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

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Chapter1: Units, Trigonometry. And Vectors
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**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.
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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