Conservation of energy of a toboggan problem

AI Thread Summary
A 305kg toboggan traveling at 4.60m/s slows down by 1.10m/s over a 9.00m rough region, prompting a discussion on calculating the average friction force exerted. The initial and final kinetic energy conditions are identified, but the average deceleration needs clarification. The average deceleration is determined to be -0.12m/s², leading to confusion regarding the calculation of friction force. The correct approach involves using the formula v²_f = v²_i + 2ax to find acceleration, confirming the average deceleration. Ultimately, the discussion emphasizes the importance of correctly applying formulas to find the average friction force.
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Homework Statement


A loaded 305kg toboggan is traveling on smooth horizontal snow at 4.60m/s when it suddenly comes to a rough region. The region is 9.00m long and reduces the toboggan's speed by 1.10m/s.


Homework Equations



What average friction force did the rough region exert on the toboggan?

The Attempt at a Solution



I know you can find the initial conditions of the kinetic energy and then the final conditions given the final velocity. I just don't know how to get the force of the rough region. Any help is appreciated.
 
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How much average deceleration of the toboggan was there? You have the initial v, final v and the distance it decelerated by.
 
-.12m/s
 
Is that negative velocity? Check your units. Also in the case that is acceleration, then where is the problem of finding the average force of friction? \bar F_f=m \bar a where a is the average deceleration experienced.
 
It would be simple but am I correct in saying the average acceleration (deceleration in this case) is -.12m/s^2? Because when I do -.12*305 it is not correct. The friction force would be -37.3N which is incorrect.
 
The correct formula to find acceleration with a distance, and 2 velocities is v^2_f=v^2_i+2ax x=distance, a=acceleration
 
Ah of course. I knew that. Thanks!
 
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