Question

A 0.504 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 5.37 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 2.31, what is the initial acceleration a of the metal cylinder?

Answer 1

Answer:

Explanation:

Given:

m=0.504kg

r=5.37mm

metal cylinde factor=2.31

$P_0=1atm$

we know that

Upward force = Downward force

$P_0A + mg=PA$

Net force$F=2.31PA-P_0A-mg\\\\ma=2.31(P_0A+mg)-P_0A-mg\\\\ma=1.31P_0A+1.31mg\\\\a=\frac{1.31P_0A}{m}+1.31g\\\\=\frac{1.31\times 1.013\times 10^5 \times \pi \times (7.24\times 10^{-3})^2}{0.504}+(1.31\times 9.8)\\\\a=56.20m/s^2$