A trough is 5 meters long, 1 meters wide, and 4 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 4 meters, and base, on top, of length 1 meters). The trough is full of water (density 1000 kg/m3 ). Find the amount of work in joules required to empty the trough by pumping the water out of an outlet that is located 3 meters above the top of the tank. See problem 21, page 464 of the text for a diagram of this trough configuration. (Note: Use g=9.8 m/s2 as the acceleration due to gravity.)
To calculate the amount of work required to empty the trough, we need to consider the weight of the water being pumped out and the height it is being lifted.
First, let's find the volume of water in the trough:
Volume = length x width x depth
Volume = 5 m x 1 m x 4 m
Volume = 20 cubic meters
Next, let's calculate the mass of the water in the trough:
Density = mass / volume
mass = density x volume
mass = 1000 kg/m^3 x 20 m^3
mass = 20000 kg
Now, let's calculate the weight of the water:
Weight = mass x gravity
Weight = 20000 kg x 9.8 m/s^2
Weight = 196000 N (Newtons)
Since the water is being pumped out to a height of 3 meters,
Work = Weight x height
Work = 196000 N x 3 m
Work = 588000 N∙m
Finally, let's convert the work to joules:
1 joule (J) = 1 N∙m
Work = 588000 J
Therefore, the amount of work required to empty the trough is 588000 joules.
To find the amount of work required to empty the trough, we can break down the problem into two parts:
1. The work required to raise the water to the top of the trough.
2. The work required to lift the water from the top of the trough to the outlet.
Let's start with the first part:
1. The work required to raise the water to the top of the trough:
- The mass of the water in the trough can be calculated by multiplying the volume of the trough by the density of water. The volume of the trough can be found by multiplying the length, width, and depth of the trough.
- In this case, the volume of the trough is 5m * 1m * 4m = 20 cubic meters.
- The mass of the water is therefore 20 cubic meters * 1000 kg/m^3 = 20,000 kg.
- To raise this mass of water to a height of 3 meters, we need to calculate the gravitational potential energy using the formula: potential energy = mass * height * gravity, where gravity is 9.8 m/s^2.
- The potential energy required to raise the water to the top of the trough is therefore: 20,000 kg * 3 m * 9.8 m/s^2 = 588,000 Joules.
Now let's move on to the second part:
2. The work required to lift the water from the top of the trough to the outlet:
- The mass of the water at the top of the trough can be found by multiplying the cross-sectional area of the trough by its height.
- The cross-sectional area of the trough can be found by multiplying the base (1 meter) by the height (4 meters) and dividing it by 2, since it is an isosceles triangle.
- The cross-sectional area is therefore: (1m * 4m) / 2 = 2 square meters.
- The mass of the water at the top of the trough is therefore: 2 square meters * 1000 kg/m^3 = 2000 kg.
- To lift this mass of water to a height of 3 meters, we need to calculate the potential energy using the formula: potential energy = mass * height * gravity, where gravity is 9.8 m/s^2.
- The potential energy required to lift the water from the top of the trough to the outlet is therefore: 2000 kg * 3 m * 9.8 m/s^2 = 58,800 Joules.
To find the total work required to empty the trough, we add the work required for both parts:
Total work = Work to raise water to the top of the trough + Work to lift water from the top of the trough to the outlet
Total work = 588,000 Joules + 58,800 Joules
Total work = 646,800 Joules
Therefore, the amount of work required to empty the trough by pumping the water out of the outlet is 646,800 Joules.
work = force * distance
force = gm
consider thin sheets of water of area 5*(y/4) where y/4 is the width across the tank at depth y
So, the weight of the sheet of volume dV is 1000kg/m^3 * 9.8N/kg * 5(y/4)dy m^3 = 12250y * dy N
Work = Int[0,4] 12250y dy
= 6125 y^2 [0,4]
= 98000N
Check my math. . .