A block of mass m is pulled by a constant powert `P` placed on a rough horizontal plane. The friction coefficient the block and surface is `mu`. The maximum velocity of the block is.

Options

A. `(muP)/(mg)`
B. `(mumg)/(P)`
C. `mumgP`
D. `(P)/(mumg)`

Answer

To find the maximum velocity of a block of mass \( m \) being pulled by a constant power \( P \) on a rough horizontal plane with a friction coefficient \( \mu \), we can follow these steps:

### Step 1: Understand the relationship between power, force, and velocity
The power \( P \) is defined as the product of force \( F \) and velocity \( v \):
\[ P = F \cdot v \]
From this, we can express velocity as:
\[ v = \frac{P}{F} \]

### Step 2: Identify the forces acting on the block
When the block is pulled on a rough surface, the forces acting on it are:
- The external force \( F_{\text{external}} \) pulling the block.
- The frictional force \( F_{\text{friction}} \) opposing the motion.

The frictional force can be expressed as:
\[ F_{\text{friction}} = \mu \cdot N \]
where \( N \) is the normal force. On a horizontal surface, the normal force \( N \) is equal to the weight of the block:
\[ N = mg \]
Thus, the frictional force becomes:
\[ F_{\text{friction}} = \mu mg \]

### Step 3: Set up the condition for maximum velocity
At maximum velocity, the net force acting on the block is zero, which means the external force must equal the frictional force:
\[ F_{\text{external}} = F_{\text{friction}} = \mu mg \]

### Step 4: Substitute the expression for external force into the power equation
Now we can substitute \( F_{\text{external}} \) into the power equation:
\[ P = F_{\text{external}} \cdot v \]
Substituting \( F_{\text{external}} = \mu mg \):
\[ P = (\mu mg) \cdot v \]

### Step 5: Solve for maximum velocity
Rearranging the equation to solve for \( v \):
\[ v = \frac{P}{\mu mg} \]

### Conclusion
Thus, the maximum velocity of the block is given by:
\[ v_{\text{max}} = \frac{P}{\mu mg} \]