The 41-kg crate is projected down an incline as shown with an initial speed of 5.2 m/s. Investigate the time t required for the crate to come to rest and the corresponding distance x traveled if 0 = 11.4°. What is the largest angle for which the block will stop? %=5.2 m/s H₂ = 0.27 41 kg Answers 0 = 11.4°, t= IN i 8 S, X = HI
The 41-kg crate is projected down an incline as shown with an initial speed of 5.2 m/s. Investigate the time t required for the crate to come to rest and the corresponding distance x traveled if 0 = 11.4°. What is the largest angle for which the block will stop? %=5.2 m/s H₂ = 0.27 41 kg Answers 0 = 11.4°, t= IN i 8 S, X = HI
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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Transcribed Image Text:### Problem Description
The 41-kg crate is projected down an incline as shown with an initial speed of 5.2 m/s. Investigate the time \( t \) required for the crate to come to rest and the corresponding distance \( x \) traveled if \( \theta = 11.4^\circ \). What is the largest angle \( \theta \) for which the block will stop?
### Diagram Explanation
The diagram illustrates a crate with the following parameters:
- Mass (\( m \)) = 41 kg
- Initial velocity (\( v_0 \)) = 5.2 m/s
- Angle of incline (\( \theta \)) = 11.4°
- Coefficient of friction (\( \mu \)) = 0.27
The crate is shown moving down an incline making an angle \( \theta \) with the horizontal. The direction of motion is indicated with an arrow.
### Answers
Given \( \theta = 11.4^\circ \):
- \( t = \square \)
- \( s, x = \square \)
Largest angle for which the block will stop:
- \( \theta_{\text{max}} = \square^\circ \)
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