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

AI Thread Summary
The discussion focuses on calculating the minimum stopping distance of a truck with locked brakes using the initial speed (Vo), gravitational acceleration (g), and the coefficient of kinetic friction (μk). The derived formula for stopping distance is d = Vo^2 / (2μkg), which surprisingly does not include mass. Participants clarify that while mass influences the forces acting on the truck, it cancels out in this specific equation due to proportional relationships. The conversation highlights the importance of understanding the underlying physics principles rather than just the formula itself. This problem illustrates the nuances of physics in real-world applications, such as vehicle stopping distances.
DrummingAtom
Messages
657
Reaction score
2

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:
d = \frac{-\frac{1}{2}mV_{o}^2}{-m\mu_{k}g}

Then after rearranging:

d = \frac{V_{o}^2}{2\mu_kg}

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.
 
Physics news on Phys.org
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top