Problem 18 - College Physics

Question

\(\cdot\) \(\cdot\) \(\cdot\) A 1.50 \(\mathrm{kg}\) book is sliding along a rough horizontal sur- face. 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: Determine the Work Done (Part a)

To find the work done on the book between points A and B, we use the work-energy principle: \[ W_{AB} = rac{1}{2}m(v_B^2 - v_A^2) \]where \( m = 1.50 \; \text{kg} \), \( v_A = 3.21 \; \text{m/s} \), and \( v_B = 1.25 \; \text{m/s} \).Substitute these values in: \[W_{AB} = \frac{1}{2} \times 1.50 \times (1.25^2 - 3.21^2) \]Solving, we find:\[W_{AB} = \frac{1}{2} \times 1.50 \times (1.5625 - 10.3041) \]\[W_{AB} = \frac{1}{2} \times 1.50 \times (-8.7416) \]\[W_{AB} = -6.5562 \; \text{J} \]

Step 2: Velocity at Point C with -0.750 J Work Done (Part b)

Using the work-energy principle again, the change in kinetic energy from B to C is equal to the work done: \[ W_{BC} = \frac{1}{2}m(v_C^2 - v_B^2) \]Given \( W_{BC} = -0.750 \; \text{J} \) and \( v_B = 1.25 \; \text{m/s} \),\[-0.750 = \frac{1}{2} \times 1.50 \times (v_C^2 - 1.25^2) \]\[-0.750 = 0.75 \times (v_C^2 - 1.5625) \]\[-1 = v_C^2 - 1.5625 \]\[v_C^2 = 0.5625 \]\[v_C = \sqrt{0.5625} = 0.75 \; \text{m/s} \]

Step 3: Velocity at Point C with +0.750 J Work Done (Part c)

If instead, +0.750 J of work is done, set up the equation:\[ W_{BC}' = \frac{1}{2}m(v_C'^2 - v_B^2) \]where \( W_{BC}' = +0.750 \; \text{J} \),\[0.750 = \frac{1}{2} \times 1.50 \times (v_C'^2 - 1.25^2) \]\[0.750 = 0.75 \times (v_C'^2 - 1.5625) \]\[1 = v_C'^2 - 1.5625 \]\[v_C'^2 = 2.5625 \]\[v_C' = \sqrt{2.5625} = 1.60 \; \text{m/s} \]

Key Concepts

Kinetic Energy Change

When objects move, they have what's called kinetic energy, which is the energy of motion. The formula for kinetic energy is \[ KE = \frac{1}{2}mv^2 \]where:

• \( KE \) is the kinetic energy
• \( m \) is the mass of the object
• \( v \) is its velocity

Kinetic energy changes when the object's speed changes. When a book slides on a surface and slows down from a speed of 3.21 m/s to 1.25 m/s, its kinetic energy decreases. This change in kinetic energy can be calculated by using its initial and final velocities. This principle is crucial because it helps us understand how energy is transferred or converted in different systems. In this example, the energy is transferred by the force of friction as it does work against the book's motion.

Velocity Calculations

Velocity tells us how fast an object is moving in a specific direction. To find out how fast the book is moving at different points, we can use the work-energy principle. This principle relates the work done on an object to its change in kinetic energy. For instance, when the book goes from point B to point C, and we know the work done (-0.750 J or +0.750 J), we can calculate the new velocity:1. **With -0.750 J of work:** We can set up our equation: - \(-0.750 = \frac{1}{2} \cdot 1.50 \cdot (v_C^2 - 1.25^2)\) - By solving this equation, we find \(v_C = 0.75 \; \text{m/s}\) at point C.2. **With +0.750 J of work:** If instead the work done was +0.750 J, we use the same method and find: - \(v_C' = \sqrt{2.5625} = 1.60 \; \text{m/s}\)These calculations show how work affects the speed of an object by changing its kinetic energy, helping us gain insights into how different forces influence motion.

Work Done on Objects

Work is one of the main ideas that helps us understand how forces act on objects. When a force is applied over a distance, it does work. The formula for work can be written as:\[ W = F \cdot d \cdot \cos(\theta) \]where:

• \( W \) is the work done
• \( F \) is the force applied
• \( d \) is the distance over which the force is applied
• \( \theta \) is the angle between the force and the direction of movement

In our exercise, work is done to the book by the force of friction, which is aligned against the movement. We calculated work when the book slowed from 3.21 m/s to 1.25 m/s by using the change in kinetic energy. Negative work indicates that the force is opposing the movement, leading to a decrease in speed. Understanding how work is calculated and how it affects motion is key to solving many problems in physics and helps us design systems like brakes, engines, and many other mechanical tools.