The load falls freely from a certain height without initial speed. Having flown 40 meters, the cargo acquired a speed of 20 m / s. On this section of the path, the ratio of the change in the kinetic energy of the load to the work of the air resistance force is?
V0 = 0 m / s.
V = 20 m / s.
g = 10 m / s2.
Δh = 40 m.
ΔEk / A -?
The kinetic energy Ek of a body of mass m, which moves with a speed V. is determined by the formula: Ek = m * V ^ 2/2.
ΔEk = Ek – Ek0 = m * V ^ 2/2 – m * V0 ^ 2/2 = m * V ^ 2/2.
When the body moves due to the presence of frictional force, part of the mechanical energy is transferred into the internal one. The change in mechanical energy ΔE is equal to the work of the friction force A: ΔE = A.
ΔЕ = ΔЕп – ΔЕк = m * g * Δh – m * V ^ 2/2 = m * (g * Δh – V ^ 2/2).
A = m * (g * Δh – V ^ 2/2).
ΔEk / A = m * V ^ 2/2 * m * (g * Δh – V ^ 2/2) = V ^ 2/2 * (g * Δh – V ^ 2/2).
ΔEk / A = (20 m / s) ^ 2/2 * (10 m / s2 * 40 m – (20 m / s) ^ 2/2) = 1.
Answer: ΔEk / A = 1.
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