A 2.0-kg rock is dropped from a height of 21.0 m. At what height is the rock's kinetic energy twice its potential energy?
To solve this problem, we need to first determine the kinetic energy and potential energy of the rock at any given height.
The potential energy (PE) of an object at a certain height is given by the equation: PE = m * g * h,
where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
In this case, the mass of the rock is 2.0 kg and the height is 21.0 m. So, the potential energy can be calculated as:
PE = 2.0 kg * 9.8 m/s^2 * 21.0 m
Next, we need to find the kinetic energy (KE) of the rock. The kinetic energy of an object is given by the equation: KE = 0.5 * m * v^2,
where m is the mass of the object and v is its velocity.
When the rock is dropped from a height, it will accelerate due to gravity and gain velocity. Let's assume the rock reaches its maximum velocity (v_max) at some height, at which its kinetic energy is twice its potential energy.
So, when KE = 2 * PE, we can equate the equations for KE and PE:
0.5 * m * v_max^2 = 2 * (m * g * h)
We can now solve this equation to find the value of h, which represents the height at which the rock's kinetic energy is twice its potential energy.
you want 1/2 mv^2 = 2mgh
so plug in your numbers and solve for h
note that the mass does not matter.