An object weighing 200 g was thrown onto a balcony located at a height of 6 m. During the flight, the object reached

An object weighing 200 g was thrown onto a balcony located at a height of 6 m. During the flight, the object reached a maximum height of 8 m from the surface of the earth. Determine the work of gravity when the object is flying up, down and all the way. Find the resulting change in potential energy.

h0 = 6 m.

m = 200 g = 0.2 kg.

g = 10 m / s2.

h = 8 m.

Av -?

An -?

A -?

ΔЕп -?

We express the work of gravity A by the formula: A = m * g * S * cosα, where m * g is the force of gravity, S is the movement of the body, ∠α is the angle between the direction of action of the force and the movement.

When the body moves upward, its movement S = h – h0, directed vertically upward, the force of gravity is always directed vertically downward. Therefore, ∠α = 180 °, cos180 ° = – 1.

Ab = – m * g * (h – h0).

Av = – 0.2 kg * 10 m / s2 * (8 m – 6 m) = – 4 J.

When the body moves downward, its movement S = h, directed vertically downward, the force of gravity is always directed vertically downward. Therefore, ∠α = 0 °, cos0 ° = 1.

An = m * g * h.

An = 0.2 kg * 10 m / s2 * 8 m = 16 J.

A = An + Av = 16 J – 4 J = 12 J.

ΔЕп = Еп – Еп0 = m * g * h * – m * g * h = m * g * (h * – h0), where h * is the height of the body at the moment of falling.

h * = 0 m.

ΔEp = – m * g * h0 = – 0.2 kg * 10 m / s2 * 6 m = -12 J.

The “-” sign indicates that the potential energy of the body is decreasing.

Answer: Av = – 4 J, An = 16 J, A = 12 J, ΔEn = -12 J.