Answer:
the height at which the rock's kinetic energy is twice its potential energy is approximately 4.5 m.
Step-by-step explanation:
When the rock is dropped from a height of 9.0 m, it initially has potential energy equal to mgh, where m is the mass of the rock, g is the acceleration due to gravity (9.8 m/s^2), and h is the height from which the rock was dropped. Therefore, the potential energy of the rock is:
U = mgh = (6.0 kg) * (9.8 m/s^2) * (9.0 m) = 529.2 J
As the rock falls, its potential energy is converted to kinetic energy, given by the expression (1/2)mv^2, where v is the velocity of the rock. At a height where the kinetic energy of the rock is twice its potential energy, we can write:
(1/2)mv^2 = 2mgh
Simplifying this expression, we get
v^2 = 4gh
At this height, the kinetic energy of the rock is given by:
K = (1/2)mv^2 = (1/2)m(4gh) = 2mgh = 2U
Substituting the values of m, g, and U, we get:
v^2 = 4gh = 4(9.8 m/s^2)h = (2 * 529.2 J) / 6.0 kg = 176 J/kg
Solving for h, we get:
h = (v^2) / (4g) = (176 J/kg) / (4 * 9.8 m/s^2) ≈ 4.5 m