A body of specific gravity 3.0 is immersed in water. What is the minimum force required to pull it out at velocity 2

To determine the minimum force required to pull the body out of water, we need to consider the buoyant force acting on the body and the force required to overcome the resistance due to the velocity of the body.

1. Buoyant force:
The buoyant force is the upward force exerted on an object immersed in a fluid. It is equal to the weight of the fluid displaced by the body. The formula for buoyant force is:
Buoyant Force = Weight of the Fluid Displaced
= Density of Fluid × Volume of Fluid Displaced × g

In this case, the fluid is water and the density of water is approximately 1000 kg/m^3. The volume of water displaced by the body is equal to the volume of the body immersed in water. Since specific gravity is the ratio of the density of a substance to the density of water, the density of the body is 3 times that of water. Therefore, the volume of the body immersed in water is equal to the volume of the body divided by 3.
Let's assume the weight of the body is W. Therefore, the buoyant force on the body is:
Buoyant Force = 1000 × (Volume of Body/3) × g
= (1000/3) × (Volume of Body × g)

2. Resistance force:
The resistance force is the force required to overcome the resistance due to the velocity of the body. It is given by Newton's second law of motion:
Resistance Force = Mass of Body × Acceleration due to Velocity
= Mass of Body × Velocity/Time

Since we know that F = ma, and the velocity is given as 2 m/s, we can write:
Resistance Force = Mass of Body × (2/1)
= 2 × Mass of Body

The minimum force required to pull the body out is the sum of the buoyant force and resistance force:
Minimum Force = Buoyant Force + Resistance Force
= (1000/3) × (Volume of Body × g) + 2 × Mass of Body

To calculate the minimum force, we need to know the volume of the body and its mass.

Please provide the volume and mass of the body to proceed with the calculation.