An object of mass (m) is located on the horizontal plane at the origin O. The body acquires horizontal velocity V. The mean power developed by the frictional force during the whole time of motion is ( `mu=` frictional coefficient )

Options

A. `mumgV`
B. `1/2mumgV`
C. `mumgV/g`
D. `3/2mumgV`

Answer

To solve the problem step by step, we will analyze the motion of the object and the forces acting on it.

### Step 1: Identify the forces acting on the object
The object of mass \( m \) is on a horizontal plane and is subjected to a frictional force as it moves. The frictional force \( F_f \) can be expressed as:
\[ F_f = \mu \cdot N \]
where \( \mu \) is the coefficient of friction and \( N \) is the normal force. Since the object is on a horizontal plane, the normal force \( N \) is equal to the weight of the object:
\[ N = mg \]
Thus, the frictional force becomes:
\[ F_f = \mu mg \]

### Step 2: Determine the acceleration of the object
According to Newton's second law, the frictional force will cause a deceleration (negative acceleration) on the object:
\[ F_f = ma \]
Substituting the expression for the frictional force:
\[ \mu mg = ma \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ a = \mu g \]

### Step 3: Calculate the time taken to reach velocity \( V \)
The object starts from rest (initial velocity \( u = 0 \)) and reaches a velocity \( V \). Using the equation of motion:
\[ V = u + at \]
Substituting \( u = 0 \) and \( a = \mu g \):
\[ V = 0 + \mu g t \]
Rearranging gives us:
\[ t = \frac{V}{\mu g} \]

### Step 4: Calculate the work done by the frictional force
The work done \( W \) by the frictional force is equal to the change in kinetic energy of the object. The initial kinetic energy \( KE_i \) is:
\[ KE_i = 0 \quad (\text{since the object starts from rest}) \]
The final kinetic energy \( KE_f \) when the object reaches velocity \( V \) is:
\[ KE_f = \frac{1}{2} m V^2 \]
Thus, the work done by the frictional force is:
\[ W = KE_f - KE_i = \frac{1}{2} m V^2 - 0 = \frac{1}{2} m V^2 \]

### Step 5: Calculate the mean power developed by the frictional force
Mean power \( P \) is defined as the work done per unit time:
\[ P = \frac{W}{t} \]
Substituting the expressions for \( W \) and \( t \):
\[ P = \frac{\frac{1}{2} m V^2}{\frac{V}{\mu g}} = \frac{1}{2} m V^2 \cdot \frac{\mu g}{V} \]
Simplifying gives:
\[ P = \frac{1}{2} \mu m g V \]

### Final Answer
Thus, the mean power developed by the frictional force during the whole time of motion is:
\[ P = \frac{1}{2} \mu m g V \]