If the block starts from rest, how many revolutions does it make before the tube breaks?

Question

A 600 g steel block rotates on a steel table (μk = 0.6) while attached to a 1.0 m long hollow tube. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.2 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50 N. If the block starts from rest, how many revolutions does it make before the tube breaks?

Answer 1

Friction force Ff = mgμk = 3.528 N
Net accelerating force F = 4.2-Ff = 0.672 N
Acceleration a = F/m = 1.12 m/s^2
Centripetal force Fc = mv^2/r, so v = sqrt(Fc*r/m)
When the tube breaks, Fc = 50, so v = sqrt(50r/m) = 9.1287 N
Distance traveled s = at^2/2 = v^2/(2a) = 37.2024 m
Revolutions = s/(2pi*r) = 5.921
EDIT: Since steve has given you the numeric answer, I thought I'd check my solution, so I appended the numbers also.

Answer 2

Net F = F1 - µmg = 4.2 - .6*.6*9.8 = .672 N

a = Fn/m = .672/.6 = 1.12 m/s

α = a/r = 1.12/1.0 = 1.12 rad/sec²

Breaking force = Fb = 50 = mrω² where ω² = 2αΘ→

50/(2mrα) = Θ = 37.202 radians = 5.92 rev