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))/(2"gt"_(0))`
B. `(v_(0))/(4"gt"_(0))`
C. `(3v_(0))/(4"gt"_(0))`
D. `(v_(0))/("gt"_(0))`

Explanation

To solve the problem step by step, we can follow these steps:

### Step 1: Determine Initial and Final Kinetic Energy
The initial kinetic energy (KE_initial) of the particle can be expressed as:
\[ KE_{\text{initial}} = \frac{1}{2} m V_0^2 \]
According to the problem, three-fourths of the kinetic energy is lost due to friction, meaning that the final kinetic energy (KE_final) is:
\[ KE_{\text{final}} = KE_{\text{initial}} - \frac{3}{4} KE_{\text{initial}} = \frac{1}{4} KE_{\text{initial}} \]

### Step 2: Express Final Kinetic Energy
Substituting the expression for KE_initial into the equation for KE_final:
\[ KE_{\text{final}} = \frac{1}{4} \left(\frac{1}{2} m V_0^2\right) = \frac{1}{8} m V_0^2 \]

### Step 3: Relate Final Kinetic Energy to Final Velocity
The final kinetic energy can also be expressed in terms of the final velocity \(V'\):
\[ KE_{\text{final}} = \frac{1}{2} m V'^2 \]
Setting the two expressions for KE_final equal gives:
\[ \frac{1}{2} m V'^2 = \frac{1}{8} m V_0^2 \]
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ \frac{1}{2} V'^2 = \frac{1}{8} V_0^2 \]

### Step 4: Solve for Final Velocity
Multiplying both sides by 2:
\[ V'^2 = \frac{1}{4} V_0^2 \]
Taking the square root:
\[ V' = \frac{V_0}{2} \]

### Step 5: Determine the Acceleration
The acceleration \(a\) due to friction can be expressed as:
\[ a = -\mu g \]
where \(\mu\) is the coefficient of friction and \(g\) is the acceleration due to gravity.

### Step 6: Use the Equation of Motion
Using the equation of motion:
\[ V' = V_0 + at \]
Substituting the known values:
\[ \frac{V_0}{2} = V_0 - \mu g t_0 \]
Rearranging gives:
\[ \mu g t_0 = V_0 - \frac{V_0}{2} = \frac{V_0}{2} \]

### Step 7: Solve for the Coefficient of Friction
Now, we can solve for \(\mu\):
\[ \mu g t_0 = \frac{V_0}{2} \]
\[ \mu = \frac{V_0}{2 g t_0} \]

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

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