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

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The discussion clarifies that average thermal power is not negative in the friction scenario because it measures thermal power produced rather than work done on the rock. The friction force acts in the same direction as the rock's velocity, which influences the power calculation. The average power is derived from the product of the friction force and the average velocity. This distinction between thermal power and work is crucial for understanding the scenario. Overall, the confusion arises from the interpretation of forces and their directions in relation to power calculations.
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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