Posted by **Maria** on Tuesday, March 18, 2014 at 7:09pm.

A 5.00-kg 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?

- PHYSICS -
**Damon**, Tuesday, March 18, 2014 at 7:17pm
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

- PHYSICS -
**Maria**, Tuesday, March 18, 2014 at 7:20pm
thank you.

- PHYSICS -
**Damon**, Tuesday, March 18, 2014 at 7:23pm
You are welcome. As always, check my arithmetic.

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