A particle moves move on the rough horizontal ground with some initial velocity `V_(0)`. If `(3)/(4)` of its kinetic enegry lost due to friction in time `t_(0)`. The coefficient of friction between the particle and the ground is.

Options

A. `(V_(0))/(2g t_(0))`
B. `(V_(0))/(4g t_(0))`
C. `(3V_(0))/(4g t_(0))`
D. `(V_(0))/(g t_(0))`

Explanation

To solve the problem step by step, we need to analyze the given information about the particle's motion and the energy lost due to friction.

### Step 1: Understand the Initial Conditions
The particle has an initial velocity \( V_0 \) and it loses \( \frac{3}{4} \) of its kinetic energy due to friction in time \( t_0 \).

### Step 2: Calculate Initial Kinetic Energy
The initial kinetic energy (KE_initial) of the particle can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m V_0^2 \]

### Step 3: Determine Final Kinetic Energy
Since the particle loses \( \frac{3}{4} \) of its kinetic energy, the final kinetic energy (KE_final) will be:
\[ KE_{\text{final}} = KE_{\text{initial}} - \frac{3}{4} KE_{\text{initial}} = \frac{1}{4} KE_{\text{initial}} = \frac{1}{4} \left(\frac{1}{2} m V_0^2\right) = \frac{1}{8} m V_0^2 \]

### Step 4: Relate Final Kinetic Energy to Final Velocity
The final kinetic energy can also be expressed in terms of the final velocity \( V_f \):
\[ KE_{\text{final}} = \frac{1}{2} m V_f^2 \]
Setting the two expressions for \( KE_{\text{final}} \) equal gives:
\[ \frac{1}{8} m V_0^2 = \frac{1}{2} m V_f^2 \]
Cancelling \( m \) from both sides and solving for \( V_f \):
\[ \frac{1}{8} V_0^2 = \frac{1}{2} V_f^2 \implies V_f^2 = \frac{1}{4} V_0^2 \implies V_f = \frac{V_0}{2} \]

### Step 5: Apply the Equation of Motion
Using the first equation of motion:
\[ V_f = V_0 - a t_0 \]
Where \( a \) is the retarding acceleration due to friction. The retarding force due to friction is given by:
\[ F_{\text{friction}} = \mu m g \]
Thus, the acceleration \( a \) is:
\[ a = \frac{F_{\text{friction}}}{m} = \mu g \]
Substituting this into the equation of motion gives:
\[ \frac{V_0}{2} = V_0 - \mu g t_0 \]

### Step 6: Rearranging the Equation
Rearranging the equation to solve for \( \mu \):
\[ \mu g t_0 = V_0 - \frac{V_0}{2} = \frac{V_0}{2} \]
\[ \mu = \frac{V_0}{2 g t_0} \]

### Final Answer
The coefficient of friction \( \mu \) between the particle and the ground is:
\[ \mu = \frac{V_0}{2 g t_0} \]