Answer :
otal energy= PEmax on the spring = KEmax of the oject
Total Energy= KE+PE at v or x not maximum.
Using the second line
E= 3/4 E + PE
PE = 1/4 E
PEmax =E = 1/2 kA^2
PE=E/4 = (1/2)kA^2 / 4= kA^2/8
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Total Energy= KE+PE at v or x not maximum.
Using the second line
E= 3/4 E + PE
PE = 1/4 E
PEmax =E = 1/2 kA^2
PE=E/4 = (1/2)kA^2 / 4= kA^2/8
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
System's potential energy when its kinetic energy is equal to 34E will be PE=1/8KA^2
What is the system's potential energy?
The potential energy is the energy possessed by the body due to its position with respect to the mean point or equilibrium point.
An object of mass m attached to a spring of force constant K oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.
[tex]P.E = \dfrac{1}{8}KA^2[/tex]
From conservation of energy, the total energy in the system is given as the sum of potential and kinetic energy.
Thus,
Total Energy; E = K.E.+P.E.
In simple harmonic motion, the total energy is given by;
[tex]E = \dfrac{1}{2}KA^2[/tex]
We are told that kinetic energy is [tex]\dfrac{3}{4}E[/tex]
Thus,
[tex]\dfrac{1}{2}KA^2 = \dfrac{3}{4}(\dfrac{1}{2}KA^2) + P.E[/tex]
[tex]P.E = \dfrac{1}{2}KA^2 - \dfrac{3}{8}KA^2[/tex]
[tex]P.E = \dfrac{1}{8}KA^2[/tex]
System's potential energy when its kinetic energy is equal to 34E will be PE=1/8KA^2
To know more about Potential energy follow
https://brainly.com/question/24933254