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A 0.200-kilogram ball sits on a shelf 2.00 meters from the floor and then falls to a point 1.00 meter from the floor. Ignoring any affects due to friction, at this position _____.

its mechanical energy is 3.92 J
its potential energy is 3.92 J
its kinetic energy is 3.92 J
its potential energy is 1.96 J
its kinetic energy is 1.96 J
its mechanical energy is 1.96 J

Respuesta :

Mechanical energy of the ball will remain conserved

so here we have

[tex]E = mgh + \frac{1}{2}mv^2[/tex]

[tex]E = 0.200(9.8)(2) + 0[/tex]

[tex]E = 3.92 J[/tex]

so mechanical energy will remain same at all positions

Now when ball comes to position of 1 m height then potential energy is given as

[tex]U = mgh[/tex]

[tex]U = (0.200kg)(9.8 m/s^2)(1 m)[/tex]

[tex]U = 1.96 J[/tex]

Now since total mechanical energy is conserved so we will say

[tex]KE + U = E[/tex]

[tex]KE + 1.96 = 3.92 [/tex]

[tex]KE = 1.96 J[/tex]

so we have

its mechanical energy = 3.92 J

its potential energy = 1.96 J

its kinetic energy = 1.96 J

Answer:

its mechanical energy = 3.92 J

its potential energy = 1.96 J

its kinetic energy = 1.96 J

Explanation:

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