I solving three problems (2 for friction, 1 in hooke's law)

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SUMMARY

This discussion focuses on solving three physics problems related to friction and Hooke's Law, specifically from the MCAT Physics Answers document. The participants analyze problem 4, which involves calculating force using Hooke's Law with a spring constant (k) of 8000 and a distance of 0.05 meters, leading to a force of 400 N. For problem 7, the correct approach involves using the centripetal force equation Fc = mv²/r, while problem 14 can be solved without mass by applying the relationship F = ma = μmg, allowing for acceleration to be determined through kinematic equations.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of centripetal force and its equations
  • Familiarity with friction coefficients and their application in physics
  • Basic kinematics and the ability to manipulate equations
NEXT STEPS
  • Study the application of Hooke's Law in various contexts
  • Learn how to derive and apply the centripetal force equation in different scenarios
  • Explore the relationship between friction coefficients and motion in physics problems
  • Review kinematic equations and their use in solving for acceleration and velocity
USEFUL FOR

Students preparing for the MCAT, physics educators, and anyone looking to strengthen their understanding of mechanics, particularly in the context of friction and Hooke's Law.

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1. http://www.prep101.com/MCAT/102MCATPhysicsAnswers.pdf Numbers 4 7 and 14



2. As you can see the answers are already available, I'm just having trouble trying to get to them.



3. For number 4 I've tried plugging in the k value (8000) and distance (.05) into get a force value. I'm kind of lost from there as I've tried. Number 7 and 14 are kind of the same. However, for 7 I did get the answer when I plugged I plugged it into a Fc = v/r. So I did in fact get 10m/s but when I ignored the squared sign from the original equation (Fc = v^2/r). If I did do something correct then why did the squared sign disappear? For number 14 I don't really know how to tackle this if I don't have mass.

Once again if you can tell me how to reach the answers to one of these questions that would be great.
 
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Welcome to PF.

In 4 you have determined the force between the 2 blocks. That means that m2 is being accelerated with 400 N. Doesn't that give you then the acceleration for the whole system?

For 7 I'd say it's coincidence. Work out the equations for mv²/r = μmg. Solve for v and then plug in your numbers.

For 14 you don't need the mass. F = ma = μmg ... hence a = μg then solve by ordinary kinematic means.
 
Dude thanks so much...I have trouble remembering all the little substitutions you've showed me.

Thanks
 

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