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CALC An object has several forces acting on it. One of

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 98P Chapter 6

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 98P

CALC An object has several forces acting on it. One of these forces is a force in the x- direction whose magnitude depends on the position of the object, with ? = 2.50 N/m? . Calculate the work done on the object by this force for the following displacements of the object: (a) The object starts at the point (x = 0, y = 3.00 m) and moves parallel to the x- axis to the point (x = 2.00 m, y = 3.00 m). (b) The object starts at the point (x = 2.00 m, y = 0) and moves in the y- direction to the point (x = 2.00 m, y = 3.00 m). (c) The object starts at the origin and moves on the line y = 1.5 x to the point (x = 2.00 m, y = 3.00 m).

Step-by-Step Solution:

Solution 98P Introduction Force as a function displacement is given, we have to calculate the work done by the force. Also we can see that the force is only in the x direction so only the x component of the displacement will have non zero work done. Step 1 In the first case the the particle is moving in the x direction hence y remains constant at y = 3.00 m. Hence the work done is given by Hence the work done by the force in this case is 15.0 J. Step 2 In the second case the displacement is in the y direction and the force is in the x direction. Hence the force and displacement are perpendicular. So the work done by the force is zero.

Step 3 of 3

Chapter 6, Problem 98P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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