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

The initial acceleration of the metal cylinder is approximately 43.497 m/s^2.

Step-by-step explanation:

To find the initial acceleration of the metal cylinder, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this case, the net force is the difference between the pressure exerted on the metal cylinder by the plunger and the pressure outside the tube, multiplied by the cross-sectional area of the cylinder.

The equation for the net force is:

F = (P_plunger - P_outside) * A_cylinder

Where F is the net force, P_plunger is the pressure exerted by the plunger, P_outside is the pressure outside the tube, and A_cylinder is the cross-sectional area of the cylinder.

Given that the pressure between the plunger and the cylinder is increased by a factor of 2.55, we can write:

P_plunger = 2.55 * P_outside

Substituting this into the equation for net force:

F = (2.55 * P_outside - P_outside) * A_cylinder

F = 1.55 * P_outside * A_cylinder

Finally, using the equation for force and mass, we can find the acceleration:

F = m * a

a = F / m

Plugging in the values:

a = (1.55 * P_outside * A_cylinder) / m

Now we can calculate the acceleration using the given values:

a = (1.55 * (1.013 * 10^5 Pa) * (7.41 * 10^-3 m)^2) / 0.412 kg

a ≈ 43.497 m/s^2