A 400 g steel block rotates on a steel table while attached to a 1.20 m -long ho

Question

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 4.91 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? Figure 1 of 1 234 SubmitPrev ious Answers Request Answer Air 1.2 m X Incorrect; Try Again; 4 attempts remaining Tube Pivot Provide Feedback

Explanation / Answer

Net force on the block will be given by:

Fnet = Fthrust - Ff

Fnet = m*a

Fthrust = 4.91 N

Ff = uk*N = uk*m*g = 0.600*0.400*9.81 = 2.354 N

a = (Ft - Ff)/m = (4.91 - 2.354)/0.400 = 6.39 m/sec^2

Given that max force that tube can withstand without breaking = 50.0 N

Fc = 50 N = m*v^2/r

v = sqrt (Fc*r/m)

v = sqrt (50*1.20/0.400) = 12.25 m/sec

Now distance traveled will be

V^2 = Vi^2 + 2*a*d

Vi = 0 m/sec

d = V^2/2a = 12.25^2/(2*6.39)

d = 11.74 m

Now angular displacement will be

d = r*theta

theta = d/r = 11.74/1.20 = 9.78 rad = 1.56 rev

1 rad = 1/(2*pi) rev

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