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]

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 energyWork done = Initial potential energy – final potential energyInsert 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]