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Physics · WORK, ENERGY AND POWER
हिन्दीA running athlete of mass \( 70\text{ kg} \) runs with a speed of \( 10\text{ m s}^{-1} \) having kinetic energy \( K_1 \). A stone of mass \( 1\text{ kg} \) dropped from a height of \( 10\text{ m} \) reaches the ground with kinetic energy \( K_2 \approx 10^2\text{ J} \). The ratio \( \frac{K_1}{K_2} \) is equal to:
Correct answer · Option 3
\( 35 \)
Explanation
Kinetic energy of athlete \( K_1 = \frac{1}{2}(70)(10)^2 = 3500\text{ J} = 3.5 \times 10^3\text{ J} \). Kinetic energy of stone \( K_2 = 10^2\text{ J} = 100\text{ J} \). Thus, \( K_1 / K_2 = 3500 / 100 = 35 \).
