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

AI Thread Summary
The discussion focuses on solving three physics problems related to friction and Hooke's law from an MCAT practice set. The user is struggling to derive the correct answers, particularly for problems 4, 7, and 14. For problem 4, it is clarified that knowing the force allows for calculating the acceleration of the system. In problem 7, a misunderstanding about the squared term in the equation is addressed, suggesting a method to solve for velocity using the correct formula. Problem 14 is explained by indicating that mass is not necessary for the calculation, as acceleration can be derived from the coefficient of friction and gravitational force.
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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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