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Chapter 1: 34P (page 1)

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 x=0tox=8.0m?

Short Answer

Expert verified

The net work done on the brick by variable force W=800J

Step by step solution

01

Understanding the concept

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

Formula:

  1. W=F×x
  2. A=12base×height

Given:

  1. The mass of brick,m=10kg
  2. The scale of acceleration is set as, as=10m/s2
02

Calculate the work done

We have,

W=F.xW=ma.xW=ma×x

Here, the product a×xis given from the graph, as the area under the curve of ax

So, W=m×A

Where

A=12×base×height=12×8.0m×20.0m=80m2

Hence,

W=10kg×80m2W=800J

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

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