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Fundamentals Of Physics
Found in: Page 173

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Short Answer

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’s vertical axis is set by Fs=10.0 N. How much work is done by the force as the block moves from the origin to x=8.0 cm?

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

See the step by step solution

Step by Step Solution

Step 1: Given

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

Step 2: 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=xixfF(x)dxA=12×base×heightA=length×breadth

Step 3: 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=10 N×2 m+12×10 N×2 m+0+-12×2 N×5mW=25J

Therefore, the net work done is 25J

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