The Work Done in Pumping Fuel from a Tank into a Truck
What is the work done in pumping the entire contents of a cylindrical gasoline tank into a truck if the tank is lying on the ground and the fuel opening on the truck is 2 feet above the top of the tank?
In this scenario, we have a cylindrical gasoline tank with a diameter of 4 feet and a length of 3 feet lying on the ground horizontally. A truck is refueling from the tank with the fuel opening 2 feet above the top of the tank. The weight density of gasoline is 42 pounds per cubic foot. To calculate the work done in pumping the entire contents of the fuel tank into the truck, we need to determine the potential energy difference between the initial and final states. The initial state is when the tank is on the ground, and the final state is when all the fuel is in the truck. The height difference between the initial and final states is 2 feet, which is the height of the fuel opening on the truck. The volume of the cylindrical tank can be calculated using the formula V = πr²h, where r is the radius (2 feet) and h is the length of the tank (3 feet). Therefore, the volume of the tank V = π(2²)(3) = 12π cubic feet. The weight of the fuel W can be calculated using the formula W = density × volume, where the density of gasoline is 42 pounds per cubic foot. Hence, W = 42 × 12π = 504π pounds. The work done in pumping the fuel can be calculated as W = mgh, where m is the mass of the fuel (W divided by the acceleration due to gravity) and g is the acceleration due to gravity (32.2 feet per second squared). By substituting the values, we get W = (504π / 32.2) × 2 = 31.29π foot-pounds. Therefore, the work done in pumping the entire contents of the fuel tank into the truck is approximately 31.29π foot-pounds.Calculating Work Done in Pumping Fuel