A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 ...

Question

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. What average power is produced by friction as the rock stops?

Accepted Answer

Hey everybody. So today we're being introduced to power and we're being asked to calculate the average power generated by friction on a plate as it comes to a stop. Now, we know a few things about this plate in the system as a whole. The plate itself has a weight of two kg. The horizontal first surface that's moving on has a coefficient of kinetic friction of 0.1. And it is moving at a rate of 5m/s until the friction caused it to stop. So it slows down to a stop. So there's a few things we need to know here. The average power generated by the friction on the plate as it comes to stop. Well, we need to know that the force of friction produces heat, right? The force of friction force of friction, which is right here, which is the uh force of kinetic or the coefficient of kinetic friction multiplied by the normal force which is defined as M. G. On a horizontal surface. And the power which is an average or the average power here, P average will be the average rate of the work done, which can be defined as the force of uh forest perpendicular, multiplied by the average velocity or the force of friction. Rather my bad, the force of friction multiplied by the average velocity. Now, we can also recall that average velocity Is simply equal to the uh sum of two velocities divided by two in this case, or however many velocities there are, so the initial velocity or the initial velocity is five m per second is five m per second, and the final is zero meters per second Divided by two. gives us a answer of 2.5 m/s. So that is the average velocity with all this in line. We can plug this into our average power formula and we get that the power, the average power, which is the average work of average rate of work done by the friction in this case, no utilizing our definition of the force of friction, It'll be 0.1 Multiplied by two kg into 9.81 m/s squared because that is the force of acceleration due to gravity, multiplied by 2.5 m/s. Our average velocity. So here's our average velocity. Here is our G. This is our mass and our force due to kinetic or our coefficient of kinetic friction. Anyways solving this and simplifying, we get a final answer of 4.9 watts. This means that the average power generated by the force of friction on the plate as it came to a stop With answer choice, a 4.9 watts. I hope this helps. And I look forward to seeing you all in the next one

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