.6 newtons; force of friction: 4.9 newtons; force on system: 19.6 - 4.9 = 14.7 newtons; acceleration of system: 2.1 m/s^2;

confidence rating #$&*:32; 3

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Given Solution:

Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object.

The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.

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Self-critique (if necessary):OK

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Self-critique rating: OK

#*&!

&#This looks very good. Let me know if you have any questions. &#