In the figure, a 3.6 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is vo = 5.9 m/s, the height difference is h = 1. m, and µK = 0, 625. Find d. R0.625 l = 0- 5.9 1.1
In the figure, a 3.6 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is vo = 5.9 m/s, the height difference is h = 1. m, and µK = 0, 625. Find d. R0.625 l = 0- 5.9 1.1
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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
Transcribed Image Text:**Problem Description:**
In the figure, a 3.6 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There, a frictional force stops the block in a distance \( d \).
Given:
- The block's initial speed is \( v_0 = 5.9 \, \text{m/s} \).
- The height difference is \( h = 1.1 \, \text{m} \).
- The coefficient of kinetic friction is \( \mu_k = 0.625 \).
Find the stopping distance \( d \).
**Diagram Explanation:**
1. **Track Level:**
- The track starts at one level and transitions smoothly through a valley before rising to a higher level where friction is present.
2. **Block Movement:**
- The block begins with an initial speed \( v_0 \) and travels along the frictionless part of the track.
- After ascending the height \( h \), the block encounters a frictional section where it stops due to kinetic friction.
3. **Key Points:**
- \( \mu = 0 \) indicates the frictionless portion of the track.
- \( \mu_k = 0.625 \) represents the coefficient of kinetic friction on the higher level.
- Distance \( d \) is the stopping distance on the higher level.
- Height \( h \) is marked between the initial and final levels of the track.
This scenario provides a practical application of energy conservation and friction principles, allowing for calculations of energy lost to friction and motion equations to determine the stopping distance.
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