A 0.389 kg metal cylinder is placed inside the top of a plastic tube, the lower

Question

A 0.389 kg metal cylinder is placed inside the top of a plastic tube, the lower end of which is sealed off by an adjustable plunger. The cylinder comes to rest some distance above the plunger. The plastic tube has an inner radius of 7.41 mm and is frictionless. Neither the plunger nor the metal cylinder allow any air to flow around them. If the plunger is suddenly pushed upwards, increasing the pressure between the plunger and the metal cylinder by a factor of 1.83, what is the initial acceleration a of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm. as m/s

Explanation / Answer

Solution:

Initial pressure in chamber

p = 1atm + mg/A = 1.013*105Pa + 0.389kg*9.8m/s² / ?(0.00741m)² = 2.21*104 Pa

And Final pressure p' = 1.83*p = 4.04*104 Pa

And so, The increase of pressure ?p = 1.83*104 Pa

This increase of pressure translates to an increase of force

F = ?p*A = 1.83*104Pa * ?(0.00741m)² = 3.164 N

Now, Instantaneous acceleration,

a = 3.164N / 0.389kg

a = 8.134 m/s².

I hope you understood the problem, If yes rate me!! or else comment for a better solution.

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