A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.
Question
Question
A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.
Answer:
The work done by gravity is [tex]4.975 \: Joules[/tex]
Explanation:
The data given in the question is :
Mass is [tex]0.245 kg[/tex]
Height from ground is [tex]2.07 m[/tex]
As we know , the work done is state function , it depends on initial and final position not on the path followed.
So, work done by gravity = change in potential energy
Work done = Initial potential energy – final potential energy
Insert values from question
Work done = [tex]mass \times gravity \times (change \: in \: height)[/tex]
Work done = [tex]0.245 kg \times 9.81 m/s^{2} \times 2.07 m[/tex]
So, work done = [tex]4.975 Joules[/tex]
Hence the work done by gravity is [tex]4.975 \: Joules[/tex]