An iron nail of mass m is dropped from a height h from level of a sand bed. If it penetrates through a distance x in the sand before coming to rest. Calculate the average force exerted by the sand on the nail. (Take g=`10m//s^(2)`)

Options

A. `mg(h/x+1)`
B. `mg(x/h+1)`
C. `mg(h/x-1)`
D. `mg(x/h-1)`

Answer

To solve the problem, we will use the work-energy theorem, which states that the work done by all forces acting on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Nail**:
- The gravitational force acting downwards: \( F_g = mg \)
- The drag force exerted by the sand acting upwards: \( F_d \)

2. **Determine the Work Done by Gravity**:
- The nail falls a total distance of \( h + x \) (where \( h \) is the height from which it was dropped and \( x \) is the distance it penetrates into the sand).
- The work done by gravity (which acts downwards) is given by:
\[ W_g = F_g \cdot d = mg(h + x) \]

3. **Determine the Work Done by the Drag Force**:
- The drag force acts upwards while the nail penetrates a distance \( x \).
- The work done by the drag force is:
\[ W_d = -F_d \cdot x \]
- The negative sign indicates that the drag force is acting in the opposite direction to the displacement.

4. **Apply the Work-Energy Theorem**:
- According to the work-energy theorem:
\[ W_g + W_d = \Delta KE \]
- Since the nail starts from rest and comes to rest, the change in kinetic energy (\( \Delta KE \)) is zero:
\[ mg(h + x) - F_d \cdot x = 0 \]

5. **Solve for the Drag Force \( F_d \)**:
- Rearranging the equation gives:
\[ F_d \cdot x = mg(h + x) \]
- Thus, we can express the drag force as:
\[ F_d = \frac{mg(h + x)}{x} \]

6. **Simplify the Expression**:
- This can be simplified to:
\[ F_d = mg \left(1 + \frac{h}{x}\right) \]

### Final Answer:
The average force exerted by the sand on the nail is:
\[ F_d = mg \left(1 + \frac{h}{x}\right) \]