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

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A box dropped onto a conveyor belt moving at 1.7 m/s experiences friction with a coefficient of 0.7, which affects its movement. The horizontal force required to move the box is determined by the frictional force, leading to an acceleration of 6.86 m/s². The calculations show that the box travels approximately 0.2106 meters before it moves without slipping. An alternative method of solving the problem using different equations yields the same result. The discussion confirms the calculations are correct, with minor variations in approach.
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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 alot.
 
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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 = Vot - 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 alot
 
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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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