There if a force applied on a block such that it moves with constant speed v in the presence of friction. There are 2 different people, one is on the ground, stationary, and there other is moving relative to the block very quickly. To the observer moving quickly, the work done by friction on the block would be very large, due to the large displacement. But work done by friction should translate directly to heat in this situation, and the heat generated from each reference frame. Is there an intuitive way to understand this?
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Work done by friction force does not really obey "should translate directly to heat" in general; a part of work of friction force on a body can translate to both increase and decrease of kinetic energy of the body.
In your example, the block has constant kinetic energy, so no part of work of friction force goes to change of kinetic energy, it goes to something else. But even then, it need not go all into heat; part can go to other moving bodies. Which bodies move and thus can receive work depends on the frame.
Consider a person pushing with force $F$ along distance $\delta$ with respect to the ground.
In the frame of the ground, the push force by the person does positive work $F\delta$ on the block (the friction force does negative work of the same magnitude on the block, but let's focus on the positive work by the push force). In this frame, no other body receives positive work, so all positive work done by the push force goes to heat generated on the block-ground interface (some of this heat goes into the block, some into the ground).
Consider a frame in which the block and the person move with much higher speed in the same direction. The ground moves in the same direction as well, with somewhat smaller speed. The person acts on the block with the same push force $F$ as in the ground frame, but the block does much greater displacement $L$, so the positive work done on it is much greater. However, since also the ground moves in this frame, the friction force due to block acting on the ground (also $F$) does positive work on the ground. If the block undergoes displacement $L$ and the ground undergoes displacement $L-\delta$, then the block receives positive energy $FL$ from the push force, and the ground receives positive energy $F(L-\delta)$ from the friction force due to the block. So net work the block receives is $F\delta$. This net work does not depend on $L$, and thus does not depend on which frame we use to describe the situation. This net work is, if kinetic energy of the block remains constant, what "translates into heat".
Ján Lalinský
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All forces, including the kinetic friction force, are reference frame independent.
The work done by the kinetic friction force is due to the relative (sliding) motion between the surfaces. The relative motion is reference frame independent.
The kinetic friction work, and the heat generated as a result of it, is the kinetic friction force times the relative displacement between the surfaces.
Hope this helps.
Bob D
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