What is the system's potential energy when its kinetic energy is equal to 34e?
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
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System's potential energy when its kinetic energy is equal to 34E will be PE=1/8KA^2
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.
\(P.E = \dfrac{1}{8}KA^2\)
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;
\(E = \dfrac{1}{2}KA^2\)
We are told that kinetic energy is \(\dfrac{3}{4}E\)
Thus,
\(\dfrac{1}{2}KA^2 = \dfrac{3}{4}(\dfrac{1}{2}KA^2) + P.E\)
\(P.E = \dfrac{1}{2}KA^2 - \dfrac{3}{8}KA^2\)
\(P.E = \dfrac{1}{8}KA^2\)
System's potential energy when its kinetic energy is equal to 34E will be PE=1/8KA^2
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