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suhe total energy of a block-spring system is 0.2 J. The block oscillates with an angulal frequency of 20 rad/s and amplitude of 10 cm. Find (a) Find the mass of the block, (b) Find the maximum speed, (c) Find the kinetic energy when the displacement is at 6 cm, (d) Find the positions when kinetic energy equals potential energy. 0.0G m

suhe total energy of a block-spring system is 0.2 J. The block oscillates with an angulal frequency of 20 rad/s and amplitude of 10 cm. Find (a) Find the mass of the block, (b) Find the maximum speed, (c) Find the kinetic energy when the displacement is at 6 cm, (d) Find the positions when kinetic energy equals potential energy. 0.0G m

College Physics
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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SThe total energy of a block-spring system is 0.2 J. The block oscillates with an angulal frequency of 20 rad/s and amplitude of 10 cm. Find (a) Find the mass of the block, (b) Find the maximum speed, (c) Find the kinetic energy when the displacement is at 6 cm, (d) Find the positions when kinetic energy equals potential energy. 0.0G M
Transcribed Image Text:SThe total energy of a block-spring system is 0.2 J. The block oscillates with an angulal frequency of 20 rad/s and amplitude of 10 cm. Find (a) Find the mass of the block, (b) Find the maximum speed, (c) Find the kinetic energy when the displacement is at 6 cm, (d) Find the positions when kinetic energy equals potential energy. 0.0G M
Expert Solution
Step 1

Given E=0.2 J

           A= 10 cm = 0.1 m

           ω= 20 rad/s

 

a)  ω = (k/m)0.5

           E= kinetic energy + potential energy

              when the displacement of the body is maximum kinetic enrgy is zero 

         Therfore potential enrgy = 0.2 J when x=A

                      P.E= 0.5kx2

                           0.2= 0.5*k*0.1*0.1

                                k= 0.2/0.5*0.01

                            k =40 N/m

                       m= k/ω2

                                 = 40/20*20

                           m= 0.1 Kg

b) For max speed x=0

            E= 0+K.E

                  K.E= 0.5mv2

                       0.2=0.5*0.1*v2

                      v2= 4

                   V(max)= 2m/sec

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