acceleration and stopping distance...friction question?

Question

You launch your physics textbook along a horizontal desk with an initial speed of 5.0 m/s. The kinetic coefficient of friction is .15. Find the acceleration of the textbook and the stopping distance.

Answer 1

f = ma = kN = kmg; so a = kg = .15*9.81; where a is the deceleration, k = .15 the friction coefficient and g = 9.81 m/sec^2 the acceleration due to gravity at Earth's surface.

Work = fd = kNd = 1/2 m v^2 = kmgd; where friction force f = kN times distance (d) is work (energy in Joules), which in this case comes from kinetic energy 1/2 m v^2 where v = 5 m/sec and m is the mass of my damn physics book. Upon rearranging factors to find d, we have:

d = 1/2 v^2/(kg) = v^2/(2kg); where d is the stopping distance, v = 5 m/sec, k = .15 the friction coefficient, and g = 9.81 m/sec^2 from gravity.

You can do the arithmetic. The important physics lesson learned here is that neither answer (a or d) depends on the mass of that textbook...and both depend in large measure on k in that larger k means greater deceleration and smaller k means longer stop distance.

Answer 2

a = -0.15*9.80662 = -1.35993 m/sec^2
0 - 5.0^2 = - 2*-1.35993s
s = 9.19 m