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)
I know that you said it is F * X and I'm just confused. Can you help me walk through this?
Work is defined as force * distance. The force here is weight, or mg. Work done by gravity is then mgh.
Now, that work results in kinetic energy, 1/2 mv^2
1/2 m v^2=mgh
1/2 v^2=gh
v= sqrt 2gh
Certainly! To calculate the final velocity of the object using the work-energy principle, we need to consider the initial potential energy and the final kinetic energy of the object.
Let's start by calculating the initial potential energy (PE) of the object. The potential energy is given by the formula PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
Given:
Mass (m) = 20 kg
Height (h) = 30 m
Acceleration due to gravity (g) = 9.8 m/s²
So, the initial potential energy (PE) of the object is:
PE = m * g * h
= 20 kg * 9.8 m/s² * 30 m
= 5880 J
Next, we need to calculate the final kinetic energy (KE) of the object at the surface. The kinetic energy is given by the formula KE = 1/2 * m * v², where m is the mass of the object and v is the final velocity.
Since the object is falling freely, all of its potential energy is converted into kinetic energy at the surface. So, we can equate the initial potential energy (PE) to the final kinetic energy (KE).
PE = KE
5880 J = 1/2 * 20 kg * v²
Now, let's solve for the final velocity (v).
Dividing both sides of the equation by 1/2 * 20 kg yields:
5880 J = 10 kg * v²
Dividing both sides of the equation by 10 kg gives:
588 J/kg = v²
Taking the square root of both sides of the equation gives:
v = √(588 J/kg)
v ≈ 24.25 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, is approximately 24.25 m/s.