Question

A trough has a trapezoidal cross section with a height of 5 m and horizontal sides of width 5/2 m and 5 m. Assume the length of the trough is 10 m. Complete parts? (a) and? (b) below.

a) How much work is required to pump the water out of the trough? (to the level of the top of the? trough) when it is? full? Use 1000 kg/m3 for the density of water and9.8 m/s2 for the acceleration due to gravity. Draw a? y-axis in the vertical direction? (parallel to? gravity) and use the midpoint of the bottom of one edge of the trough as the location of the origin. For 0?y?5?, find the? cross-sectional area? A(y) in terms of y.

b) Set Up the integral that gives the work required to pump the water out of the tank.

Answer 1

Answer:

Step-by-step explanation:

• Considering Volume of water inside this = Base area x Height

• Base is trapezoid = 0.5 (2.5 + 5) X 5 = 18.75 m2

• Height = 10 m

• Volume = 10 * 18.75 = 187.5 m3

Total mass of water = density X volume = 1000 kg/m3 x 187.5 m3

• = 187500kg

• Work done to pump out the water = change in potential energy of system of water

• Work = mg(H2 - H1)

where H2 = initial position of centre of mass from ground

where H1 = final position of centre of mass from ground

• here H2 = 2.78 m (from bottom)

• H1 = 0

• Work = 187500kg X 9.8 X (2.78 - 0) = 5.11 x 106 J
• The detailed analysis of the (b) part is as shown in the attachment

Answer 2

The work required to pump the water out of the trough is $5.11\times 10^6$ J and this can be determine by using the formula of work done.

Given :

• A trough has a trapezoidal cross section with a height of 5 m and horizontal sides of width 5/2 m and 5 m.
• Assume the length of the trough is 10 m.

a) The work required to pump the water out of the trough is given by:

$\rm W=mg(H_2-H_1)$    --- (1)

where m is the total mass of the water $\rm H_1$ is the initial position and $\rm H_2$ is the final position of the center of mass.

So. the mass of the water is given below:

$=1000\times 187.5$

= 187500 Kg

Now, the value of the initial position of the center of mass is zero and the value of the final position of the center of mass is 2.78 m.

Now, substitute the known terms in the expression (1).

$\rm W = 187500\times 9.8\times (2.78-0)$

$\rm W = 5.11\times 10^{6}\;J$

b) The width of the cylinder is given by:

$\rm w-5=\dfrac{5-2.5}{0-5}(x-0)$

Simplify the above expression.

w = -0.5x + 5

Now, the volume of the cylinder is given by:

$\rm dv = 10wdx$

Substitute the value of w in the above expression.

$\rm dv = 10(-0.5x+5)dx$

dv = (50-5x) dx

Now, the value of the force is given by:

$\rm dF = 9800dv$

Substitute the value of dv in the above expression.

dF = 9800(50 - 5x) dx

Now, the expression of work is given by:

dW = x dF

Substitute the value of force in the above expression.

dW = x(9800(50 - 5x)) dx

Now, integrate the above expression.

$\rm W = 9800\int^5_0(50x-5x^2)dx$

$\rm W = 9800 \times \left(25x^2-\dfrac{5x^3}{3}\right)^5_0$

$\rm W= 9800\left(625-\dfrac{625}{3} \right)$

W = 4083333.34 J

For more information, refer to the link given below:

https://brainly.com/question/3902440