The initially stationary 24-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 to at which the block comes to rest. H₂=0.43 Hs = 0.51 Answe: ts= i 24 kg 4.181 P 26° P. N 173 0 5 t, s
The initially stationary 24-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 to at which the block comes to rest. H₂=0.43 Hs = 0.51 Answe: ts= i 24 kg 4.181 P 26° P. N 173 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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![**Problem Statement:**
The initially stationary 24-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 \(t_s\) at which the block comes to rest.
**Given Values:**
\[
\mu_k = 0.43 \quad \mu_s = 0.51 \quad 26^\circ\quad 24\, \text{kg}
\]
**Diagrams:**
1. **Diagram of Block with Forces:**
- A block weighing 24 kg is placed on a surface inclined at an angle of 26 degrees.
- The coefficient of kinetic friction (\(\mu_k\)) is 0.43.
- The coefficient of static friction (\(\mu_s\)) is 0.51.
- There is a force \(P\) applied upwards along the ramp.
2. **Plot of Force \(P\) Against Time \(t\):**
- A graph is shown with Force \(P\) (in N) on the vertical axis and Time \(t\) (in s) on the horizontal axis.
- From \(t = 0\) to \(t = 5\) seconds, the force \(P\) increases linearly from 0 N to 173 N.
- For times greater than 5 seconds, the force \(P\) is constant at 0 N.
**Answer:**
\[
t_s = \boxed{4.181} \, \text{s}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b7dd8f1-dbab-4607-8b33-bde9974f18c9%2Fda51d93a-56b1-4c44-8797-a2e3d6f4cfcf%2F9t1ql6f_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
The initially stationary 24-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 \(t_s\) at which the block comes to rest.
**Given Values:**
\[
\mu_k = 0.43 \quad \mu_s = 0.51 \quad 26^\circ\quad 24\, \text{kg}
\]
**Diagrams:**
1. **Diagram of Block with Forces:**
- A block weighing 24 kg is placed on a surface inclined at an angle of 26 degrees.
- The coefficient of kinetic friction (\(\mu_k\)) is 0.43.
- The coefficient of static friction (\(\mu_s\)) is 0.51.
- There is a force \(P\) applied upwards along the ramp.
2. **Plot of Force \(P\) Against Time \(t\):**
- A graph is shown with Force \(P\) (in N) on the vertical axis and Time \(t\) (in s) on the horizontal axis.
- From \(t = 0\) to \(t = 5\) seconds, the force \(P\) increases linearly from 0 N to 173 N.
- For times greater than 5 seconds, the force \(P\) is constant at 0 N.
**Answer:**
\[
t_s = \boxed{4.181} \, \text{s}
\]
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