Work Done in Pumping Gasoline

How to calculate the work done in pumping gasoline that weighs 6600 newtons per cubic meter from a truck to a tractor tank located 5 meters higher? The work done in pumping the entire contents of a cylindrical gas tank of diameter 3m and length 6m from a truck to a tractor tank 5m higher, with gasoline weighing 6600 newtons per cubic meter, is 13.94 Megajoules.

First, let's find the volume of a cylindrical fuel tank since the capacity of the tank is required to calculate the amount of fuel pumped. The formula to calculate the volume of a cylinder is π x (radius)^2 x height. In this case, the radius of the tank is 1.5m (half of 3m) and the length of the tank is 6m. So, the volume of the tank is π x (1.5)^2 x 6 = 42.41 cubic meters.

Now, we know that work is basically the energy required to move a certain amount of object by a certain distance. In this case, it's the energy required to pump the gasoline from the truck tank to the tractor tank. The formula for work done is weight x height x distance. As the entire content of the tank has to be pumped, the weight is the weight of gasoline which is the volume of gasoline x density of gasoline. The height is the distance to the tractor's tank which is 5 meters higher than the truck's tank.

Substituting the given values, the work done is 6600 N/m^3 x 42.41 m^3 x 5m = 13,937,550 Joules or 13.94 Megajoules.

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