A knife of mass m is at a height x from a large wooden block. The knife is allowed to fall freely, strikes the block and comes to rest after penetrating - distance y. The work done y the wooden block to stop the knife is

Options

A. `mgx`
B. `-mgy`
C. `-mg(x+y)`
D. `mg(x-y)`

Answer

To solve the problem, we need to calculate the work done by the wooden block to stop the knife after it penetrates a distance \( y \). Here’s a step-by-step solution:

### Step 1: Determine the initial potential energy of the knife
The knife is dropped from a height \( x \). The potential energy (PE) of the knife at this height can be calculated using the formula:
\[ PE = mgh \]
where \( h = x \) and \( g \) is the acceleration due to gravity. Therefore, the initial potential energy of the knife is:
\[ PE = mgx \]

### Step 2: Calculate the velocity of the knife just before impact
When the knife falls freely from height \( x \), it converts its potential energy into kinetic energy (KE) just before it strikes the block. The kinetic energy can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
At the point of impact, all the potential energy has been converted into kinetic energy:
\[ mgx = \frac{1}{2} mv^2 \]
From this equation, we can solve for \( v \):
\[ v^2 = 2gx \quad \Rightarrow \quad v = \sqrt{2gx} \]

### Step 3: Apply the work-energy principle
When the knife penetrates the block and comes to rest after penetrating a distance \( y \), the work done by the block to stop the knife is equal to the change in kinetic energy of the knife. The initial kinetic energy just before impact is:
\[ KE_{initial} = \frac{1}{2} mv^2 = \frac{1}{2} m(2gx) = mgx \]
The final kinetic energy when the knife comes to rest is:
\[ KE_{final} = 0 \]
Thus, the work done by the block \( W \) is:
\[ W = KE_{final} - KE_{initial} = 0 - mgx = -mgx \]

### Step 4: Work done by the block
The work done by the block to stop the knife is equal to the force exerted by the block times the distance \( y \) (in the opposite direction of motion). Therefore, we can express the work done by the block as:
\[ W = -F_{block} \cdot y \]
where \( F_{block} \) is the force exerted by the block to stop the knife.

### Step 5: Relate the work done to the penetration distance
The force exerted by the block can be approximated as the weight of the knife:
\[ F_{block} = mg \]
Thus, we can write:
\[ -mgy = -mgx \quad \Rightarrow \quad mg(y) = mg(x) \]
This implies that:
\[ mg(y) = mg(x - y) \]
Rearranging gives:
\[ W = mg(x - y) \]

### Final Answer
The work done by the wooden block to stop the knife is:
\[ W = mg(x - y) \]

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