In the figure, a 3.9 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 \(v_0 = 6.0 \, \text{m/s}\), the height difference is \(h = 1.1 \, \text{m}\), and \(\mu_k = 0.560\). Find \(d\).**Diagram Description:**The diagram shows a block on the left side with an initial velocity \(v_0\) indicated by an arrow. The block moves along a frictionless curved track that rises up to a higher horizontal level. At the higher level, a frictional force (\(\mu_k\)) acts over a distance \(d\) until the block comes to a stop. The height difference between the initial and higher level is \(h\).**Input Fields:**- Number: [Input Box]- Units: [Dropdown Box]

mass of block (m) = 3.9 kg initial velocity (vo) = 6 m/s Height difference (h) = 1.1 m coefficient of kinetic friction (μk) = 0.56 Using work energy theorem :Net work done = change in kinetic energy work done by gravitational force + work done by friction = final kinetic energy - initial kinetic energy (-m×g×h) + (-μk×m×g×d) = 0 - (12×m×vo2)(-9.8×1.1) + (-0.56×9.8×d) = - (6)2210.78 + 5.488d = 185.488 d = 7.22d = 7.225.488d = 1.315 m Thus, distance moved by block on rough surface before coming to rest = 1.315 m

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In the figure, a 3.9 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 = 6.0 m/s, the height difference is h = 1.1 m, and Hk = 0.560. Find d. µ = 0- Number i Units

In the figure, a 3.9 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 = 6.0 m/s, the height difference is h = 1.1 m, and Hk = 0.560. Find d. µ = 0- Number i Units

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter6: Energy Of A System

Section: Chapter Questions

Problem 60P: Why is the following situation impossible? In a new casino, a supersized pinball machine is...

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Question

In the figure, a 3.9 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 \(v_0 = 6.0 \, \text{m/s}\), the height difference is \(h = 1.1 \, \text{m}\), and \(\mu_k = 0.560\). Find \(d\).

**Diagram Description:**

The diagram shows a block on the left side with an initial velocity \(v_0\) indicated by an arrow. The block moves along a frictionless curved track that rises up to a higher horizontal level. At the higher level, a frictional force (\(\mu_k\)) acts over a distance \(d\) until the block comes to a stop. The height difference between the initial and higher level is \(h\).

**Input Fields:**

- Number: [Input Box]
- Units: [Dropdown Box]

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