A 2.2 kg rock initially at rest loses 412 J of potential energy while falling to the ground. Assume that air resistance is negligible.
Calculate the kinetic energy that the rock gains while falling.
What is the rock's speed just before it strikes the ground?
To calculate the kinetic energy gained by the rock while falling, we can use the principle of conservation of energy. The potential energy lost by the rock is equal to the kinetic energy gained.
Given:
Mass of the rock (m) = 2.2 kg
Potential energy lost (ΔPE) = 412 J
The formula for potential energy is given by:
PE = m * g * h
Where:
m is the mass of the object (in kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
h is the height from which the object falls (in meters)
Since the rock starts at rest, its initial kinetic energy is zero (KEinitial = 0).
Therefore, the kinetic energy gained by the rock (KEfinal) can be calculated as:
KEfinal = ΔPE
Substituting the given values:
KEfinal = 412 J
To calculate the speed of the rock just before it strikes the ground, we can use the formula for kinetic energy:
KEfinal = (1/2) * m * v^2
Where:
v is the velocity or speed of the object (in m/s)
Rearranging the formula, we can solve for v:
v^2 = (2 * KEfinal) / m
v = sqrt((2 * KEfinal) / m)
Substituting the given values:
v = sqrt((2 * 412) / 2.2)
Calculating the value:
v ≈ sqrt(376) ≈ 19.39 m/s
Therefore, the rock's speed just before it strikes the ground is approximately 19.39 m/s.
To calculate the kinetic energy gained by the rock while falling, we need to use the principle of conservation of energy. The potential energy lost by the rock is converted into kinetic energy.
The formula for potential energy is given by:
Potential energy = mass * acceleration due to gravity * height
Given that the potential energy lost by the rock is 412 J, and the mass of the rock is 2.2 kg, we can rearrange the equation to solve for the height:
Height = Potential energy / (mass * acceleration due to gravity)
The acceleration due to gravity is approximately 9.8 m/s^2. Plugging in the values, we get:
Height = 412 J / (2.2 kg * 9.8 m/s^2)
Height ≈ 19.3 meters
Now, we can determine the final kinetic energy of the rock just before it strikes the ground using the equation:
Kinetic energy = 1/2 * mass * velocity^2
Since the rock was initially at rest, we can write the equation as:
Kinetic energy = 1/2 * mass * velocity_final^2
To solve for the velocity_final, we can rearrange the equation:
velocity_final^2 = (2 * Kinetic energy) / mass
Plugging in the mass of the rock (2.2 kg) and the height calculated earlier, we can solve for the final velocity:
velocity_final^2 = (2 * 412 J) / 2.2 kg
velocity_final^2 ≈ 376.36 m^2/s^2
Taking the square root of both sides, we find:
velocity_final ≈ √376.36 m^2/s^2
velocity_final ≈ 19.4 m/s
Therefore, the rock's speed just before it strikes the ground is approximately 19.4 m/s.
KE=ΔPE=412 J
mv²/2=mgh
v=sqrt(2gh)