1. A person throws a 1-kg stone upward at 2 m/s while standing on the edge of a cliff so that the stone can fall to the base of the cliff 40 meters below. What is the agbara kainetik of stone at 10 meters above the ground? Isare nitori walẹ g = 10 m/s2.
A mọ̀:
ibi (m) = 1 kg
Iyara akọkọ (v)o) = 2 m/s
The change in height = 40 – 10 = 30 meters
Ìyára tí ó jẹ́ nítorí òòfà ilẹ̀ (g) = 10 m/s2
A fẹ́: kinetic energy of stone at 10 meters above the ground
Ojutu:
Agbara ẹrọ akọkọ
The initial gravitational potential energy (EP) = m g h = (1)(10)(30) = 300 Joule
The initial kinetic energy (KE) = ½ m vo2 = ½ (1)(2)2 = ½ (4) = 2 Joule
awọn ni ibẹrẹ darí agbara = the initial agbara agbara gravitational + the initial kinetic energy = 300 + 2 = 302 Joule.
The final mechanical energy
The final mechanical energy = the final kinetic energy = the initial gravitational potential energy + the initial kinetic energy = 300 + 2 = 302 Joule.
2. A person throws a 1-kg object upward into the air with an initial velocity of 10 m/s. Determine (a) the gravitational potential energy at the maximum height (b) the maximum height.
Acceleration due to gravity is 10 m/s2
A mọ̀:
Ìwọ̀n (m) = 1 kg
Iyara akọkọ (v)o) = 10 m/s
Ìyára tí ó jẹ́ nítorí òòfà ilẹ̀ (g) = 10 m/s2
fe : the gravitational potential energy at the maximum height and the maximum height
Ojutu:
(a) the gravitational potential energy at the maximum height
The initial mechanical energy :
The initial mechanical energy (ME) = the initial kinetic energy (KE) = ½ m vo2 = ½ (1)(10)2 = ½ (100) = 50 Joule.
The final mechanical energy :
The final mechanical energy (ME) = the gravitational potential energy (PE)
Ìlànà ìpamọ́ agbára ẹ̀rọ:
Agbara ẹrọ akọkọ = agbara ẹrọ ikẹhin
KE = PE
50 = PE
The gravitational potential energy is 50 Joule.
(b) The maximum height
PE = m g h
50 = (1)(10) h
50 = 10 wakati
h = 50 / 10 = 5 meters
The maximum height is 5 meters above the ground.
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