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A person tosses his car keys on a shelf with an initial velocity of $2\:\mathrm{m/s}$. How far will the keys slide if the coefficient of kinetic friction between the surfaces is $0.3$?

How would you guys solve this question, I believe you use $d=v_it+\frac{a}{2} t^2$ but I am not sure how to solve for acceleration with the coefficient of friction.

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The definition of coefficient of friction $\mu$ is: $F = \mu N$ where $N$ is the normal force (the force that keeps the key from falling through the table like a ghost), and $F$ is the force due to fiction that is always opposite to the velocity. In this case since the key is lying flat on the table (i.e. not accelerating up or down) the sum of the forces in the vertical direction is zero so $N +(-mg) = 0$ so $N= mg$ and $F=\mu mg$. $F = ma$ by newton's second law, $m$ is the mass of the key and $a$ is its acceleration.

Solve for $a$ and use the equation you already mentioned and then you're golden.

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