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$W=2.2 \times 10^2J$

Giancoli 6th Edition, Chapter 6, Problem 14

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Chapter 6, Problem 14 is solved.

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Transcript for this Giancoli solution

Here's a graph, showing what they've told us in the question. They've told us that the force exerted on the particle increases linearly from 0 to 24, over 3 meters. And the word linear means straight line. So you can draw a straight line from 0 up to the 24 newtons, at 3 meters, then the force is constant until 8 meters, and then decreases back to 0 in a straight line until, 13 meters. In the fifth edition, this 13 is actually 11. We'll talk about that in a minute. So our job is going to be to find the total area underneath this curve. So we have 'A1' plus 'A2' plus 'A3'. 'A1' and 'A3' are triangles and 'A2' is a rectangle. And the work done is going to be a sum of those three areas. So that's going to be one half, times 3 meters is the base of triangle 'A1' and then it's 24 newtons high, plus the rectangle 'A2', 5 meters times 24 Newtons, plus the area three, one half times 5 meters, that's the difference between 13 and 8, times 24 Newtons. And the answer is SOLUTION joules of work done. In the fifth edition 13 is actually 11 meters instead. So this 5 instead become 3 meters, the difference between 11 and 8 is 3. And otherwise it's the same and the fifth edition answer is SOLUTION Joules.