3. 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?

a) W=2.4*1.5*cos(0)= 3.6J

b) W=0.600*1.5*cos(180)= -0.9J

c) W=0J because the displacement and the normal force is perpendicular to each other so cos(90)=0

d) W=0J for the same reason as part (c), but with the gravity

e) W(net)= 3.6+(-0.9)+0+0= 2.7J
you are just adding all the works together

I will be happy to critique your thinking.

(a) Well, if you push your physics book with a force of 2.40 N, you're definitely doing some work. But how much? Well, work is equal to force times distance, so if you push the book 1.50 m, the work you do is 2.40 N × 1.50 m. Math, making physics fun since forever!

(b) Ah, the pesky friction force. It's always trying to ruin our fun! The work done by friction can be calculated in the same way: force times distance. In this case, the friction force is 0.600 N, and since it acts in the opposite direction of the book's motion, the work it does is -0.600 N × 1.50 m. Remember, friction brings negative vibes!

(c) The normal force from the tabletop is perpendicular to the book's motion, so it doesn't do any work. It's just hanging around, keeping the book steady and feeling...normal.

(d) Gravity, my dear friend. It's hard to escape its pull, but luckily, it doesn't do any work in this scenario either. Since the book is moving horizontally, gravity's force is acting perpendicular to the displacement, so its work is 0 J. Guess gravity just wants to be a spectator this time.

(e) To find the net work done on the book, we just need to add up all the work done by each force. So, net work = work by your push + work by friction + work by normal force + work by gravity. And, as we calculated before, the last two are 0 J. So the net work is just the sum of the first two, which is 2.40 N × 1.50 m plus (-0.600 N × 1.50 m). Crunching the numbers gives us...wait for it...0 J! Yup, no net work done on the book. It's a tie, folks!

To determine the work done by each force and the net work done on the book, we can use the formula:

Work = Force x Distance x Cos(θ)

Where:
- Work is the work done by the force (in joules, J)
- Force is the magnitude of the force applied (in newtons, N)
- Distance is the distance over which the force is applied (in meters, m)
- θ is the angle between the direction of the force and the direction of displacement.

(a) Work done by the 2.40 N push:
The 2.40 N push is applied horizontally, so the angle (θ) between the force and the displacement is 0°.

Work (push) = 2.40 N x 1.50 m x cos(0°)

Since cos(0°) = 1, the work done by the push is:

Work (push) = 2.40 N x 1.50 m = 3.60 J

(b) Work done by the friction force:
The friction force opposes the motion, so the angle (θ) between the force and the displacement is 180°.

Work (friction) = -0.600 N x 1.50 m x cos(180°)

Since cos(180°) = -1, the work done by the friction force is:

Work (friction) = -0.600 N x 1.50 m = -0.90 J

The negative sign indicates that the friction force does negative work, meaning it takes energy away from the system.

(c) Work done by the normal force:
The normal force is perpendicular to the displacement, so the angle (θ) between the force and the displacement is 90°.

Work (normal force) = 0 N (perpendicular to displacement)

The normal force does no work since it acts perpendicular to the displacement.

(d) Work done by gravity:
The work done by gravity depends on the direction of the displacement. If the displacement is vertical, then the work done by gravity will be non-zero.

Work (gravity) = m x g x h

Where:
- m is the mass of the object (in kg)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- h is the vertical displacement of the object (in meters)

Without information about the vertical displacement, we cannot calculate the work done by gravity.

(e) Net work done on the book:
The net work done on the book is the sum of the work done by each force:

Net Work = Work (push) + Work (friction)

Net Work = 3.60 J + (-0.90 J)

Net Work = 2.70 J

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

To calculate the work done by each force, we will use the formula:

Work = Force * Distance * cos(theta)

Where:
- Force is the magnitude of the force applied in Newtons.
- Distance is the displacement of the object in meters.
- Theta is the angle between the force and the direction of displacement, which is 0 degrees for horizontal displacements.

(a) Work done by the 2.40-N push:
Since the push is applied horizontally, the angle between the force and displacement is 0 degrees.
Work = 2.40 N * 1.50 m * cos(0)
Work = 2.40 N * 1.50 m * 1
Work = 3.60 Joules

(b) Work done by the friction force:
Friction acts opposite to the direction of motion, so the angle between the force and displacement is 180 degrees.
Work = 0.600 N * 1.50 m * cos(180)
Work = 0.600 N * 1.50 m * -1
Work = -0.900 Joules

(c) Work done by the normal force:
The normal force is perpendicular to the direction of displacement, so the angle between the force and displacement is 90 degrees.
Work = 0 N * 1.50 m * cos(90)
Work = 0 Joules

(d) Work done by gravity:
Gravity acts vertically downwards, while the displacement is horizontal. Therefore, there is no displacement in the direction of gravity, so no work is done by gravity.
Work = 0 Joules

(e) Net work done on the book:
The net work done on an object is the sum of the work done by all the forces acting on it.
Net Work = Work(a) + Work(b) + Work(c) + Work(d)
Net Work = 3.60 Joules + (-0.900 Joules) + 0 Joules + 0 Joules
Net Work = 2.70 Joules

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