The truck, the box and the slide

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The discussion revolves around a physics problem involving a truck, a box, and the sliding motion of the box. The key points include the maximum acceleration of the box and how far it moves relative to the ground. Dr. D's answer indicates that the box will slide when the truck's deceleration exceeds the coefficient of friction multiplied by gravity, leading to a stopping distance of approximately 40.77 meters for the box. The truck must stop within 37.77 meters to allow the box to stop safely. Participants are encouraged to utilize hints and their understanding of Newton's laws to solve the problem independently.
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problem is attached as image...

i asked on yahoo answers, received a reply but is wrong...

it is very challenging (in my opinion)..
I spent over an hour without reaching even a wrong option..
 

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What do you think? Show what you've done so far.

Hints: What's the maximum acceleration of the box? Figure out how far the box moves with respect to the ground.
 
http://answers.yahoo.com/question/i...nkoSvGAazKIX;_ylv=3?qid=20090925002613AAbpiRt

Dr D's answer is correct:

The block will slide once the deceleration of the truck exceeds μg.
Once the block begins to slide, its absolute deceleration is constant at μg.
So the stopping distance of the block is v^2 / (2μg) = 40.77 m.

And the block is allowed to stop within 3 m more than the truck. So the minimum stopping distance of the truck is 37.77 m.

If you use g = 9.81, you get 37.77 m
If g = 10, then you get 37 m.

but his answer is a bit short, and hard to understand...
 
yoyobarn said:
but his answer is a bit short, and hard to understand...
Rather than hunting for answers, your time might be better spent trying to figure it out on your own. Use the hints I gave and your knowledge of Newton's laws.
 

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