Newton's laws and pulley mass system

AI Thread Summary
The discussion centers on solving a physics problem involving a pulley system with two masses, m1 and m2, where m1 is on a surface with friction. The coefficient of friction is given, and the task is to find the speed of m2 after it falls a specified distance. The user calculated an acceleration of 4.89 m/s² and tension of 24.45 N but expressed uncertainty about the calculations. A suggested approach involves using kinematic equations to relate acceleration, initial velocity, and distance to find the final velocity. The conversation emphasizes the importance of drawing free body diagrams to analyze forces acting on both masses for a comprehensive solution.
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Homework Statement



The coefficient of friction between the block of mass m1 = 3.00 kg and the surface in the figure below is μk = 0.335. The system starts from rest. What is the speed of the ball of mass m2 = 5.00 kg when it has fallen a distance h = 1.45 m?

8-p-022.gif


Homework Equations



Ff = UsN
T - mg = ma

Y = Yo + Vot + 1/2at^2 (maybe, I'm not sure)

The Attempt at a Solution




I found that a = 4.89m and that the tension in the string is 24.45N. I don't know how to then take teh distance 0f 1.45m traveled into consideration and work with this.
 
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are you sure from the value you got for a? I feel that there is a mistake there? (please show that steps you followed to get the acceleration)


in case you are sure from your answer, then to get the speed I suggest you use the following equation:

vf^2 = vo^2 + 2ax, where vf: the final velocity, vo: the initial velocity, a:acceleration, x: the distance

you have vo =0 (from the question "the system starts from rest"), a (you got it), and x (given in the question)..
 
I'm using a simulation to solve acceleration and Tension ( g = 9.8 m/s^2)

Newton-Pulley-system1.JPG


Newton-Pulley-system.JPG


a = 4.89 m/s^2

vo = 0
h = 0.5*a*t^2
t = root(2*h/a) = root(2*1.45/4.89) = 0.77 s

spreed of the system after 0.77 s is

vt = vo + at = 0 + (0.77 s)*(4.89 m/s^2) = 3.7653 m/s

_____________________
http://janggeng.com/Newtons-laws-aplication-pulley-system-part-1/"
 
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That looks like some pretty blatant self promotion :p although it does look like it would have been a handy application to help with learning mechanics.
 
I have a hunch your teacher probably wants you to be able to draw FBD for both masses allowing you to get two equations with T and a being your unknowns. Two equations, two variables, solve for a.

The rest is just simple kinematics. If the teacher had angled the plane the application that one poster wrote would be unusable. Its for one of many types of problems that could be given with simple statics or dynamics. I would think you would want to put pencil to paper and do it yourself. Then play with simulations. Or write your own.

So what are all the forces on the mass on the plane in the vertical and horizontal and all the forces in the vertical for the mass hanging... start there. a and T must be the same magnitude for both
 
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