Question

let h be the the height above the pump of the water reservoir and h the height where the water is to end up. also assume m is the total mass of the water leaving the reservoir and m is the mass that reaches the final height, moving with a speed v. use conservation of energy to derive a relationship for the mass of water, m that can reach the final height in terms of m, h, and h. assume no energy loss due to friction. what percentage of m must be expended so that some water, m just reaches the higher reservoir (v = 0)?

Explanation:

Step1: Calculate initial energy

The initial energy \(E_{i}\) of the water in the reservoir is gravitational potential energy \(E_{i}=MgH\) (since it starts from rest).

Step2: Calculate final energy

The final energy \(E_{f}\) of the water that reaches height \(h\) is \(E_{f}=mgh+\frac{1}{2}mv^{2}\) (using the formula for gravitational potential energy \(U = mgh\) and kinetic energy \(K=\frac{1}{2}mv^{2}\)).

Step3: Apply conservation of energy

Since there is no energy loss (\(E_{i}=E_{f}\)), we have \(MgH=mgh+\frac{1}{2}mv^{2}\).

For the case when \(v = 0\):

Step4: Simplify the equation

When \(v = 0\), the energy - conservation equation becomes \(MgH=mgh\).

Step5: Solve for \(m\)

We can solve for \(m\) as \(m=\frac{H}{h}M\).

Step6: Calculate the percentage

The percentage of \(M\) is \(\frac{m}{M}\times100\%=\frac{H}{h}\times 100\%\)

Answer:

When \(v = 0\), the percentage of \(M\) that must be expended is \(\frac{H}{h}\times100\%\)