An open knife edge of mass $'m' $ is dropped from a height $'h'$ on a wooden floor. If the blade penetrates upto the depth $ 'd' $ into the wood, the average resistance offered by the wood to the knife edge is
An open knife edge of mass $'m' $ is dropped from a height $'h'$ on a wooden floor. If the blade penetrates upto the depth $ 'd' $ into the wood, the average resistance offered by the wood to the knife edge is
$mg$
$mg\left( {1 - \frac{h}{d}} \right)$
$mg\left( {1 + \frac{h}{d}} \right)$
$mg{\left( {1 + \frac{h}{d}} \right)^2}$
An open knife edge of mass $'m' $ is dropped from a height $'h'$ on a wooden floor. If the blade penetrates upto the depth $ 'd' $ into the wood, the average resistance offered by the wood to the knife edge is
Let the blade stops at depth d into the wood.
${v^2} = {u^2} + 2aS$
$⇒$ $0 = {(\sqrt {2gh} )^2} + 2(g - a)d$
by solving $a = \left( {1 + \frac{h}{d}} \right)g$
So the resistance offered by the wood $ = mg\left( {1 + \frac{h}{d}} \right)$
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