A 500 g steel block rotates on a steel table while attached to a 1.20 m -long ho

Question

A 500 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 60.0 N . Assume the coefficient of kinetic friction between steel block and steel table is 0.60. (Figure 1)

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

Explanation / Answer

Here

mass , m = 500 gm = 0.50 Kg

L = 1.20 m

F = 4.91 N

let coefficient of kinetic friction is u

let the speed when the string beaks is T

T + u *mg = m * v^2/L

60 + 0.60 * 0.500 * 9.8 = 0.500 * v^2/1.20

solving for v

v = 12.3 m/s

Now , using second law of motion

F = m * a

4.91 = 0.5 * a

a = 9.82 m/s^2

let the distance moved is d

Using third equation of motion

v^2 - u^2 = 2 * a * d

12.3^2 = 2 * 9.82 * d

d = 7.7 m

number of revolutions = d/(2 * pi * r)

number of revolutions = 7.7/(2pi * 1.20)

number of revolutions = 1.02

the number of revolutions it make tube breaks is 1.02

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