A body weighing 1 kg, thrown from a tower 7 m high at a speed of 8 m / s, fell on the Earth’s surface at a speed of 14 m / s, calculate the change in the total mechanical energy of the body at the moment of its fall on the Earth’s surface.
To find the losses of the total mechanical energy of a thrown body, we use the equality: Ep0 + Ek0 = Ek1 + A, whence we express: A = Ep0 + Ek0 – Ek1 = m * g * hb + m * Vn ^ 2/2 – m * Vk ^ 2 / 2 = m * (g * hb + Vn ^ 2/2 – Vk ^ 2/2).
Variables and constants: m – body weight (m = 1 kg); g – acceleration due to gravity (g ≈ 10 m / s2); hb – tower height (hb = 7 m); Vн – initial speed (Vн = 8 m / s); Vк – falling speed (Vк = 14 m / s).
Calculation: A = m * (g * hb + Vn ^ 2/2 – Vk ^ 2/2) = 1 * (10 * 7 + 8 ^ 2/2 – 14 ^ 2/2) = 4 J.
Answer: At the moment of the fall, the total mechanical energy decreased by 4 J.
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