How Is Terminal Velocity Calculated in a Submerged Block System?

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SUMMARY

The terminal velocity of a submerged block system involving two masses, m1 (3 kg) and m2 (4 kg), is calculated considering forces such as gravitational force, upthrust, and drag force. The drag force is given by FD = 7v, where v is the velocity. The correct calculation leads to a terminal velocity of 1.68 m/s for m2, after resolving the forces acting on the system. The negative value obtained during the calculation indicates a direction error in the force analysis.

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  • Understanding of Newton's laws of motion
  • Knowledge of fluid dynamics, specifically buoyancy and drag forces
  • Familiarity with basic algebra and solving equations
  • Concept of terminal velocity in physics
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  • Study the principles of buoyancy and Archimedes' principle
  • Learn about drag force calculations in fluid mechanics
  • Explore the dynamics of connected masses and pulley systems
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trevo100
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Homework Statement
Consider two blocks connected by a light, inextensible string over a light pulley
as follows:

dpzu2x.jpg


The slope is frictionless and the masses are released from rest. The mass m1 is
3 kg and the mass m2 is 4 kg where m2 is submerged completely (from the
beginning) in a liquid of density 1000 kg/m3. Take the density of m2 to be
8000kg/m3 and the drag force (in Newtons) to be FD = 7v, find the terminal
velocity of m2. (Hint: Don’t forget to include upthrust.)

The attempt at a solution

m1*g*sin50 + Fupthrust - fdrag - m2*g = (m1+m2)*a, when it reaches terminal velocity, acceleration becomes 0

22.5 + 4.9 - 39.2 -Fdrag = 0
Fdrag = -11.8
7v = -11.8
v = -1.68 m/s

The solution from my prof is 1.68m/s, but I can't figure out why I got minus in the equation.
 
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I suggest you take a closer look at your values within your calculation.
 
You need to think what the general direction you use in this question..
 

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