Question

why to empty h meter deep fulfilled well, the height is used h/2?

why to raise water from a h meter deep fulfilled well, the height is used h?

why to empty the half of a fulfilled well which is h meter deep, the height is used h/4?

why to raise water from a h meter deep well which has water level at h/4 ,the height is used 3h/4?

what is the main concept? please xplain

Answer 1

Answer on Question #50816-Physics-Mechanics-Kinematics-Dynamics

A)_Why to empty h meter deep fulfilled well, the height is used h/2?

B) Why to raise water from a h meter deep fulfilled well, the height is used h?

C) Why to empty the half of a fulfilled well which is h meter deep, the height is used h/4?

D) why to raise water from a h meter deep well which has water level at h/4 ,the height is used 3h/4?

What is the main concept?

Solution

The main concept

To up the mass $dm = \rho Adh$ of water from the depth $h$, where $A$ is area, we need energy

$$dE = dmgh = \rho Adhgh = (\rho gA)hdh.$$

If we want to raise the mass from level 1 to level 2, we need energy

$$E_2 - E_1 = mg(h_2 - h_1).$$

To empty the part of a fulfilled well we should use integration:

$$E = \int_{h_1}^{h_2} (\rho gA)hdh.$$

A) To fulfill the well we need

E = \int_0^H (\rho gA)hdh = (\rho gA) \frac{h^2^H}{2} = (\rho gA) \frac{H^2}{2} = \frac{mgH}{2}.

The volume is $V = AH$.

B) If we want to raise the mass $m$ of water from the depth $h$ we need energy

$$E = mg(h - 0).$$

C) To empty the half of a fulfilled well which is $h$ meter deep, we need energy

E = \int_0^{\frac{H}{2}} (\rho gA)hdh = (\rho gA) \frac{h^2^{\frac{H}{2}}}{2} = (\rho gA) \frac{\left(\frac{H}{2}\right)^2}{2} = \frac{mgH}{4}.

The volume is $V = A \frac{H}{2}$.

D) To raise water from a h meter deep well which has water level at $\frac{H}{4}$, we need energy

$$E = mg\left(H - \frac{H}{4}\right) = mg \cdot \frac{3}{4}H.$$

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