PHYSICS
posted by Maria .
A 5.00kg box slides 7.00 m 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 3.00 m/s?

Initial speed = 3 m/s
acceleration constant during braking
therefore average speed during stop = 3/2 = 1.5 m/s
time to stop = t =7 m / 1.5 m/s = 4.67 seconds
v = Vi +a t
0 = 3 + a(4.67)
a = 0.643 m/s^2
F = m a
F = 5 (.643)
F = 3.21 Newtons
F = mu (weight)
3.21 = mu (5)(9.81)
mu =  .0655 
thank you.

You are welcome. As always, check my arithmetic.
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