A 17-kg rock is on the edge of a 103-m cliff.
(a) What potential energy does the rock possess relative to the base of the cliff?
. J
(b) The rock falls from the cliff. What is its kinetic energy just before it strikes the ground?
J
we don't have textbooks, and my teacher dosent teach, he just gives us really hard homework:(
a) mass*gravity*height= gravitational potential energy.
17*103*9.8= x Joules
b) as the object falls the potential energy is converted to kinetic. so just before it hits the ground all of the potential energy has become kinetic.
A 20 kg rock is at the edge of a cliff 100 m above the ground. What is the potential energy of the rock?
To solve these problems, we need to apply the principles of potential energy and kinetic energy.
(a) To find the potential energy of the rock relative to the base of the cliff, we can use the formula:
Potential Energy = Mass x Gravitational Acceleration x Height
Given:
Mass = 17 kg
Height = 103 m
Gravitational acceleration on Earth = 9.8 m/s^2 (approximately)
Substituting the given values into the formula, we can calculate the potential energy:
Potential Energy = 17 kg x 9.8 m/s^2 x 103 m
Potential Energy ≈ 17 x 9.8 x 103 ≈ 16,898 J
Therefore, the potential energy of the rock relative to the base of the cliff is approximately 16,898 J.
(b) To find the kinetic energy of the rock just before it strikes the ground, we need to consider the conservation of energy. The potential energy at the top of the cliff is converted entirely into kinetic energy as the rock falls.
Kinetic Energy = Potential Energy
The potential energy was already calculated to be 16,898 J.
Therefore, the kinetic energy just before the rock strikes the ground is also approximately 16,898 J.
Yeah! Right!
(a) Surely your text material incudes the formula for gravitational potential energy. Use it.
(b) The kinetic energy increase at the bottom equals the potential energy calculated in part (a)