How Far Does a Box Slide Before Stopping If Pushing Force Is Removed?

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A box with a mass of 11.1 kg is pushed at a constant speed of 3.80 m/s, requiring an applied force of 18.5 N to overcome kinetic friction with a coefficient of 0.170. When the pushing force is removed, the box experiences a deceleration of 1.66666667 m/s², which should be noted as negative. Using the kinematic equation, the box slides approximately 4.3 meters before coming to rest. It's important to remember to include the negative sign for acceleration in calculations to avoid errors in grading. Understanding both the force required to maintain motion and the distance slid after the force is removed is crucial for solving similar physics problems.
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Homework Statement



A stockroom worker pushes a box with mass 11.1 kg on a horizontal surface with a constant speed of 3.80 m/s. The coefficient of kinetic friction between the box and the surface is 0.170.


Homework Equations


F=ma
w=mg


What horizontal force must be applied by the worker to maintain the motion?

I found that the Fapplied is 18.5N


If the force calculated in part A is removed, how far does the box slide before coming to rest?

This question I'm having problems on. Here is what I did

18.5N = 11.1kga

a=1.66666667 m/s^2

Then I used the equation of

Vfinal^2=Vinitial^2+2a\Deltax

\Deltax = 0 - 14.44m/s
--------------
2(1.66666667)


\Deltax=4.3m ?
 
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Yes that is correct.

Also you could have used conservation of energy.

Note that your a=1.66666667 m/s2 should really be a=-1.66666667 m/s2

But I believe you put it as negative while working it out, so just remember to write the - sign else your teacher might give you it wrong.
 
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