Calculate the work done in pumping gasoline

The subject of the query revolves around the concept of work in physics, and how it applies to the pumping of gasoline in this context. The work done in pumping gasoline into the tractor can be calculated using the principles of physics - specifically, the principle of work and energy.

Work done is typically defined as the force applied over a certain distance. In this case, it means moving a certain volume of gasoline a certain vertical distance (from the truck to the tractor, which is 5 meters).

If we define the mass of gasoline moved, m, as the volume of the tank multiplied by the density of gasoline, the gravitational force, g, as about 9.8 m/s², and the vertical distance, h, as 5 m (the height of the gasoline is pumped), we can calculate the work done using the equation W = mgh.

First, we can find the volume of the cylindrical tank using the formula for the volume of a cylinder, V = πr²h, where r is the radius (half the diameter) and h is the height (or length of the cylinder in this case). With a 3-meter diameter, the radius is 1.5 meters. The length or height of the tank is 2 meters. Therefore, the volume of the tank is 1.5² x 2 x π = 4.5π m³.

Therefore, the work done, W, in pumping the gasoline from the tank in the truck to the tank in the tractor, is (6600 N/m³ x 4.5π m³)(9.8 m/s²)(5 m) = 69,300π Joules of work.

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