A 0.366 kg metal cylinder is placed inside the top of a plastic tube, the lower

Question

A 0.366 kg metal cylinder is placed inside the top of a plastic tube, the lower end of which is sealed off by an adjustable plunger, and comes to rest some distance above the plunger. The plastic tube has an inner radius of 6.90 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 2.55, what is the initial acceleration of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm.

Explanation / Answer

The outside pressure is 1.00 atm.
Since the system was at equilibrium initially, the initial pressure inside must have been 1.00 atm.

By pushing the plunger, the pressure increases by 2.55 times, becoming 2.55 atm.

Now, the pressure difference is 2.55 - 1.00 = 1.55 atm
Area = pi*radius^2
= 3.1415*5.71

=17.13 mm
So,
force = pressure difference * area
= 1.55* 17.13

=26.55
Acceleration = force / mass
= 26.55/ 0.366kg
=72.54

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