Please answer this ....
And explain the last part pls
Answers
Answer:
(I) When object is located at 4m above the ground
PE = mgh = 20 x 10 x 4
= 800J
KE = mv²/2 = 20 x 0² /2
= 0J
TE = PE + KE = 800J (Total Energy = PE + KE )
(II) When object is located at 3m above the ground
PE = mgh = 20 x 10 x 3
= 600J
KE = mv²/2 = 20 x 20/2
= 200J
TE = PE + KE = 800J
(III) When object is located at 2m above the ground
PE = mgh = 20 x 10 x 2
= 400J
KE = mv²/2 = 20 x 40/2
= 400J
TE = PE + KE = 800J
(IV) When object is located at 1m above the ground
PE = mgh = 20 x 10 x 1
= 200J
KE = mv²/2 = 20 x 60/2
= 600J
TE = PE + KE = 800J
(V) When object is located just above the ground
We will assume that the height is ≈ 0
PE = mgh = 20 x 10 x 0
= 0J
KE = mv²/2 = 20 x 80/2
= 800J
TE = PE + KE = 800J
To take out the velocity, use the formula v² - u² = 2as. Take u = 0m/s
Abbreviation Glossary
TE - Total Energy
PE - Potential Energy
KE = Kinetic Energy