Question

You grab a book and give it a quick push across the top of a horizontal table. After a short push, the book slides across the table, and because of friction, comes to a stop. (a) Draw an FBD of the book while you are pushing it. (b) Draw an FBD of the book after you have stopped pushing it, while it is sliding across the table. (c) Draw an FBD of the book after it has stopped sliding. (d) In which of the preceding cases is the net force on the book not equal to zero? (e) If the book has a mass of \(0.50 \mathrm{kg}\) and the coefficient of friction between the book and the table is \(0.40,\) what is the net force acting on the book in part (b)? (f) If there were no friction between the table and the book, what would the free-body diagram for part (b) look like? Would the book slow down in this case? Why or why not?

Short answer

Answer: The net force acting on the book while sliding across the table is 1.962 N, opposing the motion.

Step-by-step solution

(a) FBD while pushing the book

First, draw the FBD of the book while it is being pushed. The forces acting on the book are: 1. Gravitational force (weight) acting downward: \(F_g = mg\) 2. Normal force acting upward: \(F_n = mg\) (since there's no vertical acceleration) 3. Applied force (push) in the horizontal direction: \(F_{applied}\) 4. Frictional force opposing the applied force: \(F_{friction} = \mu F_n\) Draw a diagram with these forces acting on the book.

(b) FBD while sliding across the table

Now, draw the FBD of the book after being pushed and sliding across the table: 1. Gravitational force (weight) acting downward: \(F_g = mg\) 2. Normal force acting upward: \(F_n = mg\) 3. Frictional force opposing the motion: \(F_{friction} = \mu F_n\) Draw a diagram with these forces acting on the book.

(c) FBD after the book has stopped sliding

Next, draw the FBD of the book when it comes to rest: 1. Gravitational force (weight) acting downward: \(F_g = mg\) 2. Normal force acting upward: \(F_n = mg\) Draw a diagram showing these forces acting on the book.

(d) Identify when the net force is not equal to zero

The net force on the book is not equal to zero when there's an unbalanced force acting on it. This occurs in part (a) when you are pushing the book, and in part (b) while it's sliding due to the frictional force.

(e) Calculate the net force in part (b)

Given the mass of the book \(m = 0.50 kg\) and the coefficient of friction between the book and the table \(\mu = 0.40\). First, find the normal force: \(F_n = mg = (0.50 kg)(9.81 m/s^2) = 4.905 N\) Now calculate the frictional force: \(F_{friction} = \mu F_n = (0.40)(4.905 N) = 1.962 N\) The net force acting on the book in part (b) is the frictional force, which is \(1.962 N\) opposing the motion.

(f) FBD and motion without friction

If there were no friction between the table and the book, the FBD for part (b) would have only two forces: gravitational force acting downward and the normal force acting upward, both equal in magnitude and opposite in direction, which results in a balanced scenario. In this case, the book would not slow down because there's no opposing force to decelerate it. It would continue to slide with constant velocity according to Newton's first law of motion.