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

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Homework Help Overview

The problem involves determining the velocity of a steel block after it has moved a distance of 1.20 meters. The block has a mass of 2.00 kg and is subjected to a tension force of 17.0 N, while experiencing kinetic friction with a coefficient of 0.600.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the block, including static and kinetic friction. There is an exploration of how to calculate the net force acting on the block and its implications for acceleration and velocity.

Discussion Status

Some participants have provided insights into the calculations of friction forces and the net force acting on the block. There is an acknowledgment of the steps taken to analyze the problem, but no consensus or final solution has been reached.

Contextual Notes

Participants are working with the coefficients of static and kinetic friction, and there is a mention of the need to consider the normal force in the context of friction calculations. The original poster expresses uncertainty about the relevant equations for the problem.

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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 [itex]F_k=\mu_kN[/itex] 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 a lot! Your a life saver! haha
 

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