Question

Calculate the work (in joules) required to pump all of the liquid out of a full tank. Assume a=6 m, b=12 m, liquid exits through the spout, c=3 m, and density of the liquid is 1000 kg/m3. (Use decimal notation. Give your answer to two decimal place.)

Answer 1

Answer:

$3.18*10^{5}J$

Explanation:

From the expression of workdone, we no that

Workdone= force * distance

also  

force=pressure *area

since pressure = density*height* gravity

we have

work done=pressure * area * distance

from the question, the workdone will be a dependent function on the height

hence we can write

$dw=pgh*A*dh\\A=a*b\\dw=pgh*(a*b)*dh$

to get the absolute workdone, we have to integrate both sides

Hence we have

$\int\limits^ {} dw\, dx =pgab\int\limits^3_0 {hdh} \, dx \\w=1000*9.8*6*12(\frac{h^{2}}{2} )|_{0}^{3}\\w=1000*9.8*6*12(\frac{3^{2}}{2} )\\w=3175200J\\W=3.18*10^{5}J$

Answer 2

The work done required to empty the tank is 3.17×10⁶J

Given that the dimensions of the tank are:

a = 6m and b = 12m.

So, area A of the tank is :

A = 6×12 = 72m²

and the density of the liquid ρ =  1000 kg/m³

According to the question, as the liquid is pumped out of the tank its level or height in the tank will decrease.

Work Done:

The work done W is the product of force F and displacement h. Now, the force F acting on the liquid can be determined by the pressure P multiplied by area A.

F = ρghA

Since the height h keeps decreasing so does the work done, so we take infinitesimal displacement  and integrate over the provided range:

$\int dw=\rho gA\int\limits^3_0 {h} \, dh \\\\int dw={\rho gA}[h^2/2]_0^3 \\\\W=1000\times9.8\times72[4.5]J\\\\W=3.17\times10^6J$

Learn more about work done:

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