Question

A trough is 10 meters long, 2.5 meters wide, and 2 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 2 meters, and base, on top, of length 2.5 meters). The trough is full of water (density 1000 kg/m^3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g=9.8 m/s^2as the acceleration due to gravity.)

Answer 1

To solve this problem, we are going to follow few mathematical steps.

Firstly, we need to find out the volume of the water in the trough. We know that the volume V of a triangular prism which is the shape of the trough is given by the formula `V = 1/2 * Base * Height * Length`.

Substituting the given values in the formula:
Base (width of the trough) = 2.5 meters
Height (depth of the trough) = 2 meters
Length (length of the trough) = 10 meters

We have `V = 1/2 * 2.5m * 2m * 10m = 25m^3`. So, the volume of the water in the trough is 25 cubic meters.

Secondly, we need to calculate the mass of the water in the trough. This can be done by multiplying the volume with the density of the water. From the question, the density of the water is 1000 kg/m^3.

Substituting the given values in the formula `Mass = Volume * Density`, we get `Mass = 25m^3 * 1000kg/m^3 = 25000kg`. So, the mass of the water in the trough is 25000 kilograms.

Lastly, we are required to calculate the work done to pump the water over the top of the trough. The work done on an object is given by the formula `Work = Mass * Gravity * Height`.

We know that Mass = 25000 kg, Gravity = 9.8 m/s^2 and height is same as the depth of the trough which is 2 meters.

Substituting these in the formula, we get `Work = 25000kg * 9.8m/s^2 * 2m = 490000 Joules`.

So, to empty the trough by pumping the water over the top requires 490000 Joules of work.