What is the minimum stopping distance of a truck with locked brakes?

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SUMMARY

The minimum stopping distance of a truck with locked brakes can be calculated using the formula d = Vo2 / (2μkg), where Vo is the initial speed, μk is the coefficient of kinetic friction, and g is the acceleration due to gravity. The mass of the truck does not appear in the final equation because all forces involved are proportional to mass, thus canceling it out. This result highlights the relationship between friction and stopping distance, regardless of vehicle mass.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with kinetic friction and its coefficient (μk)
  • Basic knowledge of work-energy principles
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the implications of kinetic friction in vehicle dynamics
  • Learn about the effects of air resistance on stopping distances
  • Explore advanced topics in physics related to momentum and energy conservation
  • Investigate real-world applications of stopping distance calculations in automotive safety
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Physics students, automotive engineers, safety analysts, and anyone interested in understanding vehicle stopping dynamics and the effects of friction.

DrummingAtom
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Homework Statement


A truck is moving on a level street with speed Vo when the brakes lock (the tires don't roll and skid). Find the minimum stopping distance of the truck in terms of Vo, g, and the coefficient of kinetic friction μk.


Homework Equations


W = F*d
W = K2-K1
K = 1/2mv2


The Attempt at a Solution


Ok, I think I got this one because want to make sure because when I'm plugging in numbers at the end it seems a little strange.

I started with W = K2-K1 and K2 will be zero because it's stopped at that point.

W = -K1

I took that -g = n and applied it to the left side, was that step right?

-F*d = -K1

-F*d = -1/2mVo2

Solving for d gives:
[tex]d = \frac{-\frac{1}{2}mV_{o}^2}{-m\mu_{k}g}[/tex]

Then after rearranging:

[tex]d = \frac{V_{o}^2}{2\mu_kg}[/tex]

Now here's where I'm a little confused, if what I did was right. How can the final equation not have mass? A semi would certainly take a longer distance to stop than a Geo Metro. That's why this equation is so bizarre to me. I'm guessing this is why my Professor assigned this problem, haha. Thanks for any help.
 
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Hi DrummingAtom! :smile:
DrummingAtom said:
Now here's where I'm a little confused, if what I did was right. How can the final equation not have mass? A semi would certainly take a longer distance to stop than a Geo Metro. That's why this equation is so bizarre to me. I'm guessing this is why my Professor assigned this problem, haha. Thanks for any help.

No, your professor thinks this result is perfectly natural …

all the forces are proportional to m, so m doesn't matter. :wink:

(btw, air resistance is proportional to size, not mass, so if you include air resistance, the mass does matter)

I'm afraid you're the bizarre one! :biggrin:
 
Haha, thanks.
 

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