Question

A 0.25 kg ball bearing is released from rest at the surface of the water in the tank of water.

After falling 2 m in the water, its speed is 5 m/s.

What is the amount of work done by the water on the ball bearing?

Answer 1

amount of work done = -1.77 J

Step-by-step explanation:

given data

mass = 0.25 kg

speed = 5 m/s

height = 2m

to find out

amount of work done by the water on the ball bearing

solution

we get here potential energy when ball fall

potential energy = mgh

potential energy = 0.25 × 9.8 × 2

potential energy = 4.9 J

and that lose potential energy gain by kinetic energy that is express as

kinetic energy = 0.5 × m × v²

kinetic energy = 0.5 × 0.25 × 5²

kinetic energy = 3.13 J

so we get now work done that is

work done = kinetic energy - potential energy

work done = 3.13 J - 4.9 J

amount of work done = -1.77 J