**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:**\[t_s = \boxed{4.181} \, \text{s}\]

To find: The time at which the block comes to the rest. Given: The mass of the block is m=24 kg. The…

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

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Question

**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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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, at which the block comes to rest. P, N 24 kg 26° T Mk=0.43 Hg = 0.51 Answe: ts = i 173 S 5 t, s