Question

An iron nail is dropped from a height h from the level of a sand bed. If it penetrates through a distance x in the sand before coming to rest, the average force exerted by the sand on the nail is,

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

Answer 1

The correct option is A $mg(\frac{h}{x}+1)$

The nail hits the sand with a speed $V_0$ after falling through a height h
$\Rightarrow V_0^2=2gh\Rightarrow V_0=\sqrt{2gh}$ …(1)
The nail stops after sometime say t, penetrating through a distance, x into the sand. Since its velocity decreases gradually the sand exerts a retarding upward force, R (say). The net force acting on the nail is given as
$\sum F_y=R-mg=ma$
$\Rightarrow R=m(g+a)$ …(2)

Where a = deceleration of the nail. Since the nail penetrates a distance x
$0-V_0^2=-2ax$ …(3)
Putting $V_0$ from (1) and ‘a’ from (2) in (3), we obtain
$2gh=2\left(\frac{R-mg}{m}\right)x\Rightarrow R=\frac{mg(h+x)}{x}$