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

Answer 1

Answer: (A)

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

Answer 2

The nail hits the sand with a speed V_o after falling through a height h

⇒ $V_{0}^{2}\mspace{6mu} = \mspace{6mu} 2gh\mspace{6mu} \Rightarrow \mspace{6mu} V_{0}\mspace{6mu} = \mspace{6mu}\sqrt{2gh}$ …(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

ΣF_y = R – mg = ma

⇒ R = m(g + a) …(2)

Where a = deceleration of the nail. Since the nail penetrates a distance x

0 − V²₀ = − 2a x …(3)

Putting V₀ from (1) and ‘a’ from (2) in (3), we obtain

$2gh\mspace{6mu} = \mspace{6mu} 2\left( \frac{R - mg}{m} \right)\mspace{6mu} x\mspace{6mu} \Rightarrow \mspace{6mu} R = \frac{mg(h + x)}{x}$

⇒ $R = \mspace{6mu} mg\left( \frac{h}{x} + 1 \right)\mspace{6mu}$, Hence (1) is the correct choice.