You have a spring with k = 640 n/m connected to a mass with m = 10 kg. you start the system oscillating and measure the velocity of the mass as it moves through the spring's equilibrium position to be 4 m/s. at what position is the kinetic energy of the system equal to the potential energy?

The position at which the kinetic energy of the system equal to the potential energy is at; x = 0.25m

According to the question;

The potential energy of the system is given mathematically as;

The kinetic energy of the system is given mathematically as;

When; P.E = K.E

Therefore, The position at which the kinetic energy of the system equal to the potential energy is at; x = 0.25m.

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X= 0.5 m

Explanation:

Given that

K= 640 N/m

m=10 kg

v= 4 m/s

We know that

kinetic energy of the mass

KE=1/2 m v²

Potential energy

PE=1/2 k X²

Given that kinetic and potential energy are equal

1/2 k X² = 1/2 m v²

k X² =  m v²

Now by putting the values

640  X² = 10 x 4²

X= 0.5 m

So at 0.5 m  kinetic and potential energy will be equal.

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr