Question

The deepest mine ever drilled has a depth of 12300 meters. suppose you drop a rock with a mass of .12 kilograms down the shaft of this mine. What would the rock's kinetic energy be after falling 3200 meters? What would the potential energy associated with the rock be at that same moment? Assume no air resistance and a constant free-fall acceleration. (please show the equations used and why you used these equations.) Thank you.​

Answer 1

Correct answer: Ek = 384,000 J = 0.384 · 10⁶ J = 0.384 MJ

Ep = 1,092,000 J = 1.092 · 10⁶ J = 1.092 MJ

Et = 1,476,000 J = 1.476 · 10⁶ J = 1.476 MJ

Step-by-step explanation:

The total energy at all times is equal to the sum of kinetic and potential energy.

Et = Ep + Ek

We will take the bottom of the mine as a reference point

We can solve this problem combined with the help of kinetics and energy conservation laws, or just with energy conservation laws, which I will do.

Before we drop a rock the potential energy is maximum and the kinetic energy is equals to zero.

The formula for calculating potential energy is:

Ep = m g h , we will take that g = 10 m/s²

This is the maximum potential energy (or total energy)

Epmax = Et = 12 · 10 · 12,300 = 1,476,000 J = 1.476 · 10⁶ J = 1.476 MJ

After a fall of 3,200 m the rock will be on height of 9,100 m and its potential energy will be:

Ep = m g h₁ = 12 · 10 · 9,100 = 1,092,000 J = 1.092 · 10⁶ J = 1.092 MJ

The difference between the maximum potential energy (or total energy) and the potential energy at height h₁ is actually kinetic energy.

Ek = ΔEp = Et - Ep = 1,476,000 J - 1,092,000 J

Ek = 384,000 J = 0.384 · 10⁶ J = 0.384 MJ

God is with you!!!