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A body of mass $2 k g$ is dropped from a height of $10 m$ on to the ground. If the body penetrates $4 c m$ in to the ground, the average resistance offered by the ground on to the body is $(g = 10 m s^{- 2})$

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$5020 N$

$50.2 N$

$2500 N$

$2510 N$

answer is B.

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Detailed Solution

$F = M g (1 + \frac{h}{s}) = 2 \times 10 (1 + \frac{10}{4 \times 10^{- 2}}) = 20 \times 251 = 5020 N$

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