Problem 39 - College Physics

Question

Six identical books, \(4.0 \mathrm{~cm}\) thick and each with a mass of \(0.80 \mathrm{~kg}\), lie individually on a flat table. How much work would be needed to stack the books one on top of the other?

Step-by-Step Solution

Answer

The total work needed is 4.71 J.

Step 1: Understand the Problem

We need to calculate the total work done to stack six books from an initial state where they lie flat individually to a final state where they are one on top of the other.

Step 2: Define the Work Formula

Work done against gravity (lifting) is given by the formula: \( W = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (9.81 m/s²), and \( h \) is the height to which the mass is lifted.

Step 3: Calculate the Mass Lifting Required for Each Book

When each book is lifted, it needs to be at a specific height: - 1st book is at 0 cm (no lifting) - 2nd book is lifted 4 cm (1 book height) - 3rd book is lifted 8 cm (2 book heights) - 4th book is lifted 12 cm (3 book heights) - 5th book is lifted 16 cm (4 book heights) - 6th book is lifted 20 cm (5 book heights) Convert these heights to meters for our calculations.

Step 4: Calculate the Work Done for Lifting Each Book

For each book, apply the formula \( W = mgh \):- 2nd book: \( W = 0.80 \times 9.81 \times 0.04 \)- 3rd book: \( W = 0.80 \times 9.81 \times 0.08 \)- 4th book: \( W = 0.80 \times 9.81 \times 0.12 \)- 5th book: \( W = 0.80 \times 9.81 \times 0.16 \)- 6th book: \( W = 0.80 \times 9.81 \times 0.20 \)

Step 5: Sum the Work Done

Add all the individual works calculated above to get the total work done:\[ W_{total} = 0.80 \times 9.81 \times (0.04 + 0.08 + 0.12 + 0.16 + 0.20) \]

Step 6: Solve for Total Work Done

Compute the total work:- Total height = 0.04 + 0.08 + 0.12 + 0.16 + 0.20 = 0.60 m- Total work: \( W_{total} = 0.80 \times 9.81 \times 0.60 \)- \( W_{total} = 4.7088 \) J (Joules)

Key Concepts

Gravitational Work

Gravitational work is the work done by or against the force of gravity to move an object vertically. In our problem, we lifted books to stack them, which means we are doing work against the gravitational force.
When you lift a book from the table to a higher position, gravity resists this move. We calculate this work using the formula \[ W = mgh \]where:

• \( W \) is the work done,
• \( m \) is the mass of the object (in kilograms),
• \( g \) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\) on Earth),
• \( h \) is the height of the lift above the original position (in meters).

Through this exercise, we learn how each book's mass and the height we lift it to play crucial roles in calculating gravitational work done.

Mechanics

Mechanics is the branch of physics that deals with motion and the forces that produce motion. In our scenario, we deal with a subset of mechanics that focuses on statics and dynamics - using work to move objects vertically.
Whenever you apply a force on an object causing it to move, you're dealing with mechanics. Even lifting a book from a table involves mechanics because you're using a force larger than the gravitational pull against the book.
By understanding these concepts through practical problems like stacking books, you can develop a strong grasp of how forces, like gravity, affect objects at rest and in motion. This knowledge forms the foundation for more advanced topics in physics.

Kinetic Energy

Kinetic energy refers to the energy of motion. While the main concern in our exercise involves gravitational work, kinetic energy joins the conversation once you start lifting a book.
When a book is in motion, even momentarily as it’s lifted, it possesses kinetic energy. However, when calculating the work done in stacking books, kinetic energy primarily illustrates potential conversion into different energy forms, not directly used in the calculation of total work done.
Understanding kinetic energy helps us comprehend how energy transitions through different forms in real-world applications. You'll find this topic very crucial while delving deeper into energy transformations and mechanical processes.

Energy Conservation

Energy conservation is a principle stating that energy in an isolated system remains constant. It’s neither created nor destroyed but transformed between forms.
In our context of stacking books, initially, all books possess potential energy while lying flat. As each book is lifted, chemical energy (human effort) gets converted into gravitational potential energy. The work done doesn’t disappear; it’s stored as potential energy due to the book’s higher position.
This principle is vital for understanding how energy transitions in a system, helping us build efficient systems minimizing energy loss in various applications. It underpins many mechanical systems, ensuring energy is utilized effectively, aligning back to energy conservation laws in physics.