Chapter 10, Problem 97A

To determine

The number of boxes to carry on each trip and the time required to carry the boxes on each trip.

Answer to Problem 97A

The number of boxes to carry on each trip is $3\text{ boxes}$ .

The time required for each trip is $12\text{ min}$ .

Explanation of Solution

Given:

The total mass of the 30 boxes is $m=150\text{ kg}$ .

The distance above the ground is $d=12\text{ m}$ .

The number of boxes is $n=30$ .

The given graph is shown below.

  

Formula used:

The expression for the work done is,

  $W=mgd$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $d$ is the distance.

The expression for the power is,

  $P=\frac{W}{t}$

Here, $W$ is the work done and $t$ is the time.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

The work done is,

  $\begin{array}{l}W=mgd\\ W=\left(150\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\times \left(12\text{ m}\right)\\ W=1.76\times {10}^4\text{ J}\end{array}$

The mass of a box is,

  $\begin{array}{l}{m}_{\text{box}}=\frac{m}{n}\\ m_{\text{box}}=\frac{150\text{ kg}}{30}\\ m_{\text{box}}=5\text{ kg}\end{array}$

From the graph, the maximum power is $P=25\text{ W}$ at the mass of $15\text{ kg}$ .

The number of boxes to carry on each trip is,

  $\begin{array}{l}\text{Number of boxes on each trip}=\frac{15\text{ kg}}{5\text{ kg}}\\ \text{Number of boxes on each trip}=3\text{ boxes}\end{array}$

The time required to carry the boxes on each trip is,

  $\begin{array}{l}P=\frac{W}{t}\\ t=\frac{W}{P}\\ t=\frac{1.76\times {10}^4\text{ J}}{25\text{ W}}\\ t=704\text{ s}\times \frac{\text{1 min}}{\text{60 s}}\end{array}$

  $t=12\text{ min}$

Conclusion:

Thus, the number of boxes to carry on each trip is $3\text{ boxes}$ . Also, the time required to carry the boxes on each trip is $12\text{ min}$ .