When lifting a bucket of 10 tons loaded with coal from the mine, the work was performed at 6400 kJ. How deep is the mine?

Given: m (mass of a bucket loaded with coal) = 10 t; A (work done to lift the bucket) = 6400 kJ; vertical lift of the bucket (cosα = 1).

Reference values: g (acceleration due to gravity) = 9.81 m / s2 ≈ 10 m / s2.

SI: m = 10 t = 10 * 10 ^ 3 t; A = 6400 kJ = 6400 * 10 ^ 3 J.

Solution:

The formula from which we express the depth of the mine: A = F * s * cosα = m * g * h, whence h = A / (m * g).

Calculation: h = 6400 * 10 ^ 3 / (10 * 10 ^ 3 * 10) = 640 m.

Answer: The depth of the mine is 640 meters.