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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?

Respuesta :

Answer:

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.

adioabiola

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

According to the question;

  • We are required to determine At what position is the kinetic energy of the system equal to the potential energy.

The potential energy of the system is given mathematically as;

  • P.E = (1/2) × k × X²

The kinetic energy of the system is given mathematically as;

  • K.E = (1/2) ×m × v².

When; P.E = K.E

  • (1/2) × k × X² = (1/2) ×m × v²

  • (1/2) × 640 × X² = (1/2) × 10 × 4²

  • 320x = 80

  • x = 80/320

  • x = 0.25m.

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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