**Question:** A group of particles of total mass 42 kg has a total kinetic energy of 363 J. The kinetic energy relative to the center of mass is 87 J. What is the speed of the center of mass?**Answer Box:** [Input field] m/s**Explanation:**The problem involves calculating the speed of the center of mass of a system of particles given the total kinetic energy and the kinetic energy relative to the center of mass. The given values are:- Total mass: 42 kg- Total kinetic energy: 363 J- Kinetic energy relative to the center of mass: 87 JThe kinetic energy of the center of mass (KE_cm) is found by subtracting the given relative kinetic energy from the total kinetic energy:\[ \text{KE}_{\text{cm}} = 363 \, \text{J} - 87 \, \text{J} = 276 \, \text{J} \]The speed of the center of mass \( v_{\text{cm}} \) can be found using the formula for kinetic energy:\[ \text{KE}_{\text{cm}} = \frac{1}{2} m v_{\text{cm}}^2 \]Where:- \( \text{KE}_{\text{cm}} \) is the kinetic energy of the center of mass.- \( m \) is the total mass (42 kg).- \( v_{\text{cm}} \) is the speed of the center of mass.Solving for \( v_{\text{cm}} \):\[ 276 = \frac{1}{2} \times 42 \times v_{\text{cm}}^2 \]\[ v_{\text{cm}}^2 = \frac{276 \times 2}{42} \]\[ v_{\text{cm}}^2 = \frac{552}{42} \]\[ v_{\text{cm}} \approx \sqrt{13.14} \]\[ v_{\text{cm}} \approx 3.62 \, \text{m/s} \]The speed of the center of mass is approximately 3.62 m/s.

Name: **Question:** A group of particles of total mass 42 kg has a total ki
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Description: **Question:** A group of particles of total mass 42 kg has a total kinetic energy of 363 J. The kinetic energy relative to the center of mass is 87 J. What is the speed of the center of mass?**Answer Box:** [Input field] m/s**Explanation:**The probl

Thus the total kinetic energy (Ttotal) of a group of particles is equal to the sum of the kinetic energy the center of mass (Tcm) plus the the kinetic energy individual particles relative to the center of mass (T). Ttotal = Tcm + T or, Ttotal = (1/2)MV2 + T M is the mass of the group of particles, V is velocity of the center of mass. Mass of the group of particles, M = 42 kg Total kinetic energy of the group of particles, Ttotal = 363 J Kinetic energy relative to center of mass, T = 87 J. Therefore, Ttotal = (1/2)MV2 + T, gives, 363 = (1/2)(42)V2 + 87 or, (1/2)(42)V2 = 363 - 87 or, 21V2 = 276 or, V2 = 276/21 or, V = 3.62 m/s So, the velocity of the center of mass is 3.62 m/s.

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A group of particles of total mass 42 kg has a total kinetic energy of 363 J. The kinetic energy relative to the center of mass is 87 J. What is the speed of the center of mass? m/s

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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