Why Is Average Thermal Power Not Negative in This Friction Scenario?

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SUMMARY

The discussion clarifies that average thermal power in a friction scenario is not negative because it refers to the thermal power produced rather than the work done on the object. The friction force acts in the same direction as the velocity of the rock, which leads to positive thermal power output. The calculation method used involves multiplying the friction force by the average velocity, specifically (Friction Force * (v/2)), confirming that the average power remains positive.

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yashboi123
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Homework Statement
A 20.0 kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200.
Relevant Equations
P = F . v
P = Fvcos(x)
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I'm curious as to why the average power is not negative in this scenario, since I thought the friction force is in the opposite direction of velocity. As far as I see friction force is the only acting force in the problem, but I may be wrong. I solved this by simply doing (Friction Force * (v/2)), the average velocity.
 
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yashboi123 said:
I'm curious as to why the average power is not negative in this scenario
Because it asks for the thermal power produced, not the work per unit time done on the rock. The friction force the rock exerts on the surface is in the same direction as its velocity.
 

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