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# You push your physics book 1.50 m along a horizontal tabletop with a horizontal push of 2.40 N while 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, and (c) gravity? (e) What is the net work done on the book?

### 1 Answer

- 4 weeks ago
The work is equal to force times displacement traveled by the object; and the force is zero when the direction of the force is perpendicular to the direction of motion.

a) Knowing the force of your push and the distance the book was displaced; then the work done by your push is thus,

W_push = 2.4 * 1.50 = 3.6 J

b) knowing the friction force, and the distance which the book traveled you can find the work; where W_fr = - 0.5 *1.50 = - 0.75 JThe negative sign indicates that the force opposes the direction of motion (this work is exerted to stop the motion ).c) The gravity is pointing always downward, and the dot product between the direction of the motion and the direction of the force is zero, therefor is no work done by gravity in the given horizontal motion. In other words, since the direction of motion is perpendicular to the force of gravity, then the work done by gravity is zero. Thus,

W_gr = 0

e) The net work done on the book, is the net force acting on the book time the distance which the book have traveled; therefor,

W_net = (2.40 -0.5 ) * 1.50 = 1.9 * 1.5 = 2.85 J

Also, If you observed the obtained results, you would notice that the net work is the sum of the work done by your push and the work done by the friction, i.e.

W_net = W_push + W_fr = 3.6 - 0.75 = 2.85 J