A 0.60-kg book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.8 N. (a) How much work is done on the book by friction during a displacement of 3.0 m to the left? (b) The book now slides 3.0 m to the right, returning to its starting point. During this second 3.0-m displacement, how much work is done on the book by friction? (c) What is the total work done on the book by friction during the complete round trip? (d) On the basis of your answer to part (c), would you say that the friction force is conservative or nonconservative? Explain.

Solution 29E Work done by a force is given by the mathematical relation Work = Force × Displacement. Given, the kinetic friction force isfF = 1.8 N (a) The displacement of the book is = 3.0 m Work done by friction on the book is W 1= 1.8 N × 3.0 m = 5.4 J (b) Again, when the book is displaced to the right, work done by friction on the book is W = 1.8 N × 3.0 m = 5.4 J 2 The negative sign, ‘-’ appears in each case to indicate that the force of friction always acts in a direction opposite to the applied force. (c) T herefore, total work done by the force of friction on the book is = 5.4 J 5.4 J = 10.8 J (d) A force is said to be non conservative if total work done by the force along a closed path is non-zero. Work done by the force depends on the path taken between the initial and final points. Therefore, friction is a nonconservative force.