A 250 kg rock falls off a cliff and comes to a rest on the ground, which is 40m below the cliff. At what point is the rock’s kinetic energy at maximum?
just before it lands.
m g h + (1/2) m v^2 = constant
when h is minimum, v^2 is maximum
It is at its maximum kinetic energy when it is falling until it hits the ground which then it would be considered the maximum potential energy.
To determine at what point the rock's kinetic energy is at its maximum, we need to understand the concept of potential energy and the relationship between potential and kinetic energy.
Initially, the rock at the top of the cliff possesses gravitational potential energy due to its position above the ground. As it falls, this potential energy is converted into kinetic energy. At some point, the rock will reach its maximum kinetic energy before eventually coming to a rest on the ground.
To find this point, we can equate the initial gravitational potential energy to the final kinetic energy. The formula for gravitational potential energy is:
Potential Energy = mass × gravity × height
Given that the mass of the rock is 250 kg and the height is 40 m, we can calculate the initial potential energy:
Initial Potential Energy = 250 kg × 9.8 m/s^2 × 40 m
Next, we equate this value to the final kinetic energy. The formula for kinetic energy is:
Kinetic Energy = (1/2) × mass × velocity^2
At its maximum kinetic energy, the rock will have zero potential energy left. Therefore, we have:
Initial Potential Energy = Final Kinetic Energy
250 kg × 9.8 m/s^2 × 40 m = (1/2) × 250 kg × velocity^2
Simplifying this equation:
9.8 m/s^2 × 40 m = (1/2) × velocity^2
392 m^2/s^2 = (1/2) × velocity^2
Diving both sides by (1/2):
784 m^2/s^2 = velocity^2
Taking the square root of both sides:
velocity ≈ √784
velocity ≈ 28 m/s
Hence, the rock's kinetic energy is at its maximum when its velocity is approximately 28 m/s during its descent.