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 7.24 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 1.59, what is the initial acceleration of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm.

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aristocles aristocles · Answer 1 · 2019-11-03T03:08:27+01:00

Answer:

[tex]a = 32.6 m/s^2[/tex]

Explanation:

As we know that pressure between the cylinder and plunger is increased by 1.59 times

So this will make a net force upwards on the cylinder which is given as

[tex]F = \Delta P A[/tex]

now we will have

[tex]\Delta P = P_2 - P_1[/tex]

Here initial pressure is given as

[tex]P_1 = P_o + \frac{mg}{A}[/tex]

now new pressure is given as

[tex]P_2 = 1.59 P_1[/tex]

so we have force on the cylinder given as

[tex]F = P_2A - mg - P_oA[/tex]

[tex]F = 1.59(P_0 + \frac{mg}{A})A - (mg + P_0A)[/tex]

[tex]F = 0.59(1.01 \times 10^5 \times \pi(7.24 \times 10^{-3})^2 + 0.366(9.81))[/tex]

[tex]F = 11.93 N[/tex]

now the acceleration is given as

[tex]F = ma[/tex]

[tex]11.93 = 0.366 a[/tex]

[tex]a = 32.6 m/s^2[/tex]