Solving Submersible Acceleration: A Physics Challenge

[hank09]

Homework Statement

A submersible with a mass of 2.75 x10^3 kg remains at a constant depth beneath the ocean surface. To make the submersible rise, 1.50 x10^2 kg of ballast is released. Ignoring friction, what is the upward acceleration of the submersible? (D = 1.025 x 10 ^3 kg/m^3 )

Homework Equations

Fnet=Fb-Fg
Fb=weight-apparent weight.
Fg/Fb=Density of object/Densitiy of fluid
Fb=mass of fluid x gravity

The Attempt at a Solution

First, I tried using Fnet=Fb-Fg. I multiplied 150x9.81. I thought this would be the buoyant force. Then, i multiplied 2.75 x10^3 kg by 9.81 as well. I subtracted the two values, and then divided the given net force by the sum of two masses. I think my answers are way off. I am so lost. Don't I have to calculate the Fg of the object and then calculate the required acceleration of the buoyant force that will cancel off the Fg?

 

[tiny-tim]

Welcome to PF!

Hi hank09! Welcome to PF!

A submersible with a mass of 2.75 x10^3 kg remains at a constant depth beneath the ocean surface. To make the submersible rise, 1.50 x10^2 kg of ballast is released. Ignoring friction, what is the upward acceleration of the submersible? (D = 1.025 x 10 ^3 kg/m^3 )
…
i multiplied 2.75 x10^3 kg by 9.81 as well. I subtracted the two values

why?

F = ma, so what is that F?

 

[nlightner]

I can understand your confusion and frustration in trying to solve this physics challenge. It is important to approach any problem with a clear understanding of the relevant equations and principles involved. In this case, we are dealing with the forces of buoyancy and gravity, which are crucial in understanding the acceleration of the submersible.

First, let's review the given information. We know that the submersible has a mass of 2.75 x10^3 kg and is at a constant depth beneath the ocean surface. To make the submersible rise, 1.50 x10^2 kg of ballast is released. We are also given the density of the fluid, which is 1.025 x 10 ^3 kg/m^3. From this information, we can determine the buoyant force acting on the submersible.

Using the equation Fb = mg, where Fb is the buoyant force, m is the mass of the fluid displaced, and g is the acceleration due to gravity, we can calculate the buoyant force acting on the submersible. Since the submersible displaces a volume of fluid equal to its own volume, we can use the given density of the fluid to calculate the mass of the fluid displaced.

m = ρV, where ρ is the density of the fluid and V is the volume of the submersible. Therefore, the mass of the fluid displaced by the submersible is equal to 2.75 x10^3 kg.

Substituting this value into the equation Fb = mg, we get Fb = 2.75 x10^3 kg x 9.81 m/s^2 = 26977.5 N.

Next, we need to calculate the weight of the submersible before and after the ballast is released. The weight of the submersible before the ballast is released is equal to its mass multiplied by the acceleration due to gravity, which is 2.75 x10^3 kg x 9.81 m/s^2 = 26977.5 N.

After the ballast is released, the weight of the submersible will decrease by the weight of the ballast, which is equal to 1.50 x10^2 kg x 9.81 m/s^2 = 1471.5 N.

Using the equation Fnet = Fb - Fg, where Fnet