MY NOTESASK YOUR TEACHERA block on a horizontal frictionless surface is attached to a spring. The block is exhibiting SHM and has an amplitude of 0.94 m, Additionally, the block is moving et 29 m/s whenit's at the equilibrium position.A.) Determine how far from equilibrium the block is when its PE is two-fourths of the total energy.(-B.) Determine the angular frequency.@=

Name: MY NOTESASK YOUR TEACHERA block on a horizontal frictionless surface
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Description: MY NOTESASK YOUR TEACHERA block on a horizontal frictionless surface is attached to a spring. The block is exhibiting SHM and has an amplitude of 0.94 m, Additionally, the block is moving et 29 m/s whenit's at the equilibrium position.A.) Determine

The amplitude of the body is given as: Δxmax=0.94 m At the equilibrium position, the velocity of the body is given as: vmax=29 m/s In a system exhibiting SHM, the total energy periodically changes from being completely kinetic to completely potential. The total energy of the system is hence:Etotal=12kΔxmax2where k is the spring constant.(a) The potential energy stored in a spring is given by the formula: PE=12kΔxmax2where k is the spring constant. It is given that at position x24, the energy of the system is: 24Etotal Equate the energies to calculate the position.12kx242=24Etotal12kx242=2412kΔxmax2x242=24Δxmax2x24=24Δxmax2 Substitute the values:x24=24Δxmax2=240.94 m2=24×0.8836=0.66 m Hence, at 0.66 m the potential energy of the block is equal to two-fourths of the total energy.(b) The kinetic energy of the block is given by the equation: KE=12mvmax2 The total energy of the system remains constant. It changes from being completely kinetic to completely potential. Hence, KE=PE Equate the energies to obtain the expression for the ratio km KE=PE12mvmax2=12kΔxmax2kΔxmax2=mvmax2km=vmax2Δxmax2The angular frequency of a spring is given by the formula: ω=km Substitute the relation into the equation to simplify the equation. ω=km=vmax2Δxmax2=vmaxΔxmax Substitute the values into the equation. ω=vmaxΔxmax=29 m/s0.94 m=30.85 rad/s Hence, the angular frequency is 30.85 rad/s.

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A block on a horizontal frictionless surface is attached to a spring. The block is exhibiting SHM and has an amplitude of 0.94 m, Additionally, the block is moving at 29 it's at the equilibrium position. A.) Determine how far from equilibrium the block is when its PE is two-fourths of the total energy (= 8.) Determine the angular frequency. @=

A block on a horizontal frictionless surface is attached to a spring. The block is exhibiting SHM and has an amplitude of 0.94 m, Additionally, the block is moving at 29 it's at the equilibrium position. A.) Determine how far from equilibrium the block is when its PE is two-fourths of the total energy (= 8.) Determine the angular frequency. @=

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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