Q. 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 edge is
A) $mg$
B) $mg\left( 1 - \frac{h}{d} \right)$
C) $mg\left( 1 + \frac{h}{d} \right)$
D) $mg\left( 1 + \frac{h}{d} \right)^{2}$
Explanation
(c) Let the blade stops at depth $d$ into the wood
$v^{2} = u^{2} + 2aS$
$\Rightarrow 0 = \left( \sqrt{2gh} \right)^{2} + 2(g - a)d$
By solving $a = \left( 1 + \frac{h}{d} \right)g$
So the resistance offered by the wood $= mg\left( 1 + \frac{h}{d} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.