## Solution

what is the system’s potential energy when its kinetic energy is equal to 34e?

**The answer to the above question is (KA ^{2})/8**

### Explanation

#### Law of Conservation of Energy

Law of conservation of energy states that energy can neither be created nor destroyed but can be transferred from one medium to another.

Additionally, the law states that in an isolated or closed system, the total amount of energy is always constant, although energy may be changed from one form to another.

#### Data

The net energy in a system is

Net Energy (E_{net}) = Potential Energy (P.E) + Kinetic Energy (K.E)

E_{net} = P.E + K.E

We identify the net energy in a simple harmonic motion as

E_{net} = (1/2) Kx^{2}

Hence, we can insert the above equation into Net Energy (E_{net}) = Potential Energy (P.E) + Kinetic Energy (K.E) to obtain

(1/2) Kx^{2} = P.E + K.E

and from our question, the kinetic energy, K.E = (3/4)E

Let’s assume that E = Net Energy = E_{net} = (1/2) Kx^{2}

We can now rewrite the kinetic energy as

K.E = (3/4) x E = (3/4) x E_{net} = (3/4) x (1/2) Kx^{2} = (3/8) Kx^{2}

Thus, the kinetic energy is K.E = (3/8) Kx^{2}

#### Solution

we can now insert the above expression into (1/2) Kx^{2} = P.E + K.E to get

(1/2) Kx^{2} = P.E + (3/8) Kx^{2}

It is now time to make the potential energy (P.E) subject of the formula

P.E = (1/2) Kx^{2} – (3/8) Kx^{2}

The above equation can be simplified into

P.E = (1/2) Kx^{2} – (3/8) Kx^{2} = (1/2 – 3/8) Kx^{2} = ((4-3) /8 ) Kx^{2} = (1/8) Kx^{2}

**Therefore, the system’s potential energy is (1/8) Kx ^{2}**

Drop a question in the comment section if you have a challenge with the question: “what is the system’s potential energy when its kinetic energy is equal to 34e?”

*You may also like to read:*

A Small Rock with Mass 0.20 kg is Released

a photon with an energy of 1.33

The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle.

what is the object’s velocity when its potential energy is 23e?

*Sources:*