More Power Calculations

A forklift lifts a crate of mass \(\text{100}\) \(\text{kg}\) at a constant velocity to a height of \(\text{8}\) \(\text{m}\) over a time of \(\text{4}\) \(\text{s}\). The forklift then holds the crate in place for \(\text{20}\) \(\text{s}\). Calculate how much power the forklift exerts in lifting the crate? How much power does the forklift exert in holding the crate in place?

Solution

Step 1: Determine what is given and what is required

We are given:

• mass of crate: m=\(\text{100}\) \(\text{kg}\)

• height that crate is raised: h=\(\text{8}\) \(\text{m}\)

• time taken to raise crate: \({t}_{r}=4~\text{s}\)

• time that crate is held in place: \({t}_{s}=20~\text{s}\)

We are required to calculate the power exerted.

Step 2: Determine how to approach the problem

We can use:

\(P=Fv=F\frac{\Delta x}{\Delta t}\)

to calculate power. The force required to raise the crate is equal to the weight of the crate.

Step 3: Calculate the power required to raise the crate

\begin{align*} P& = F\frac{\Delta x}{\Delta t}\\ & = m·g\frac{\Delta x}{\Delta t}\\ & = \left(100 \text{kg}\right)\left(9,8 \text{m}·{\text{s}}^{-2}\right)\frac{8 \text{m}}{4 \text{s}}\\ & = 1960 \text{W} \end{align*}

Step 4: Calculate the power required to hold the crate in place

While the crate is being held in place, there is no displacement. This means there is no work done on the crate and therefore there is no power exerted.

Step 5: Write the final answer

\(\text{1 960}\) \(\text{W}\) of power is exerted to raise the crate and no power is exerted to hold the crate in place.

What is the power output for a \(\text{60.0}\) \(\text{kg}\) woman who runs up a\(\text{3.00}\) \(\text{ m}\)high flight of stairs in \(\text{3.50}\) \(\text{s}\), starting from rest but having a final speed of\(\text{2.00}\) \(\text{m·s$^{-1}$}\)?Her power output depends on how fast she does this.

Step 2: Calculate power

Substituting the expression for \(W\) into the definition of power given in the previous equation, \(P=\frac{W}{t}\) yields:

\begin{align*}P &=\frac{W}{t}\\ &=\frac{\frac{1}{2}mv_f^2+mgh}{t} \\ &=\frac{\frac{1}{2}(\text{60.0})(\text{2.00})^2+(\text{60.0})(\text{9.80})(\text{3.00})}{\text{3.50}} \\ & = \frac{120 +1764}{\text{3.50}} \\ & = \text{538.3}\text{ W}\end{align*}

Step 3: Quote the final answer

The power generated is \(\text{538.0}\) \(\text{W}\).

The woman does \(\text{1 764}\) \(\text{J}\) of work to move up the stairs compared with only \(\text{120}\) \(\text{J}\) to increase her kinetic energy; thus, most of her power output is required for climbing rather than accelerating.

What is the power required to pump water from a borehole that has a depth\(h=\text{15.0}\text{ m}\) at a rate of\(\text{20.0}\) \(\text{l·s$^{-1}$}\)?

Solution

Step 1: Analyse the question

We know that we will have to do work on the water to overcome gravity to raise it a certain height.If we ignore any inefficiencies we can calculate the work, and power, required to raise the mass ofwater the appropriate height.

We know how much water is required in a single second. We can first determine the mass of water:\(\text{20.0}\text{ l}\times\frac{\text{1}\text{ kg}}{\text{1}\text{ l}} = \text{20.0}\text{ kg}\).

The water will also have non-zero kinetic energy when it gets to the surface because it needs tobe flowing. The pump needs to move \(\text{20.0}\) \(\text{kg}\)from the depth of the borehole every second, we know the depth so we know the speed that the waterneeds to be moving is \(v=\frac{h}{t}=\frac{\text{15.0}}{1}=\text{15.0}\text{ m·s$^{-1}$}\).

Step 2: Work done to raise the water

We can use

\begin{align*} W_{\text{non-conservative}} & = \Delta E_k + \Delta E_p \\ & = E_{k,f} - E_{k,i} + E_{p,f} - E_{p,i} \\ & = \frac{1}{2}mv^2 -(0) + mgh - 0 \\ & = \frac{1}{2}(20)(15)^2 + (20)(\text{9.8})(15) \\ & = \text{2.25} \times \text{10}^{\text{3}} + \text{2.94} \times \text{10}^{\text{3}} \\ & = \text{5.19} \times \text{10}^{\text{3}}\text{ J} \end{align*}

Step 3: Calculate power

\begin{align*}P &= \frac{W}{t} \\ &= \frac{\text{5.19} \times \text{10}^{\text{3}}}{1} \\ & = \text{5.19} \times \text{10}^{\text{3}}\text{ W}\end{align*}

Step 4: Quote the final answer

The minimum power required from the pump is \(\text{5.19} \times \text{10}^{\text{3}}\) \(\text{W}\).