Calculating the Distance a Box Moves on a Moving Belt Without Slipping

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The discussion focuses on calculating the distance a box moves on a conveyor belt before it starts moving without slipping. The conveyor belt moves at 1.7 m/s, and the coefficient of friction between the box and the belt is 0.7. The user initially calculated the acceleration using the formula mg(0.7) = ma, resulting in an acceleration of 6.86 m/s². The final distance calculated before slipping occurs is approximately 0.2106 meters, confirming the accuracy of the calculations presented.

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I have no clue how to start solving the question as follows:
A box is dropped onto a conveyor belt moving at 1.7 m/s. If the coefficient of friction between the box adn the belt is 0.7, what distance doest the box move before it moves without slipping? (g = 9.8 m/s)

What is the horizontal force to move the box on the belt?

Thanks a lot.
 
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What is the horizontal force to move the box on the belt?
Friction.
 
Why it stop slipping? How should I start my formula? Thanks
 
What does an object do when a force acts on it?
 
acceleration

It should create acceleration. Now, I have just tried to solve it and the following is the formula I used.

mg(0.7) = ma, a = 6.86. I assumed this was the acceleration responsible for moving box. Then, I use V = Vo - at
0 = 1.7 - 6.86t, t = 0.2478, substituted into d = volt - 0.5 a t^2,
d= 1.7t - 0.5 * 6.86 * (0.2478)^2
= 0.2106 m
Is there anything (step) I missed or made mistake?

Thanks a lot
 
Looks OK to me.

Although I would have written
vf = v0 + at
and let v0=0 and vf=1.7 m/s

Then d = 0 + .5at2

Gives the same answer.
 

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