You push your physics book 1.50 m along a horizontal table-top with a horizontal push of 2.40 N while the opposing force of friction is 0.600 N. How much work does each of the following forces do on the book: (a) your 2.40-N push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

Solution 1E Introduction Here force and displacement is given, and we can also find out the angle between the force and the displacement. So we can calculate the work for each case. Solution The work is given by W = F · d = Fdcos Here F and d are the magnitude of force and displacement. (a) The book will move in the direction of the applied force, if the force is horizontal then the angle between force and displacement is = 0 rad. Hence the work is given by W = Fdcos = (2.40 N)(1.50 m)cos0 = 3.60 J 1 (b) Friction always opposes the motion, that is the direction of the frictional force and the displacement will be opposite to each other, which means the angle between the frictional force and the displacement is rad. Hence the work is W =2Fdcos = (0.600 N)(1.50 m)cos = 0.900 J (C) The normal force always act perpendicular to the surface on which the object is kept. Since in this case the surface is horizontal, the normal force is vertical. Also since the displacement is horizontal the angle between the force and displacement is /2 rad. Now if N is the magnitude of normal force, then we have W =3Fdcos = N(1.50 m)cos(/2) = 0 J (d) The gravitational force always act vertically downward, hence the angle between the gravitational force and the displacement is /2 rad. Suppose F g is the gravitational force, then the work is W =4Fdcos = F dcosg/2) = 0 J (e) Hence the net work done on the book was W net (3.60 J) (0.900 J) = 2.70 J