Why does this part not require a Sin(30) to decompose the MG(Sin(30))(.8)? The example gives a correct answer. **Section 7.1: Gravitational Potential Energy**---**Page 215:****Example 7.8: An inclined plane with friction****Scenario:**A 12-kg crate is pushed up a 2.5-meter ramp inclined at 30°. Initially, a worker estimates that pushing with a speed of 5.0 m/s to the bottom of the ramp will work, disregarding friction. Nevertheless, friction isn't negligible, causing the crate to move 1.6 meters up the ramp, stop, and then slide back down.**Objective:**Identify the friction force and calculate the crate's speed when it reaches the bottom.**Solution Steps:**1. **Identify and Set Up:** - Analyze the friction force's effect during the crate's sliding. - **Point 1:** Ground level with speed $ v_1 = 5.0 $ m/s. - **Point 2:** Crate stops momentarily. - **Point 3:** Return to the starting point.2. **Equations and Calculations:** - Initial Kinetic Energy: $ K_1 = \frac{1}{2} (12 \, \text{kg}) (5.0 \, \text{m/s})^2 = 150 \, \text{J} $ - Potential Energy at the top of ramp: $ U_{\text{grav, 2}} = (12 \, \text{kg})(9.8 \, \text{m/s}^2)(0.80 \, \text{m}) = 94 \, \text{J} $3. **Friction Force Calculation:** - Total energy decreases from 150 J to 94 J. - Friction force = 35 N over 1.6 m.4. **Conclusion:** - Crate returns with 38 J of the original energy. - $ v_3 = \sqrt{\frac{2(38 \text{ J})}{12 \text{ kg}}} = 2.5 \text{ m/s} $**Figures and Diagrams:**- **Diagram (7.11a):** Illustrates the crate’s path up and down the ramp, points 1, 2, and 3 labeled. The ramp is inclined at 30° with marked distance of 0

In the given example, The problem based on the conservation of energy.

7.11 (a) A crate slides partway up the ramp, stops, and slides back down. (b) Energy bar graphs for points 1, 2, and 3. (a) The crate slides up from point 1 to point 2, then back down to its starting position (point 3). Point 2 U2 = 0 -2.5 m: -1.6 m- 5.0 m/ %3D The crate is moving at speed v, when it returns to point 3.. 0.80 m 30° Point 1.3

7.11 (a) A crate slides partway up the ramp, stops, and slides back down. (b) Energy bar graphs for points 1, 2, and 3. (a) The crate slides up from point 1 to point 2, then back down to its starting position (point 3). Point 2 U2 = 0 -2.5 m: -1.6 m- 5.0 m/ %3D The crate is moving at speed v, when it returns to point 3.. 0.80 m 30° Point 1.3

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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Question

Why does this part not require a Sin(30) to decompose the MG(Sin(30))(.8)? The example gives a correct answer.

**Section 7.1: Gravitational Potential Energy**

---

**Page 215:**

**Example 7.8: An inclined plane with friction**

**Scenario:**

A 12-kg crate is pushed up a 2.5-meter ramp inclined at 30°. Initially, a worker estimates that pushing with a speed of 5.0 m/s to the bottom of the ramp will work, disregarding friction. Nevertheless, friction isn't negligible, causing the crate to move 1.6 meters up the ramp, stop, and then slide back down.

**Objective:**

Identify the friction force and calculate the crate's speed when it reaches the bottom.

**Solution Steps:**

1. **Identify and Set Up:**
- Analyze the friction force's effect during the crate's sliding.
- **Point 1:** Ground level with speed $ v_1 = 5.0 $ m/s.
- **Point 2:** Crate stops momentarily.
- **Point 3:** Return to the starting point.

2. **Equations and Calculations:**
- Initial Kinetic Energy: $ K_1 = \frac{1}{2} (12 \, \text{kg}) (5.0 \, \text{m/s})^2 = 150 \, \text{J} $
- Potential Energy at the top of ramp: $ U_{\text{grav, 2}} = (12 \, \text{kg})(9.8 \, \text{m/s}^2)(0.80 \, \text{m}) = 94 \, \text{J} $

3. **Friction Force Calculation:**
- Total energy decreases from 150 J to 94 J.
- Friction force = 35 N over 1.6 m.

4. **Conclusion:**
- Crate returns with 38 J of the original energy.
- $ v_3 = \sqrt{\frac{2(38 \text{ J})}{12 \text{ kg}}} = 2.5 \text{ m/s} $

**Figures and Diagrams:**

- **Diagram (7.11a):** Illustrates the crate’s path up and down the ramp, points 1, 2, and 3 labeled. The ramp is inclined at 30° with marked distance of 0

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