Question

An open knife edge of mass $m$ is dropped from a height $h$ on a wooden floor. If the blade penetrates up to the depth $d$ into the wood, the average resistance offered by the wood to the knife edge is

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

Answer 1

The correct option is C $mg(1+\frac{h}{d})$


Initialy the knife have only potential energy i.e
$U=mgh$
The point when knife is just about to penetrate the wooden block it has only kinetic energy i.e.
$K.E.=\frac{1}{2}m{v}^2$
Let it penetrates a distance $d$ in the block then total K.E. will be zero and hence it will stop i.e $v_f=0$
Using third law of motion:
$v^2-u^2=-2a_{avg}d$
$v=0$
$a_{avg}=\frac{u^2}{2d}$ --- (1)
Free body diagram of knife


$R-mg=m{a}_{avg}$

$R=m{a}_{avg}+mg$ -- (2)

From eq.(1) and eq.(2)
$R=m[g+\frac{u^2}{2d}]$
$R=m[g+\frac{2gh}{2d}]$
$R=mg[1+\frac{h}{d}]$