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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鈥檚 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脳x
  2. A=12base脳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脳x

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

So, W=m脳A

Where

A=12脳base脳height=12脳8.0m脳20.0m=80m2

Hence,

W=10kg脳80m2W=800J

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

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Most popular questions from this chapter

Figure 10-54shows a flat construction of two circular rings that have a common center and are held together by three rods of negligible mass. The construction, which is initially at rest, can rotate around the common center (like a merry-go-round), where another rod of negligible mass lies. The mass, inner radius, and outer radius of the rings are given in the following table. A tangential force of magnitude 12.0Nis applied to the outer edge of the outer ring for 0.300s.What is the change in the angular speed of the construction during the time interval?

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Figure 1-6Problem 13

A single force acts on a 3.0 kgparticle-like object whose position is given byx=3.0t-4.0t2+1.0t3, with x in meters and t in seconds. Find the work done by the force fromt=0tot=4.0s.
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