What is the velocity of a steel block after moving 1.20 meters?

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The discussion revolves around calculating the velocity of a 2.00 kg steel block moving 1.20 meters under a tension of 17.0 N, considering a coefficient of kinetic friction of 0.600. The minimum tension required to initiate movement was determined to be 15.7 N based on static friction. Once in motion, the kinetic friction force was calculated as 5.88 N, leading to a net force of 11.12 N available for acceleration. The user successfully applied these calculations to solve the problem, expressing gratitude for the assistance received. The conversation highlights the importance of understanding frictional forces in motion problems.
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Homework Statement


Their is a 2.00 kg steel block on a steel table. What is the velocity of the block after moving 1.20 meters given that the string tension being pull on the block is 17.0 N and the coefficient of kinetic friction of dry steel on steel is .600.


Homework Equations


I don't really know what these would be for this problem.


The Attempt at a Solution


So far, I've figured that the minimum string tension to move the block is 15.7 N using the coefficient of static friction of dry steel on steel is 0.800. T = Mass * Static friction * gravity. T = (2.00) * (.800) * (9.8) = 15.68 ~ 15.7
 
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When the block is moving the only friction force it experiences is that of kinetic friction. the equation for the force of kinetic friction is F_k=\mu_kN where the N is the normal force between the object and the surface.

All you have to do is work out how much much of the Tension is going into acceleating the block (i.e. the tension minus the friction force) and then use the kinematic equations.
 
So I figured out the force of kinetic friction is = .600 * 9.8 which equals 5.88. I then took the tension of 17 and subtracted 5.88 to get 11.12 N. Is that right so far?
 
Ok, I figured it out. Thanks alot! Your a life saver! haha
 

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