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A knife edge of mass M is dropped from a height h on a wooden floor with its tip pointing of downward. If the blade penetrates a distance S into the wood the average resistance offered by the wood to the blade is

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$Mg (1 + \frac{h}{s})$

$Mg (1 - \frac{h}{s})$

$Mg {(1 + \frac{h}{s})}^{2}$

answer is B.

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Detailed Solution

Velocity of knife edge is $v = \sqrt{2 gh}$
Retardation, $a = \frac{v^{2}}{2 s} = \frac{gh}{s}$
Average resistance $R - mg = ma$
$\begin{array}{l} R = mg + m (\frac{gh}{s}) \\ = mg (1 + \frac{h}{s}) \end{array}$

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