Hat is the coefficient of kinetic friction between the floor and the box

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To determine the coefficient of kinetic friction between the floor and a 5kg box that slides 3m and comes to rest from an initial speed of 3m/s, one can apply the principles of conservation of energy and frictional force. The initial kinetic energy of the box is converted into work done against friction. The equation Fk = mv^2/d is used to calculate the frictional force, leading to the expression for the coefficient of kinetic friction, μ = Fk/mg. There is a suggestion that a factor of 2 might be missing in the calculations. The multiple-choice options for the coefficient include values such as 0.306, which is derived from the calculations discussed.
Nayeli MTZ
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I have no idea of how to start solving this problem.
A 5kg box slides 3m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3m/s?
Can somebody please help me solve the problem?
 
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We respectfully request students to show their work.

Now since the box has an initial velocity, it has kinetic energy. It has mass, and therefore a weight, and therefore a frictional force (in a gravitational field).

Use conservation of energy and the fact that force applied over distance is equivalent to energy. When the box stops, it has not kinetic energy. What happened to the initial kinetic energy?
 
I have no idea, it is one of my HMK problems from Physics. It is a multiple choice question, the choices are 1.05,0.587,0.153,0.306, and 0.2
 
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I think this is how it is solved.
Fk=mv2/d
Fk=15
\mu=Fk/mg
\mu=0.306
 
Please show how one derived those relations, particularly Fk=mv2/d. And please show the values one assumes.

There appears to be a factor of 2 missing.
 
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