**Physics Problem: Rope and Block System** **Problem Statement:** A rope of negligible mass supports a block that weighs 30.0 N, as shown in the diagram. The breaking strength of the rope is 50.0 N. The largest acceleration that can be given to the block by pulling up on it without breaking the rope is most nearly... - (A) 6.7 m/s² - (B) 10 m/s² - (C) 16.7 m/s² - (D) 26.7 m/s² **Diagram Explanation:** The diagram presents a vertical setup: - A block is suspended in the air, supported by a rope. - The weight of the block is labeled as 30 N. - The rope is directed upwards and labeled "shown to the rope," indicating tension or force applied in this direction. **Solution Approach:** To find the largest acceleration: 1. **Understand Forces:** - The maximum tension (T) before the rope breaks is 50 N. - The weight of the block (W) is 30 N. 2. **Net Force:** - Tension (T) minus Weight (W) provides the net force available to accelerate the block. - Net Force = T - W = 50 N - 30 N = 20 N. 3. **Apply Newton's Second Law:** - Net Force = mass (m) × acceleration (a) - Assuming g = 9.8 m/s², calculate mass: W = m × g → m = W/g = 30 N / 9.8 m/s² ≈ 3.06 kg. - 20 N = 3.06 kg × a → a ≈ 6.54 m/s². The closest answer is (A) 6.7 m/s².
**Physics Problem: Rope and Block System** **Problem Statement:** A rope of negligible mass supports a block that weighs 30.0 N, as shown in the diagram. The breaking strength of the rope is 50.0 N. The largest acceleration that can be given to the block by pulling up on it without breaking the rope is most nearly... - (A) 6.7 m/s² - (B) 10 m/s² - (C) 16.7 m/s² - (D) 26.7 m/s² **Diagram Explanation:** The diagram presents a vertical setup: - A block is suspended in the air, supported by a rope. - The weight of the block is labeled as 30 N. - The rope is directed upwards and labeled "shown to the rope," indicating tension or force applied in this direction. **Solution Approach:** To find the largest acceleration: 1. **Understand Forces:** - The maximum tension (T) before the rope breaks is 50 N. - The weight of the block (W) is 30 N. 2. **Net Force:** - Tension (T) minus Weight (W) provides the net force available to accelerate the block. - Net Force = T - W = 50 N - 30 N = 20 N. 3. **Apply Newton's Second Law:** - Net Force = mass (m) × acceleration (a) - Assuming g = 9.8 m/s², calculate mass: W = m × g → m = W/g = 30 N / 9.8 m/s² ≈ 3.06 kg. - 20 N = 3.06 kg × a → a ≈ 6.54 m/s². The closest answer is (A) 6.7 m/s².
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:**Physics Problem: Rope and Block System**
**Problem Statement:**
A rope of negligible mass supports a block that weighs 30.0 N, as shown in the diagram. The breaking strength of the rope is 50.0 N. The largest acceleration that can be given to the block by pulling up on it without breaking the rope is most nearly...
- (A) 6.7 m/s²
- (B) 10 m/s²
- (C) 16.7 m/s²
- (D) 26.7 m/s²
**Diagram Explanation:**
The diagram presents a vertical setup:
- A block is suspended in the air, supported by a rope.
- The weight of the block is labeled as 30 N.
- The rope is directed upwards and labeled "shown to the rope," indicating tension or force applied in this direction.
**Solution Approach:**
To find the largest acceleration:
1. **Understand Forces:**
- The maximum tension (T) before the rope breaks is 50 N.
- The weight of the block (W) is 30 N.
2. **Net Force:**
- Tension (T) minus Weight (W) provides the net force available to accelerate the block.
- Net Force = T - W = 50 N - 30 N = 20 N.
3. **Apply Newton's Second Law:**
- Net Force = mass (m) × acceleration (a)
- Assuming g = 9.8 m/s², calculate mass: W = m × g → m = W/g = 30 N / 9.8 m/s² ≈ 3.06 kg.
- 20 N = 3.06 kg × a → a ≈ 6.54 m/s².
The closest answer is (A) 6.7 m/s².
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