### Problem Description for Educational Website**Title:** Time at Which a Block Comes to Rest Under a Time-Varying Force**Text:** The initially stationary 23-kg block is subjected to the time-varying force whose magnitude \( P \) is shown in the plot. Note that the force is zero for all times greater than 5 seconds. Determine the time \( t_s \) at which the block comes to rest.**Diagram Explanation:**1. **Block and Force Diagram:** - The left-side diagram shows a block with a mass of 23 kg. - A force \( P \) is applied at an angle of 25 degrees to the horizontal. - The coefficients of kinetic friction \( \mu_k \) and static friction \( \mu_s \) between the block and the surface are 0.45 and 0.59, respectively.2. **Force vs. Time Graph:** - The graph on the right plots the magnitude of the force \( P \) in Newtons (N) against time \( t \) in seconds (s). - The force starts at 0 N at \( t = 0 \) and increases linearly to 152 N at \( t = 5 \) seconds. - After \( t = 5 \) seconds, the force drops abruptly to zero and remains at zero thereafter.**Question:** - Calculate the time \( t_s \) at which the block comes to rest.**Answer:**- A placeholder is provided for the user to input the calculated time \( t_s \).\[ \text{Answer: } t_s = \_\_\_\_ \text{ s} \]

Answered: Image /qna-images/answer/239c1313-e2e4-470b-a6ba-71e859614bdc.jpg

The initially stationary 23-kg block is subjected to the time-varying force whose magnitude P is shown in the plot. Note that the force is zero for all times greater than 5 s. Determine the time ts at which the block comes to rest. Hk = 0.45 Hg = 0.59 Answe: ts: = S i 23 kg 25° P, N 152 in 0 0 5 t, s

The initially stationary 23-kg block is subjected to the time-varying force whose magnitude P is shown in the plot. Note that the force is zero for all times greater than 5 s. Determine the time ts at which the block comes to rest. Hk = 0.45 Hg = 0.59 Answe: ts: = S i 23 kg 25° P, N 152 in 0 0 5 t, s

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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Forming and Shaping

Metal forming is a category of metalworking process through which the material can be given the desired shape and size by subjecting it to plastic deformation, where the component undergoes bulk deformation process upon application of loads. Unlike the shearing and machining process, the material does not undergo up to a failure point as indicated in the stress-strain curve. Hence, the stresses induced during the forming process are greater than yield strength but less than the failure strength of the material. The process does not lead to the loss of material, so it can be said to be an economical process. During forming, the material is subjected to tensile forces, compressive forces, bending, and shearing loading conditions. The material should possess the property of ductility to undergo the forming process.

Question

### Problem Description for Educational Website

**Title:** Time at Which a Block Comes to Rest Under a Time-Varying Force

**Text:**

The initially stationary 23-kg block is subjected to the time-varying force whose magnitude \( P \) is shown in the plot. Note that the force is zero for all times greater than 5 seconds. Determine the time \( t_s \) at which the block comes to rest.

**Diagram Explanation:**

1. **Block and Force Diagram:**
- The left-side diagram shows a block with a mass of 23 kg.
- A force \( P \) is applied at an angle of 25 degrees to the horizontal.
- The coefficients of kinetic friction \( \mu_k \) and static friction \( \mu_s \) between the block and the surface are 0.45 and 0.59, respectively.

2. **Force vs. Time Graph:**
- The graph on the right plots the magnitude of the force \( P \) in Newtons (N) against time \( t \) in seconds (s).
- The force starts at 0 N at \( t = 0 \) and increases linearly to 152 N at \( t = 5 \) seconds.
- After \( t = 5 \) seconds, the force drops abruptly to zero and remains at zero thereafter.

**Question:**
- Calculate the time \( t_s \) at which the block comes to rest.

**Answer:**
- A placeholder is provided for the user to input the calculated time \( t_s \).

\[ \text{Answer: } t_s = \_\_\_\_ \text{ s} \]

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