34P - Fundamentals Of Physics

Question

A 10 kg brick moves along an x axis. Its acceleration as a function of its position is shown in Fig.7-38. The scale of the figure’s vertical axis is set by as 20.0 m/s². What is the net work performed on the brick by the force causing the acceleration as the brick moves from $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{to}\mathbf{ }\mathbf{x}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{m}$?

                      

Step-by-Step Solution

Answer

The net work done on the brick by variable force $\mathrm{W}=800 \mathrm{J}$ 

Step 1: Understanding the concept

We can use the equation of work done by the spring along with the equation for area under the curve for a graph.

 

Formula:

1.  $\mathbf{W}\mathbf{=}\mathbf{F}\mathbf{\times }\mathbf{x}$
2.  $\mathbf{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{base}\mathbf{\times }\mathbf{height}$

 

Given:

1. The mass of brick, $\mathrm{m}=10 \mathrm{kg}$ 
2. The scale of acceleration is set as, ${\mathrm{a}}_{\mathrm{s}}=10 \mathrm{m}/{\mathrm{s}}^2$ 

Step 2: Calculate the work done

We have,

$$\begin{gathered} \mathrm{W}=\mathrm{F}.\mathrm{x} \\ \mathrm{W}=\mathrm{ma}.\mathrm{x} \\ \mathrm{W}=\mathrm{m}\left(\mathrm{a}\times \mathrm{x}\right) \end{gathered}$$

Here, the product $\left(\mathrm{a}\times \mathrm{x}\right)$ is given from the graph, as the area under the curve of $\mathrm{a}\left(\mathrm{x}\right)$ 

So,  $\mathrm{W}=\mathrm{m}\times \mathrm{A}$

Where 

$\begin{array}{rcl}\mathrm{A}& =& \frac{1}{2}\times \mathrm{base}\times \mathrm{height}\\ & =& \frac{1}{2}\times 8.0 \mathrm{m}\times 20.0 \mathrm{m}\\ & =& 80 {\mathrm{m}}^2\end{array}$

Hence, 

$$\begin{gathered} \mathrm{W}=10 \mathrm{kg}\times 80 {\mathrm{m}}^2 \\ \mathrm{W}=800 \mathrm{J} \end{gathered}$$

 

Therefore, the net work performed on the brick by the force 800 J.