A stone is thrown from the surface of the Earth vertically upwards at a speed of V = 12 km / h. Determine at what height its kinetic energy will decrease by 2 times. Disregard movement resistance. The coefficient g should be taken equal to 10 N / kg.
According to the law of conservation of energy:
Eк1 + Eп1 = Eк2 + Eп2, where Eк1 and Eп1 – energy at the beginning of movement; Eк2, Eп2 – energy at the moment of decreasing kinetic energy by 2 times.
Since at the beginning of the throw h = 0, then Ep1 = 0. Then
Eк1 = Eк2 + Ep2, where Eк2 = 0.5Eк1.
Eк1 = 0.5Eк1 + Ep2, Ep2 = 0.5Eк1, where Ep2 = m * g * h (g = 10 m / s ^ 2), Eк1 = (m * v ^ 2) / 2 (v = 12 km / h = 3.33 m / s).
m * g * h = 0.5 * (m * v ^ 2) / 2;
h = (v ^ 2) / 4g = (3.33 ^ 2) (/ 4 * 10) = 11/40 = 0.28 m.
Answer. The kinetic energy will decrease by 2 times by 0.28 m.
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