Hello, can you help me with this question? **Problem P14:**A group of particles of total mass \(35 \, \text{kg}\) has a total kinetic energy of \(340 \, \text{J}\). The kinetic energy relative to the center of mass is \(85 \, \text{J}\). What is the speed of the center of mass?**Explanation:**To find the speed of the center of mass, use the formula for kinetic energy \((KE)\):\[ KE = \frac{1}{2}mv^2 \]* Let the total kinetic energy of the system be \( KE_{\text{total}} = 340 \, \text{J} \).* The kinetic energy relative to the center of mass is \( KE_{\text{relative}} = 85 \, \text{J} \).The kinetic energy of the center of mass \((KE_{\text{CM}})\) is:\[ KE_{\text{CM}} = KE_{\text{total}} - KE_{\text{relative}} \]\[ KE_{\text{CM}} = 340 \, \text{J} - 85 \, \text{J} = 255 \, \text{J} \]Using the kinetic energy formula:\[ 255 = \frac{1}{2}(35)v^2 \]Solve for \( v \) (speed of the center of mass):\[ 255 = 17.5v^2 \]\[ v^2 = \frac{255}{17.5} \]\[ v^2 = 14.57 \]\[ v = \sqrt{14.57} \]\[ v \approx 3.82 \, \text{m/s} \]Thus, the speed of the center of mass is approximately \(3.82 \, \text{m/s}\).

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Answered: • P14 A group of particles of total…

• P14 A group of particles of total mass 35 kg has a total kinetic energy of 340 J. The kinetic energy relative to the center of mass is 85 J. What is the speed of the center of mass?

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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Question

Hello, can you help me with this question?

**Problem P14:**

A group of particles of total mass \(35 \, \text{kg}\) has a total kinetic energy of \(340 \, \text{J}\). The kinetic energy relative to the center of mass is \(85 \, \text{J}\). What is the speed of the center of mass?

**Explanation:**

To find the speed of the center of mass, use the formula for kinetic energy \((KE)\):
\[ KE = \frac{1}{2}mv^2 \]

* Let the total kinetic energy of the system be \( KE_{\text{total}} = 340 \, \text{J} \).
* The kinetic energy relative to the center of mass is \( KE_{\text{relative}} = 85 \, \text{J} \).

The kinetic energy of the center of mass \((KE_{\text{CM}})\) is:
\[ KE_{\text{CM}} = KE_{\text{total}} - KE_{\text{relative}} \]
\[ KE_{\text{CM}} = 340 \, \text{J} - 85 \, \text{J} = 255 \, \text{J} \]

Using the kinetic energy formula:
\[ 255 = \frac{1}{2}(35)v^2 \]

Solve for \( v \) (speed of the center of mass):
\[ 255 = 17.5v^2 \]
\[ v^2 = \frac{255}{17.5} \]
\[ v^2 = 14.57 \]
\[ v = \sqrt{14.57} \]
\[ v \approx 3.82 \, \text{m/s} \]

Thus, the speed of the center of mass is approximately \(3.82 \, \text{m/s}\).

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