A big rock falls from a height towards water below, and hits water at a speed of 12 m/s. Rock was at a height of 15 m before falling and weighs 65 kg.

a) What was the average force of air resistance (e.g., friction) acting on the rock?
b) What is the force of friction underwater if the rock reaches a depth of 2.5 m before stopping and there is a buoyant force of 650 N [upward] acting on the rock once underwater.

Question

contestada

Respuesta :

adioabiola adioabiola · Answer 1 · 2022-07-18T17:11:55+02:00

The frictional force of air resistance is; 949N while the force of friction underwater is; 1859N.

The frictional force on the action rock can be evaluated as follows;

F - W = ma

However, the acceleration of fall a can be evaluated as follows;

a = (12² -0²)/2× 15

a = 144/30 = 4.8m/s².

a) Hence, the air resistance, frictional force, F = (65×9.8) + (65 ×4.8)

F = 949N.

b) When reaches a depth of 2.5m, it follows that the acceleration is;

a = (0²-12²)/(2×2.5)

a = 144/5 = 28.8m/s².

On this note, the force balance is;

F + 650 = 65(9.8 + 28.8)

F = 2509 - 650

F = 1859N.