A block of mass M is pulled along horizontal firctionless surface by a rop of mass m. Force P is applied at one end of rope. The force which the rope exerts on the block is

Options

A. `(P)/(M-m)`
B. `(P)/(M(m +M))`
C. `("PM")/((m + M))`
D. `("PM")/(M-m)`

Explanation

To find the force that the rope exerts on the block, we can follow these steps:

### Step 1: Understand the System
We have a block of mass \( M \) and a rope of mass \( m \) being pulled by a force \( P \) on a frictionless surface. The entire system (block + rope) will accelerate together.

### Step 2: Calculate the Total Mass
The total mass of the system is the sum of the mass of the block and the mass of the rope:
\[ \text{Total Mass} = M + m \]

### Step 3: Apply Newton's Second Law
According to Newton's second law, the net force acting on the system is equal to the total mass multiplied by the acceleration \( a \):
\[ P = (M + m) \cdot a \]

### Step 4: Solve for Acceleration
Rearranging the equation to solve for acceleration \( a \):
\[ a = \frac{P}{M + m} \]

### Step 5: Analyze the Block
Now, we need to find the tension \( T \) in the rope that is exerted on the block. The only horizontal force acting on the block is the tension \( T \).

### Step 6: Write the Equation for the Block
Using Newton's second law for the block:
\[ T = M \cdot a \]
Substituting the expression for acceleration \( a \):
\[ T = M \cdot \left(\frac{P}{M + m}\right) \]

### Step 7: Final Expression for Tension
Thus, the tension \( T \) (the force which the rope exerts on the block) can be expressed as:
\[ T = \frac{P \cdot M}{M + m} \]

### Conclusion
The force which the rope exerts on the block is:
\[ \boxed{\frac{P \cdot M}{M + m}} \]