A body weighing 2 kg was thrown from a height of 250 m vertically downward with an initial

| September 5, 2021 | education

A body weighing 2 kg was thrown from a height of 250 m vertically downward with an initial speed of 20 m / s to determine the average force of soil resistance if, when falling, the body plunged into the ground by 20 cm.

m = 2 kg.

g = 10 m / s ^ 2.

V0 = 20 m / s.

h = 250 m.

l = 20 cm = 0.2 m.

Fsopr -?

Let us express the square of the body’s velocity before hitting the ground V ^ 2 from the formula: h = (V ^ 2 – V0 ^ 2) / 2 * g.

V ^ 2 = V0 ^ 2 + 2 * h * g.

V ^ 2 = (20 m / s) ^ 2 + 2 * 250 m * 10 m / s ^ 2 = 5400 (m / s) ^ 2.

When hitting the ground, all the kinetic energy of the body Ek is spent on overcoming the work of the force of resistance to movement A.

Ek = A.

We find the work of the resistance force of movement A by the formula: A = Fcopr * l.

m * V ^ 2/2 = Fcopr * l.

We find the resistance force by the formula: Fcopr = m * V ^ 2/2 * l.

Fcopr = 2 kg * 5400 (m / s) ^ 2/2 * 0.2 m = 27000 N.

Answer: the average force of soil resistance was Fsopr = 27000 N.



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