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(1+(h)/(d))`
B. `mg(1+(h)/(d))^(2)`
C. `mg(1-(h)/(d))`
D. `mg(1+(d)/(h))`

Explanation

To solve the problem of finding the average resistance offered by the wood to the knife edge, we can follow these steps:

### Step 1: Understand the scenario
An open knife of mass \( m \) is dropped from a height \( h \) onto a wooden floor, and it penetrates a depth \( d \) into the wood. We need to find the average resistance force offered by the wood.

### Step 2: Calculate the potential energy at the height
When the knife is at height \( h \), it has potential energy given by:
\[ PE = mgh \]
where \( g \) is the acceleration due to gravity.

### Step 3: Determine the total distance fallen
When the knife hits the wooden floor and penetrates into the wood, the total distance fallen is the sum of the height \( h \) and the depth \( d \):
\[ h_{\text{total}} = h + d \]

### Step 4: Calculate the work done during penetration
The work done by the knife while penetrating into the wood is equal to the force of resistance \( F \) multiplied by the distance \( d \):
\[ \text{Work done} = F \cdot d \]

### Step 5: Set the potential energy equal to the work done
At the moment of impact, the potential energy of the knife is converted into work done against the resistance of the wood. Therefore, we can equate the potential energy to the work done:
\[ mgh = F \cdot d \]

### Step 6: Solve for the average resistance force \( F \)
Rearranging the equation to solve for \( F \):
\[ F = \frac{mgh}{d} \]

### Step 7: Consider the total distance for average resistance
Since the knife penetrates a distance \( d \), we can express the average resistance force as:
\[ F = \frac{mg(h + d)}{d} \]
This can be simplified to:
\[ F = mg \left(1 + \frac{h}{d}\right) \]

### Conclusion
Thus, the average resistance offered by the wood to the knife edge is:
\[ F = mg \left(1 + \frac{h}{d}\right) \]