As shown in the figure below, two masses $ m_1 = 4.30 \, \text{kg} $ and $ m_2 $, which has a mass 60.0% that of $ m_1 $, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. If $ m_1 $ and $ m_2 $ start from rest, after they have each traveled a distance $ h = 1.70 \, \text{m} $, use energy content to determine the following:(a) speed $ v $ of the masses _______ m/s(b) magnitude of the tension $ T $ in the cord _______ N**Diagram Description:**The diagram shows a pulley system with a cord running over a pulley. Mass $ m_1 $ is shown on the left side, hanging lower, while mass $ m_2 $ is on the right. An arrow indicates the direction of motion downward for $ m_1 $ and upward for $ m_2 $, labeled with the distance $ h $.

Answered: (a) speed v of the masses m/s (b)…

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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As shown in the figure below, two masses $ m_1 = 4.30 \, \text{kg} $ and $ m_2 $, which has a mass 60.0% that of $ m_1 $, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. If $ m_1 $ and $ m_2 $ start from rest, after they have each traveled a distance $ h = 1.70 \, \text{m} $, use energy content to determine the following:

(a) speed $ v $ of the masses
_______ m/s

(b) magnitude of the tension $ T $ in the cord
_______ N

**Diagram Description:**

The diagram shows a pulley system with a cord running over a pulley. Mass $ m_1 $ is shown on the left side, hanging lower, while mass $ m_2 $ is on the right. An arrow indicates the direction of motion downward for $ m_1 $ and upward for $ m_2 $, labeled with the distance $ h $.

Expert Solution

Step 1

(a)

Net work done is equal to change in kinetic energy

$W_{n e t} = ∆ E_{k} W_{1} + W_{2} = ∆ E_{k} m_{1} g h + m_{2} g (- h) = \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} v^{2} (m_{1} a n d m_{2} moves distanace h in opposite direction) m_{1} g h - 0.6 m_{1} g h = \frac{1}{2} m_{1} v^{2} + \frac{1}{2} (0.6 m_{1}) v^{2} 0.4 m_{1} g h = = 1.6 [\frac{1}{2} m_{1} v^{2}] v^{2} = \frac{g h}{2} v = \sqrt{\frac{g h}{2}} = \sqrt{\frac{(9.8 \frac{m}{s^{2}}) (1.70 m)}{2}} = 2.886 \frac{m}{s}$

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