# Metal falls into soft sand from a height of 2 m. ball weighing 10 kg. As a result of the fall

**Metal falls into soft sand from a height of 2 m. ball weighing 10 kg. As a result of the fall, the ball went 50 cm deep into the sand. A) What is the change in the total mechanical energy of the ball? b) what is the work of the sand resistance force equal to? c) what is the average resistance force of the sand?**

h = 2 m.

g = 9.8 m / s2.

m = 10 kg.

S = 50cm = 0.5m.

ΔЕ -?

A – ?

Fsopr -?

The change in the total mechanical energy ΔЕ is found by the formula: ΔЕ = m * g * (h + S).

ΔE = 10 kg * 9.8 m / s2 * (2 m + 0.5 m) = 245 J.

The work of the resistance force A is equal to the change in the total mechanical energy of the body, taken with a minus sign: A = – ΔE.

A = – 245 J.

The work of force is expressed by the formula: A = Fcopr * S * cosα, where Fcopr is the resistance force, S is the displacement of the body under the action of the force, ∠α is the angle between Fcopr and S.

∠α = 180 °, cos180 ° = – 1.

Fcopr = A / * S * cosα.

Fcopr = – 245 J / – 0.5 m = 490 N.

Answer: ΔE = 245 J, A = – 245 J, Fcopr = 490 N.