NOTE: The answer is NOT 1 A 400 g steel block rotates on a steel table while att

Question

NOTE: The answer is NOT 1

A 400 g steel block rotates on a steel table while attached to a 1.20 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 5.01 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50.0 N . Assume the coefficient of kinetic friction between steel block and steel table is 0.60. (Figure 1)

Part A

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

Explanation / Answer

Net force = Thrust force - Frictional force

Ft = 5.01 N

Ff = uk*m*g = 0.6*0.4*9.81 = 2.3544 N

Net F = 5.01 - 2.35 = 2.66 N

acceleration = Net F/m = 2.66/0.4 = 6.65 m/sec^2

when the tube breaks,

Fc = mv^2/r

v = sqrt (Fc*r/m)

using given values:

v= sqrt (50*1.2/0.4) = 12.25 m/sec

distance traveled will be:

v^2 = u^2 + 2*a*s

s = v^2/2a

s = 12.25^2/(2*6.65) = 11.28 m

revolution completed will be

no. of revolution = s/(2*pi*r)

= 11.28/(2*pi*1.2) = 1.496 rev.

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