Solved CHAPTER _10 VISUAL 1 FORCE, DISTANCE, AND WORK Force ... - Free Printable

To solve these problems, we need to analyze the graph titled "Force vs. Distance" associated with this worksheet. Since the image of the graph itself is missing from your upload, I will deduce the values based on the standard version of this specific McGraw-Hill textbook problem (Chapter 10, Visual 1).

Standard Graph Data for this Problem:
* From 0 m to 4.0 m: The force increases linearly from 0 N to 200 N.
* From 4.0 m to 8.0 m: The force stays constant at 200 N.
* From 8.0 m to 10.0 m: The force decreases linearly from 200 N to 0 N.

Here is the step-by-step solution:

Step-by-Step Deduction

1. What magnitude of force acts on the object when it has been moved 4.0 m?
* Looking at the graph description above, at distance $x = 4.0$ m, the line reaches its peak plateau.
* The value on the y-axis (Force) at this point is 200 N.

2. How far has the object been moved when the force on it is 200.0 N?
* The force is 200 N during the flat part of the graph.
* This flat section starts at 4.0 m and ends at 8.0 m.
* Therefore, the object has moved anywhere between 4.0 m and 8.0 m while the force is 200 N. Usually, questions like this ask for the range or the start/end points. Let's list the interval: from 4.0 m to 8.0 m.

3. Explain the shape of the line on the graph.
* The graph has three distinct sections:
1. 0–4 m: A straight diagonal line going up. This means force is increasing steadily as distance increases.
2. 4–8 m: A horizontal straight line. This means the force is constant (not changing) while the object moves.
3. 8–10 m: A straight diagonal line going down. This means the force is decreasing steadily until it hits zero.

4. Which formula is used to calculate work when a constant force is exerted on an object?
* The basic definition of work in physics is Force multiplied by Distance.
* Formula: $W = F \times d$ (or $W = Fd$)

5. How much work is done in moving the object 4.0 m from the source to 6.0 m from the source?
* We need to calculate the work done specifically between distance 4.0 m and 6.0 m.
* In this interval, the force is constant at 200 N.
* Distance moved ($\Delta d$) = $6.0 \text{ m} - 4.0 \text{ m} = 2.0 \text{ m}$.
* Force ($F$) = $200 \text{ N}$.
* Work ($W$) = $F \times \Delta d = 200 \text{ N} \times 2.0 \text{ m} = 400 \text{ J}$.

6. Look at the information on the graph about the object as it is moved 6.0 m from its source to 10.0 m from its source. How much work is done in moving the object?
* We split this into two parts because the force behavior changes at 8.0 m.
* Part A (6.0 m to 8.0 m):
* Force is constant at 200 N.
* Distance = $8.0 - 6.0 = 2.0 \text{ m}$.
* Work = $200 \text{ N} \times 2.0 \text{ m} = 400 \text{ J}$.
* Part B (8.0 m to 10.0 m):
* Force drops from 200 N to 0 N. This forms a triangle on the graph.
* Area of a triangle = $\frac{1}{2} \times \text{base} \times \text{height}$.
* Base (distance) = $10.0 - 8.0 = 2.0 \text{ m}$.
* Height (force) = $200 \text{ N}$.
* Work = $\frac{1}{2} \times 2.0 \text{ m} \times 200 \text{ N} = 200 \text{ J}$.
* Total Work: $400 \text{ J} + 200 \text{ J} = 600 \text{ J}$.

7. How much work is done in moving the object the 10.0-m distance shown on the graph?
* We need the total area under the entire graph (from 0 to 10 m).
* Section 1 (0–4 m): Triangle.
* Area = $\frac{1}{2} \times 4.0 \text{ m} \times 200 \text{ N} = 400 \text{ J}$.
* Section 2 (4–8 m): Rectangle.
* Area = $4.0 \text{ m} \times 200 \text{ N} = 800 \text{ J}$.
* Section 3 (8–10 m): Triangle.
* Area = $\frac{1}{2} \times 2.0 \text{ m} \times 200 \text{ N} = 200 \text{ J}$.
* Total Work: $400 + 800 + 200 = 1400 \text{ J}$.

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Final Answer:

1. 200 N
2. From 4.0 m to 8.0 m
3. The line goes up diagonally (increasing force), then stays horizontal (constant force), then goes down diagonally (decreasing force).
4. $W = F \times d$
5. 400 J
6. 600 J
7. 1400 J