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?

Answer 1

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

Answer 2

thank you.

Answer 3

You are welcome. As always, check my arithmetic.

Answer 4

Nonsensical