Work and friction

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  • #1
leafy
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Suppose you're a point on the floor and a block mass slide pass you. The block mass said " I have friction acting on me during my motion so the mechanical power is friction x my speed". When ask the point on the floor, it said "well, I don't have motion so there is no mechanical power". Yet the floor produce heat.

you can apply Fxv to the block, but how can you apply Fxv to the floor?

{moderator’s note: split from https://www.physicsforums.com/threads/can-static-force-do-work.1010439/#post-6578451 }
 

Answers and Replies

  • #2
anuttarasammyak
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Motion is relative. The floor or a block has relative speed v to the other . For the floor, a block moves. For a block, the floor moves.
 
  • #3
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Suppose the block is being pushed in the positive ##x## direction. Then the velocity ##\vec v## (of the block) is in the positive direction and the frictional force ##-\vec F## acting on the block is in the negative direction (with the minus sign reminding us of that fact). Then the mechanical power on the block is ##P_{block}=-\vec F \cdot \vec v<0## with the negative sign indicating that mechanical power is leaving the block.

By Newton’s 3rd law the frictional force ##\vec F## is in the positive direction. Then the mechanical power on the floor is ##P_{floor}=\vec F \cdot 0 = 0##. So no mechanical power enters the floor.

The mechanical power that leaves the block and does not enter the floor is transformed into thermal energy at a rate ##\dot Q=-(P_{floor}+P_{block})=\vec F \cdot \vec v>0##

It is incorrect to assume that all of the heat ##\dot Q## goes into the floor, so ##\vec F \cdot \vec v## is not a relevant quantity for the floor. All that you can say from mechanics is that heat is produced at the interface at a rate ##\dot Q##. Then how much heat goes into the block vs the floor depends on their thermodynamic properties.
 
  • #4
leafy
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So there is no different between motion of material or motion of the force here because they're both exactly the same. What about if you're on a train moving right at 1 m/s and pushing a mass left. When the mass reach 1m/s left. What is the mechanical power? Fx1m/s or 0?
 
  • #5
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What about if you're on a train moving right at 1 m/s and pushing a mass left. When the mass reach 1m/s left. What is the mechanical power? Fx1m/s or 0?
The same rule applies always for all mechanical forces. In that case the velocity of the material at the point of application of the force is ##\vec v= 0## so the mechanical power is ##P=\vec F \cdot \vec v=0##.
 
  • #6
leafy
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The same rule applies always for all mechanical forces. In that case the velocity of the material at the point of application of the force is ##\vec v= 0## so the mechanical power is ##P=\vec F \cdot \vec v=0##.
I see that you always take the absolute(earth) frame. But to the person on the train, the power is not 0.
 
  • #7
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I see that you always take the absolute(earth) frame. But to the person on the train, the power is not 0.
Correct. Mechanical power depends on the reference frame. It is not an invariant in Newtonian physics.

I took the Earth frame (which is not absolute in any sense) because that is how you described it, but if you use a different frame you apply the same rule and thus get a different value for mechanical power.
 
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