Stopping Distance: Car F, G, and H

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SUMMARY

In the discussion regarding the stopping distances of cars F, G, and H, it is established that all three cars, despite differing masses, will travel the same distance to skid to a stop when equipped with identical tires. The relationship between mass and stopping distance is governed by the equation mgh = mv^2/2, which leads to the conclusion that the distance each car travels before stopping is determined by the coefficient of friction and the initial velocity. The force of friction, calculated as Ffriction = μmg, plays a crucial role in dissipating energy, confirming that the stopping distance remains constant across all three vehicles.

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Three cars (cars F, G, and H) are moving with the same velocity, and slam on the brakes. The most massive car is F, and the least massive is H. Assuming all 3 cars have identical tires, which car travels the longest distance to skid to a stop?

Will they all travel the same distance in stopping?
If mgh = mv^2/2 for each car:

F – (3mv^2)/2 = 3*mgh h = v^2/2g
G – (2mv^2)/2 = 2mgh h = v^2/2g
H – mv^2 = mgh h = v^2/2g

Thanks.
 
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The same tires implies the same coefficient of friction, u or [itex]\mu[/itex].

The force of friction applies, Ffriction = [itex]\mu[/itex]mg, and the energy dissipated is Efriction = Ffriction*d, where d is the distance traveled.

Find dF = dG = dH.
 
How do I find the energy of friction to find each car's d?

To find d, it would be E/F?

Thanks again.
 

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