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17 days ago

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.

Physics

2 answers:

ValentinkaMS [3.4K]17 days ago

5 0

Response:

The change in potential energy can be expressed as $\Delta P=0.98\ J$

Explanation:

It is stated that:

The ball's mass, m = 0.5 kg

Initially it is raised to a height of, h = 2 m

After that, it bounces to a height of 1.8 meters, which gives us the final height, h' = 1.8 m

We can calculate the change in potential energy for the whole process. Let $P_i$ represent the initial potential energy, which is calculated as:

$P_i=mgh$

Now, let $P_f$ encompass the final potential energy, given by:

$P_f=mgh'$

Let $\Delta P$ denote the change in potential energy, determined by:

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

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

$\Delta P=-0.98\ J$

Thus, the change in gravitational potential energy equals 0.98 J. Therefore, this is the answer sought.

kicyunya [3.2K]17 days ago

3 0

Hello! Thanks for sharing your query here.

To determine the change in potential energy, you would utilize the formula:

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

The kinetic energy is derived from the potential energy.

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