Question

Select all that apply.

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

Answer 1

its mechanical energy = 3.92 J

its potential energy = 1.96 J

its kinetic energy = 1.96 J

Step-by-step explanation:

Answer 2

Mechanical energy of the ball will remain conserved

so here we have

`E = mgh + (1)/(2)mv^2`

`E = 0.200(9.8)(2) + 0`

`E = 3.92 J`

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

`U = mgh`

`U = (0.200kg)(9.8 m/s^2)(1 m)`

`U = 1.96 J`

Now since total mechanical energy is conserved so we will say

`KE + U = E`

`KE + 1.96 = 3.92`

`KE = 1.96 J`

so we have

its mechanical energy = 3.92 J

its potential energy = 1.96 J

its kinetic energy = 1.96 J