First slide

Work Energy Theorem

Question

An open knife edge of mass M is dropped from a height I on a wooden floor. If the blade penetrates S into the wood, the average resistance offered by the wood to
the blade is

Moderate

A

Mg
B

$Mg (1 + \frac{h}{S})$
C

$Mg (1 - \frac{h}{S})$
D

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

Solution

Let F be the resistive force of wooden plank. It should oppose the weight of the body.
Resultant force opposite to the motion = F - M g,
Work done = (F - M g)S = Kinetic energy
$Velocity of fall = \sqrt{(2 gh)}$
$∴ (F - Mg) S = \frac{1}{2} M (\sqrt{2 hg})^{2}$
$or F = \frac{Mgh}{S} + Mg = M g (1 + \frac{h}{S})$ .

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