Solving the Work-Energy Theorem Error in Block-Table System

In summary, the block and table both start radiating in the infrared. The block eventually comes to a stop, but consumes some more energy in the process.
  • #1
Leong
382
2
A block (2 kg) is moving with an initial speed of 1 m/s on a horizontal rough table comes to a stop eventually.

Applying work-energy theorem to the block-table system, we obtain

Change in the kinetic energy of the block + Change in the kinetic energy of the table = Net work done on the block-table

(-1 J) + (0) [the table doesn't move] = Net work done on the block + Net work done on the table
= [Work done by normal force on the block by table + Work done by friction on the block by table + Work done by weight on the block] + [Work done by normal force on the table by block + Work done by friction on the table by block+ Work done by weight on the table]

Work done by weight on the block is 0 because the angle between the displacement and the weight is 90 degree. Work done by weight on the table is 0 because it doesn't move and therefore has 0 displacement. Work done by the two frictions is also 0 because friction on block by table and friction on table by block are action-reaction pair and thus in general, their resultant is zero and produce zero work. The same goes to the normal force.

Thus, basically the net work done on the block-table is zero.

And we get:
-1 = 0
What is wrong?
 
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  • #2
Work done by the two frictions is also 0 because friction on block by table and friction on table by block are action-reaction pair and thus in general, their resultant is zero and produce zero work.

But you just said the table doesn't move.
 
  • #3
So, the table should move to fix the -1 = 0?
 
  • #4
The energy doesn't disappear, it becomes heat.
The block and table both start radiating in the infrared.
It is also unlikely the block will come to a complete stop, but will
"bounce" and thereby consume some more energy.
 
  • #5
Unless something's holding the table in place. In which case the table and that thing get a slight amount of energy which fixes that. If it's anchored to the Earth, assuming the Earth's perfectly rigid (which it isn't, but assuming,) then it'll have its motion changed slightly to fix that.

EDIT: Duh. Heat. *Bangs head against wall due to the sheer stupidity of the fact that I overlooked that*
 
  • #6
If we include heat in the equation, how will it fix the equation?
Or if the table is free to move, then it will move and gain kinetic energy of 1 J?
 
  • #7
Work done by weight on the block is 0 because the angle between the displacement and the weight is 90 degree.

No, the work is done by the force of friction which acts opposite to the displacement vector.
If we include heat in the equation, how will it fix the equation?

There is one joule of heat disippated. This is the work done by the force of friction stopping the motion of the block.
 
  • #8
I have done some reading to find the answer. This is what I got:

1. Work-energy theorem applies only to particles.
2. Objects subject to friction cannot be treated as particles.
3. Thus it is wrong to apply work-energy theorem to objects subject to friction.
 
  • #9
From conservation of energy, there is difference in initial and final kinetic energy.
So the energy must have been converted to other form, conservative or non-conservative.
 

FAQ: Solving the Work-Energy Theorem Error in Block-Table System

What is the Work-Energy Theorem and how does it relate to the Block-Table System?

The Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. In the context of the Block-Table System, this means that the work done on the block is equal to the change in its kinetic energy as it moves along the table.

What is the error in the Work-Energy Theorem when applied to the Block-Table System?

The error in the Work-Energy Theorem when applied to the Block-Table System is that it assumes that the work done on the block is only due to the force applied on it, neglecting the work done by the normal force of the table. This can result in an incorrect calculation of the block's kinetic energy.

How can the error in the Work-Energy Theorem be solved in the Block-Table System?

The error in the Work-Energy Theorem can be solved by including the work done by the normal force of the table in the calculation of the block's kinetic energy. This can be done by considering the displacement of the block in both the horizontal and vertical directions.

What are the potential consequences of not accounting for the normal force in the Work-Energy Theorem for the Block-Table System?

Not accounting for the normal force in the Work-Energy Theorem for the Block-Table System can result in an incorrect calculation of the block's kinetic energy, which can lead to errors in other calculations and predictions based on the system. This can also lead to a misunderstanding of the underlying principles of work and energy.

Are there any real-world applications of the Work-Energy Theorem and its error in the Block-Table System?

The Work-Energy Theorem and its error in the Block-Table System have real-world applications in fields such as physics, engineering, and mechanics. It is important to understand and account for this error in order to accurately analyze and predict the behavior of objects in motion, especially in situations where the normal force plays a significant role.

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