Question

An open knife edge of mass m is droped from a height h on a wooden floor. If the blade penetrates a distance 'd' into the wood, then the average resistance offered by wood, to the blade is?

• A. $mg\left(1-\frac{h}{d}\right)$
• B. mg
• C. $mg{\left(1+\frac{h}{d}\right)}^2$
• D. $mg\left(1+\frac{h}{d}\right)$

Answer 1

Answer: (D)

$mg\left(1+\frac{h}{d}\right)$

The resistive force $F_R$ exerted by wooden floor on the blade of the knife opposes the weight of the knife. So, the work done in penetrating a distance d is equal to the change in kinetic energy
$\therefore W=(F_R=mg)d=K_f-K_i$
$\Rightarrow (F_R-mg)d=\frac{1}{2}m{v}^2-0$
$\Rightarrow (F_R-mg)d=\frac{1}{2}m\left(2gh\right)$
$\Rightarrow F_R-mg=\frac{mgh}{d}$
$\Rightarrow F_R=mg(1+\frac{h}{d})$.