The force F, acting in a constant direction on the 19-kg block, has a magnitude which varies with the position s of the block. When s= 0 the block is moving to the right at v = 6 m/s. The coefficient of kinetic friction between the block and surface is μ = 0.3. (Figure 1) Figure F (N) F F = 50s¹/2 s (m) 1 of 1 Part A Determine how far the block must slide before its velocity becomes 15 m/s. Express your answer to three significant figures and include the appropriate units. 8 = LO Submit Value μA Provide Feedback Request Answer Units ? Next >

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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### Problem Statement

The force \(\mathbf{F}\), acting in a constant direction on the 19-kg block, has a magnitude which varies with the position \(s\) of the block. When \(s = 0\), the block is moving to the right at \(v = 6 \, \text{m/s}\). The coefficient of kinetic friction between the block and the surface is \(\mu_k = 0.3\). [(Figure 1)](#).

### Graph and Diagram Description

#### Graph
- **Axes**: The graph displays force \(F(N)\) on the vertical axis and position \(s(m)\) on the horizontal axis.
- **Curve**: The curve shows how the force \(F\) varies with position \(s\) and is described by the equation \(F = 50s^{1/2}\). The curve starts from the origin (0,0) and rises, indicating that the force increases as \(s\) increases.

#### Diagram
- **Illustration**: The diagram depicts a block on a surface being acted upon by force \(\mathbf{F}\) in the horizontal direction (to the right). The initial velocity \(v\) is also directed to the right.

### Part A

Determine how far the block must slide before its velocity becomes \(15 \, \text{m/s}\).

**Instructions**: Express your answer to three significant figures and include the appropriate units.

#### Answer Submission
- **Input Fields**: There's a space provided to input the value and units for the solution.
  
#### Additional Features

- **Submit Button**: To submit your answer.
- **Request Answer**: For solution assistance.

### Navigation

- **Provide Feedback**: Option to give feedback.
- **Next**: To proceed to the next section.
Transcribed Image Text:### Problem Statement The force \(\mathbf{F}\), acting in a constant direction on the 19-kg block, has a magnitude which varies with the position \(s\) of the block. When \(s = 0\), the block is moving to the right at \(v = 6 \, \text{m/s}\). The coefficient of kinetic friction between the block and the surface is \(\mu_k = 0.3\). [(Figure 1)](#). ### Graph and Diagram Description #### Graph - **Axes**: The graph displays force \(F(N)\) on the vertical axis and position \(s(m)\) on the horizontal axis. - **Curve**: The curve shows how the force \(F\) varies with position \(s\) and is described by the equation \(F = 50s^{1/2}\). The curve starts from the origin (0,0) and rises, indicating that the force increases as \(s\) increases. #### Diagram - **Illustration**: The diagram depicts a block on a surface being acted upon by force \(\mathbf{F}\) in the horizontal direction (to the right). The initial velocity \(v\) is also directed to the right. ### Part A Determine how far the block must slide before its velocity becomes \(15 \, \text{m/s}\). **Instructions**: Express your answer to three significant figures and include the appropriate units. #### Answer Submission - **Input Fields**: There's a space provided to input the value and units for the solution. #### Additional Features - **Submit Button**: To submit your answer. - **Request Answer**: For solution assistance. ### Navigation - **Provide Feedback**: Option to give feedback. - **Next**: To proceed to the next section.
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