Question

a 0.223 kg ball is thrown straight up from 2.20 m above the ground. its initial vertical speed is 9.20 m/s. a short time later, it hits the ground. calculate the total work done by the force of gravity during that time.

Answer 1

the total work done by the force of gravity during that time is approximately 4.808 Joules.

Step-by-step explanation:

Work = force × distance

Weight = 0.223 kg × 9.8 m/s² = 2.186 N

Change in height = Final height - Initial height

= 2.20 m - 0 m

= 2.20 m

Work = Weight × Change in height

= 2.186 N × 2.20 m

= 4.808 N·m or 4.808 J