you push a physics book 1.50m along a horizontal tabletop 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 follwing forces do on the book? A) you 2.40N push. B) the friction force. C) the normal force from the table and D) gravity? E) what is the net work done on the book?

a table exerts a 2.0 -newton force on a book lying on the table. how much force is exerted by the book on the table?

A) 2.40*1.50=3.6J

b) -0.600*1.50=-0.9
c) (1.50)cos90=0
d) same as c
E) 3.6-0.9=2.7

To calculate the work done by each force on the book, you need to use the formula:

Work = Force x Distance x Cos(theta)

where:

- Work is the work done (in joules)
- Force is the magnitude of the force applied (in newtons)
- Distance is the distance over which the force is applied (in meters)
- Theta is the angle between the direction of the force and the direction of the displacement of the book (in degrees)

A) For the 2.40N push:
The force and displacement are both horizontal, so the angle between them is 0 degrees.

Work = 2.40N x 1.50m x Cos(0)
Work = 3.60 Joules

B) For the friction force:
The force and displacement are in the opposite direction, so the angle between them is 180 degrees.

Work = 0.600N x 1.50m x Cos(180)
Work = -0.900 Joules (negative sign indicates opposite direction)

C) For the normal force from the table:
The normal force is perpendicular to the displacement, so the angle between them is 90 degrees.

Since the normal force does not cause any displacement in this case, the work done by the normal force is zero (work = force x distance x cos(90), cos(90) = 0).

D) For gravity:
Assuming the book is on a flat table, the gravitational force is perpendicular to the displacement, so the angle between them is 90 degrees.

Similar to the normal force, the work done by gravity is also zero.

E) The net work done on the book is the sum of the work done by each force.

Net work = Work (2.40N push) + Work (friction force) + Work (normal force) + Work (gravity)
Net work = 3.60 J + (-0.900 J) + 0 J + 0 J
Net work = 2.70 J

Therefore, the net work done on the book is 2.70 joules.