An object with a mass of 3kg is dropped from 56.3 meters. How far above the ground is it's kinetic energy 1073J?
1073=mg(56.3-heightabove ground)
solve
128.7
To determine the distance above the ground where the object's kinetic energy is 1073J, we need to apply the principle of conservation of energy.
The kinetic energy of an object is given by the equation:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the object, and v is its velocity.
First, we need to find the velocity of the object when its kinetic energy is 1073J. Rearranging the equation, we have:
v^2 = (2 * KE) / m
Now we substitute the known values:
v^2 = (2 * 1073J) / 3kg
v^2 = 717.33 J/kg
Taking the square root of both sides, we find:
v = √(717.33 J/kg)
v ≈ 26.8 m/s
Now that we know the velocity, we can use the equation of motion to find the distance.
The equation of motion for a dropped object is given by:
h = (v^2) / (2 * g)
where h is the height or distance, v is the initial velocity, and g is the acceleration due to gravity.
Since the object is dropped, its initial velocity is 0 m/s. Thus, the equation simplifies to:
h = (0^2) / (2 * g)
h = 0 m
Therefore, the object is at a height of 0 meters above the ground when its kinetic energy is 1073J.