A block of mass m is pulled by a force of constant power P placed on a rough horizontal plane. The friction coefficient between the block and the surface is `mu`. Then

Options

A. The maximum velocity of the block during the motion is `(P)/(mumg)`
B. The maximum velocity of the block during the motion is `(P)/(2mumg)`
C. The block's speed is never decreasing and finally becomes constant
D. The speed of the block first increases to a maximum value and then decreases

Answer

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

### Step-by-Step Solution:

1. **Understanding Power**:
The power \( P \) exerted on the block can be expressed in terms of the force \( F \) acting on it and its velocity \( v \):
\[ P = F \cdot v \]

2. **Expressing Force in Terms of Power and Velocity**:
Rearranging the formula for power, we can express the force as:
\[ F = \frac{P}{v} \]
This equation shows that the force \( F \) is inversely proportional to the velocity \( v \).

3. **Frictional Force**:
The frictional force \( F_f \) opposing the motion of the block is given by:
\[ F_f = \mu \cdot m \cdot g \]
where \( \mu \) is the coefficient of friction, \( m \) is the mass of the block, and \( g \) is the acceleration due to gravity.

4. **Condition for Maximum Velocity**:
At maximum velocity \( V_{max} \), the net force acting on the block is zero. This means that the pulling force \( F \) equals the frictional force \( F_f \):
\[ F = F_f \]
Therefore, we can write:
\[ \frac{P}{V_{max}} = \mu \cdot m \cdot g \]

5. **Solving for Maximum Velocity**:
Rearranging the equation to solve for \( V_{max} \):
\[ V_{max} = \frac{P}{\mu \cdot m \cdot g} \]

### Final Expression:
Thus, the maximum velocity \( V_{max} \) of the block is given by:
\[ V_{max} = \frac{P}{\mu \cdot m \cdot g} \]

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