contestada

A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Nm-1
. The mass is displaced 5.2 cm from its equilibrium position and then released to undergo simple
harmonic motion.
At what displacement from the equilibrium position is the potential energy equal to the kinetic energy.

Respuesta :

The conservation of energy allows to find the result for the point where the kinetic and potential energy are equal is;

       x = 0.026 m

Given parameters

  • The mass of the body m = 220 g = 0.220 kg
  • The force constant k = 7.0 N / m
  • The initial displacement or amplitude xo = 5.2 cm = 0.052 m

To find

  • The point where scientific and potential energy are equal.

 

The law of the conservation of mechanical energy is one of the most important in physics, stable that if there is no friction, the mechanical energy of the system is conserved. The mechanical energy is formed by the sum of the kinetic energy and the potential energies.

               Em = K + U

Let's write the energy in two points.

Starting point. With maximum compression.

        Em₀ = U = ½ k x²

Final point. Where the kinetic and potential energy are equal.

        [tex]Em_f = K +U[/tex]  

Since the mechanical energy is constant at this point K = U, therefore we can write the energy.

        [tex]Em_f = 2U = 2 ( \frac{1}{2} \ k \ x_f^2 )[/tex]

 

Energy is conserved.

        [tex]Em_o = Em_f \\\frac{1}{2} \ k x_o^2 = 2 ( \frac{1}{2} \ k x_f^2)[/tex]Emo = Emf

        ½ k x² = 2 (½ k xf²)

        [tex]x_f = \frac{x_o}{2}[/tex]  

       

let's calculate.

        [tex]x_f = \frac{0.052}{2} \\x_f = 0.026 m[/tex]  

In conclusion using the conservation of energy we can find the point where the kinetic and potential energy are equal is;

       x = 0.026 m

Learn more here:  brainly.com/question/14688403