20. A coconut of mass \( 2 Kg \) is falling from a coconut free of height \( 10 m \). Find the potential energy, kinetic energy and total energy at the following points. \( \left( g =10 m / s ^{2}\right) \) a) Before falling from the coconut tree. b) Just before hilting the ground. c) At a height \( 4 m \) from the ground. d) Which law is evident from this example?
20. A coconut of mass \( 2 Kg \) is falling from a coconut free of height \( 10 m \). Find the potential energy, kinetic energy and total energy at the following points. \( \left( g =10 m / s ^{2}\right) \) a) Before falling from the coconut tree. b) Just before hilting the ground. c) At a height \( 4 m \) from the ground. d) Which law is evident from this example?
1 Answers
(a) Given m = 2 kg
h = 10 m
Potential energy at before falling from the coconut tree.
u = mgh
u = 2 x 10 x 10
u = 200 joule
Kinetic energy will be zero.
Total energy = Potential energy + Kinetic energy
= 200 + 0
Total energy = 200 joule.
(b) Just before hitting the ground.
Potential energy u = 0
Kinetic energy Ek = 1/2 mv2
v2 = u2 + 2gh
v2 = 0 + 2gh (∵ u = 0)
v = \(\sqrt{2gh}\)
Ek = 1/2 x 2 x 2gh
Ek = 2 x 10 x 10
Ek = 200 joule
Total energy = 200 joule
(c) At a height 4 m from the ground.
Potential energy u = mg (hi - hf)
u = mg (10 - 6)
u = mg x 6
u = 2 x 10 x 6
u = 120 joule
Kinetic energy Ek = 1/2 mv2
∵ v = \(\sqrt{2gh}\)
v2 = 2gh
= 2 x 10 x 4
v2 = 80
Ek = 1/2 x 2 x 80
Ek = 80 joule
Total energy = Kinetic energy + Potential energy
= 120 + 80
Total energy = 200 joule
(d) The example follow conservation of energy.