Problem 14 - University Physics with Modern Physics

Question

A 1.50 -kg book is sliding along a rough horizontal surface. At point \(A\) it is moving at \(3.21 \mathrm{m} / \mathrm{s},\) and at point \(B\) it has slowed to 1.25 \(\mathrm{m} / \mathrm{s}\) (a) How much work was done on the book between \(A\) and \(B ?\) (b) If \(-0.750 \mathrm{J}\) of work is done on the book from \(B\) to \(C\) , how fast is it moving at point \(C ?\) (c) How fast would it be moving at \(C\) if \(+0.750 \mathrm{J}\) of work were done on it from \(B\) to \(C\) ?

Step-by-Step Solution

Answer

(a) -6.56 J, (b) 0.75 m/s, (c) 1.60 m/s.

Step 1: Identify Known Quantities

We are given the mass of the book as \(1.50\ \text{kg}\), the initial speed at point \(A\) as \(3.21\ \text{m/s}\), and the speed at point \(B\) as \(1.25\ \text{m/s}\). We need to calculate the work done on the book from \(A\) to \(B\).

Step 2: Calculate Initial and Final Kinetic Energy

The kinetic energy at any point is given by \( KE = \frac{1}{2} m v^2 \). Calculate the initial kinetic energy (at \(A\)) and the final kinetic energy (at \(B\)).Initial kinetic energy: \( KE_A = \frac{1}{2} \times 1.50 \times (3.21)^2 = 7.7289\ \text{J} \)Final kinetic energy: \( KE_B = \frac{1}{2} \times 1.50 \times (1.25)^2 = 1.171875\ \text{J} \)

Step 3: Calculate Work Done from A to B

The work done on an object is the change in its kinetic energy, calculated as \( W = KE_B - KE_A \).Work done: \( W = 1.171875 - 7.7289 = -6.557025\ \text{J} \)

Step 4: Determine Speed at Point C for -0.750 J Work Done

For the book moving from \(B\) to \(C\), with \(-0.750\ \text{J}\) of work done on it, the kinetic energy at \(C\) is given by \( KE_C = KE_B + (-0.750) \).\( KE_C = 1.171875 + (-0.750) = 0.421875\ \text{J} \)Solve for the speed at \(C\): \( 0.421875 = \frac{1}{2} \times 1.50 \times v_C^2 \)\( v_C^2 = \frac{0.421875 \times 2}{1.50} \) \( v_C = \sqrt{0.5625} = 0.75\ \text{m/s} \)

Step 5: Determine Speed at Point C for +0.750 J Work Done

Repeat the process for \(+0.750\ \text{J}\) of work done from \(B\) to \(C\).\( KE_C = KE_B + 0.750 = 1.171875 + 0.750 = 1.921875\ \text{J} \) Solve for the new speed at \(C\): \( 1.921875 = \frac{1}{2} \times 1.50 \times v_C^2 \) \( v_C^2 = \frac{1.921875 \times 2}{1.50} \) \( v_C = \sqrt{2.5625} = 1.60\ \text{m/s} \)

Key Concepts

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. The formula for kinetic energy (KE) is \[ KE = \frac{1}{2} mv^2 \]where:

• \( m \) is the mass of the object, and
• \( v \) is its velocity.

In the given exercise, we see how the kinetic energy of a book changes as it slides over a surface. Initially, at point A, it has a velocity of 3.21 m/s, which results in a kinetic energy value. As it reaches point B, its velocity decreases to 1.25 m/s. This decrease in speed leads to a decrease in kinetic energy, demonstrating how speed directly affects the energy a moving object possesses.

Work-Energy Principle

The Work-Energy Principle is a powerful concept that relates the work done on an object to its change in kinetic energy. According to the principle:\[ W = \Delta KE = KE_{final} - KE_{initial} \]
This formula tells us that the work done on the book, when moving from point A to point B, equals the change in its kinetic energy.
In this exercise, the book's kinetic energy decreased from 7.7289 J to 1.171875 J because of the work done by external forces, like friction. This negative work done indicates that the total energy of the book was reduced as it moved from point A to point B, helping us understand the energy loss mechanisms at play.

Friction

Friction is the resistive force that opposes the motion or attempted motion of one surface across another. It plays a significant role in the exercise when the book slides on a rough horizontal surface.

• Friction acts in the opposite direction of motion, causing the book to slow down.
• This opposing force results in negative work done on the book, evidenced by the decrease in its kinetic energy.
• The energy lost due to friction is converted into other forms, such as heat or sound.

In physics problems, like the one we've examined, understanding the impacts of friction helps in predicting how quickly or slowly an object will slow down as it moves over a rough surface. This is crucial in practical applications, such as designing efficient braking systems or understanding wear and tear on moving machinery parts.