Question

Select all that apply.

Answer 1

Mechanical energy of the ball will remain conserved

so here we have

$E = mgh + \frac{1}{2}mv^2$

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

$E = 3.92 J$

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