Solving Friction Problems: 500N Box on a Non-Frictionless Floor

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To solve the problem of moving a 500 N box across a non-frictionless floor with a constant speed of 3 m/s while applying a 200 N horizontal force, the coefficient of friction can be calculated using the formula for frictional force. When pulling at a 30-degree angle, the tension in the rope and the effective weight of the box must be considered to determine the new coefficient of friction, which is given as 0.4. The discussions highlight the importance of understanding the forces acting on the box, including gravitational and frictional forces. Participants express confusion about the concepts, indicating a need for clearer explanations and examples. The thread emphasizes the necessity of grasping the fundamentals of friction and force analysis in physics problems.
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Homework Statement



You need to move a 500 N box across the floor. You pull on the box with a rope which is also horizontal. The floor is not frictionless. You move the box so that it moves with a constant v of 3m/s by applying a 200 N force.

Coefficient of Friction?

Also,

same question except pull the box at an angle of 30 degrees from the horizontal, use 0.4 coefficient.

what is the tension in the rope

The Attempt at a Solution



I just really can't even get started on this one, i have no idea, my professor really didnt cover to much of this but gave us 4 similar questions on it, any help would help me out...
 
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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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