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Chapter 7: 36P (page 173)

A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in Fig.7-39. The scale of the figure鈥檚 vertical axis is set byFs=10.0N. How much work is done by the force as the block moves from the origin to x=8.0cm?

Short Answer

Expert verified

The net work done by the force using a graph W=25J.

Step by step solution

01

Given

  1. The mass of block is, m=5.0kg
  2. The scale for vertical axis is, Fs=10.0N
02

Understanding the concept

We can use the concept that the work done is equal to the area under the curve for a graph.

Formula:

W=鈭�xixfF(x)dxA=12脳base脳heightA=length脳breadth

03

Calculate the net work done

We know that the total area will be the net work done by the force.

So,

W=W1+W2+W3+W4

Substitute all the value in the above equation.

W=10N脳2m+12脳10N脳2m+0+-12脳2N脳5mW=25J

Therefore, the net work done is 25J

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