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.

The energy of the ball has been reduced by (0.98, 8.82, 9.8) joules after bouncing back. This reduction happens because potential energy transforms to (kinetic, light, thermal energy).

Hint: Use PE = m × g × h where g = 9.8 m/s2.

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Muscardinus Muscardinus · Answer 1 · 2019-11-26T02:45:13-05:00

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.

onyebuchinnaji onyebuchinnaji · Answer 2 · 2022-04-15T07:20:22-04:00

The reduction happens because potential energy transforms to kinetic energy.

Based on the principle of conservation of energy, the total mechanical energy of the ball will be constant at any point in time.

The change in energy of the ball is calculated as follows;

ΔP.E = -ΔK.E

mg(h₂ - h₁) = -ΔK.E

0.5 x 9.8(1.8 - 2) = -ΔK.E

-0.98 J = -ΔK.E

ΔK.E = 0.98 J

Thus, the reduction happens because potential energy transforms to kinetic energy.

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