Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
6
Work and Energy
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6-1: Work, Constant Force
6-3: Kinetic Energy; Work-Energy Principle
6-4 and 6-5: Potential Energy
6-6 and 6-7: Conservatino of Mechanical Energy
6-8 and 6-9: Law of Conservation of Energy
6-10: Power

Problem 14
A
W=2.2×102JW=2.2 \times 10^2J
Giancoli 6th Edition, Chapter 6, Problem 14 solution video poster
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VIDEO TRANSCRIPT

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.

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