Question

A 0.412 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 2.55 , what is the initial acceleration a of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm .

Answer 1

Answer:

108.18 m/s2

Explanation:

1 atm = 101325 Pa

r = 7.41 mm = 0.00741m

Since the pressure is increased by a factor of 2.55, the new inner pressure inside the system would be

P = 101325 * 2.55 = 258378.75 Pa

This pressure exerts a force on the metal cylinder, knowing the radius, we can calculate the cross-section area:

$A = \pi r^2 = \pi 0.00741^2 = 0.0001725 m^2$

Then the force exerted:

$F = A*P = 0.0001725*258378.75 = 44.57 N$

Finally we can calculate the initial acceleration in accordance with Newton's 2nd law of motion:

$a = F/m = 44.57 / 0.412 = 108.18 m/s^2$