The coefficient of kinetic friction between a floor and a box made of wood is µk = 0.13. How far does the box slide if it has an initial speed of v0 = 7.0 m/s?

s = 19.23 m

Explanation:

Since, the frictional force is given as:

f = μN = μW

f = μmg   equation (1)

but, from Newton's 2nd Law:

f = ma   equation (2)

comparing equation (1) and equation (2):

ma = μmg

a = μg

here,

a = acceleration of puck

μ = coefficient of static friction = 0.13

g = -9.8 m/s² (negative sign for slowing down of puck)

Therefore,

a = (0.13)(-9.8 m/s²)

a = -1.274 m/s²

Now, using 3rd equation of motion:

2as = Vf² - Vi²

where,

s = distance covered = ?

Vf = Final Velocity = 0 m/s (since, puck finally stops)

Vi = Initial Velocity = 7 m/s

Therefore,

2(- 1.274 m/s²)s = (0 m/s)² - (7 m/s)²

s = (- 49 m²/s²)/(- 2.548 m/s²)

s = 19.23 m

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