An open knife 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 .

Options

A. `Mg ((h)/(d))`
B. `Mg (1 + (h)/(d))`
C. `Mgh + Mgd`
D. `Mg (1 - (h)/(d))`

Answer

B

Explanation

If the body strick the sand floor with a velocity `V` then `mgh = (1)/(2) mV^(2)`
With this velocity `V` , when body passes through the sand floor it comes to rest after teraveling a distance `x`. Let `F` be the resisting force acting on the body. Net force in downwards direction`= Mg - F`
We know that work done by all the forces is equal to change in `KE: (Mg - F)x = 0 - (1)/(2) MV^(2)`
or `(MG - F) x = - Mgh or Fx = - Mgh + Mgx`
or `F = Mg (1 + (h)/(x))`