Stopping Distance: Car F, G, and H

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Cars F, G, and H, despite having different masses, will travel the same distance to skid to a stop when they slam on the brakes, assuming they have identical tires and thus the same coefficient of friction. The force of friction acting on each car is proportional to its weight, which balances out with their kinetic energy during braking. The equations provided demonstrate that the energy dissipated by friction is equal for all three cars, leading to the same stopping distance. The stopping distance can be calculated using the relationship between energy and force, specifically E/F. Therefore, all cars will stop over the same distance regardless of their mass.
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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 \mu.

The force of friction applies, Ffriction = \mumg, 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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