Question

A 25-g ball is released from rest 80 m above the surface of the Earth. Just before it hits the surface its speed is 20 m/s. What is the change in energy of the ball during its fall?

Answer 1

14.6 joule

Step-by-step explanation:

mass of ball, m = 25 g = 0.025 kg

height, h = 80 m

speed at the time as the ball hits the ground, v = 20 m/s

Potential energy at the top, U = m g h

U = 0.025 x 9.8 x 80 = 19.6 Joule

Kinetic energy at the bottom, K = 0.5 x mv²

K = 0.5 x 0.025 x 20 x 20 = 5 Joule

Change in energy

E = Potential energy - Kinetic energy

E = 19.6 - 5 = 14.6 Joule

Answer 2

The change in energy of the ball during its fall is 14.6 Joules.

Step-by-step explanation:

Given that,

Mass of the ball, m = 25 g

It is released from rest 80 meters above the surface of the Earth, h = 80 m

Just before it hits the surface its speed is 20 m/s, v = 20 m/s

We need to find the change in energy of the ball during its fall. The potential energy gets converted to its kinetic energy. The change in energy is given by :

`\Delta E=mgh-(1)/(2)mv^2\\\\\Delta E=25* 10^(-3)* 9.8* 80-(1)/(2)* 25* 10^(-3)* (20)^2\\\\\Delta E=14.6\ J`

So, the change in energy of the ball during its fall is 14.6 Joules.