Question

A ball with a mass of 0.5 kilograms is lifted to a height of 2.0 meters and dropped. It bounces back to a height of 1.8 meters.

Answer 1

Answer:

Change in potential energy, $\Delta P=0.98\ J$

Explanation:

It is given that,

Mass of the ball, m = 0.5 kg

It is lifted to a height of, Initial height, h = 2 m

It bounces back to a height of 1.8 meters, final height, h' = 1.8 m

It can be assumed to find the change in potential energy in the entire process. Let $P_i$ is the initial potential energy. It is given by :

$P_i=mgh$

Let $P_f$ is the final potential energy. It is given by :

$P_f=mgh'$

Let $\Delta P$ is the change in potential energy. It is equal to :

$\Delta P=mg(h'-h)$

$\Delta P=0.5\ kg\times 9.8\ m/s^2\times (1.8-2)\ m$

$\Delta P=-0.98\ J$

So, the change in gravitational potential energy is 0.98 J. Hence, this is the required solution.

Answer 2

Hi, thank you for posting your question here at Brainly.

To compute for the change in potential energy, the equation would be:

delta PE =  mg*delta h
delta PE = 0.5*9.81*(2-1.8)
delta Pe = 0.98 J

The potential energy is converted to kinetic energy.