Answer:
part(a): The potential energy is
.
part(b): The potential energy is
.
part(c): The potential energy is
.
Step-by-step explanation:
Given:
The mass of the ball,
.
The distance from the ceiling to the center of the ball,
![d = 1.0~m](https://img.qammunity.org/2021/formulas/physics/high-school/jy5qyupw5cr2b5wub8sqrcui1caxezuoca.png)
The height of the room,
.
The potential energy of any particle of mass
situated at a height
from ground is given by
![V = mgh](https://img.qammunity.org/2021/formulas/physics/high-school/rgcaq2qshn4b6b15zryl4r9ebc8r6wayil.png)
where
is the acceleration due to gravity.
(a) The distance of the ball relative to ceiling is
. Thus, the potential energy will be
![V &=& (2~Kg)(9.8~m/s^(2))(1.0~m)\\~~~&=& 19.6~J](https://img.qammunity.org/2021/formulas/physics/high-school/1lnznznh3pewnudtsx9owgg1w65eeid4c6.png)
(b) The distance of the ball from the ground is
![h' = (3.0 - 1.0)~m = 2.0~m](https://img.qammunity.org/2021/formulas/physics/high-school/sg25cx1tqe1x0suria77qrj767qj82v0b1.png)
Thus, the potential energy will be
![V = (2~Kg)(9.8~m/s^(2))(2.0~m)\\~~~= 39.2~J](https://img.qammunity.org/2021/formulas/physics/high-school/kp5pnh1r6s3xxmty1q696unhp889p00nej.png)
(c) The distance of the ball relative to a point at the same elevation as the ball is
.
Thus, the potential energy will be
![V = (2~Kg)(9.8~m/s^(2))(0~m)\\~~~= 0](https://img.qammunity.org/2021/formulas/physics/high-school/9xavzvhs67abw865kek15nqs4hto2a5wfb.png)