If a 20 kg object fell to the surgace of the Earth from a height of 30m, what would its final velocity be? (work from the energy standpoint)
the PE is mgh
the KE is 1/2 m v^2
set equal, and solve for v.
Could you explain. I'm stuck. I got 30 v^2 so far but I don't think that is right.
To determine the final velocity of the object, we can use the principle of conservation of energy.
The potential energy of an object at a height h is given by the equation: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
The potential energy is converted into kinetic energy as the object falls, where kinetic energy is given by the equation: KE = (1/2)mv², where v is the final velocity of the object.
According to the conservation of energy, the initial potential energy of the object is equal to the final kinetic energy. Therefore, we can set up an equation: mgh = (1/2)mv².
Now, let's plug in the given values into the equation:
m = 20 kg
g = 9.8 m/s²
h = 30 m
20 kg * 9.8 m/s² * 30 m = (1/2) * 20 kg * v²
Simplifying the equation:
5880 joules = 10 kg * v²
Now, rearrange the equation to solve for v:
v² = 5880 joules / 10 kg
v² = 588 m²/s²
Finally, take the square root of both sides to find the final velocity v:
v = √(588 m²/s²)
v ≈ 24.2 m/s
Therefore, the final velocity of the 20 kg object, when it falls to the surface of the Earth from a height of 30 m, will be approximately 24.2 m/s.