A box of mass m is initially at rest on a horizontal surface. A constant horizontal force of mg/2 is applied to the box directed to the right. The coefficient of friction of the surface changes with the distance pushed as `mu=mu_(0)x` where x is the distance fr4om the initial location . For what distance is the box pushed until it comes to rest again?

Options

A. `(2)/(mu_(0))`
B. `(1)/(mu_(0))`
C. `(1)/(2mu_(0))`
D. `(1)/(4mu_(0))`

Explanation

To solve the problem step by step, we will analyze the forces acting on the box and apply the principles of mechanics.

### Step 1: Identify the forces acting on the box
The box has the following forces acting on it:
- Weight (downward): \( mg \)
- Normal force (upward): \( N \)
- Applied force (to the right): \( F = \frac{mg}{2} \)
- Frictional force (to the left): \( f \)

### Step 2: Determine the normal force
Since the box is on a horizontal surface and there are no vertical accelerations, the normal force \( N \) is equal to the weight of the box:
\[ N = mg \]

### Step 3: Write the expression for the frictional force
The frictional force \( f \) is given by:
\[ f = \mu N \]
Given that the coefficient of friction changes with distance \( x \) as \( \mu = \mu_0 x \), we can substitute this into the frictional force equation:
\[ f = \mu_0 x \cdot mg \]

### Step 4: Apply Newton's second law
According to Newton's second law, the net force acting on the box can be expressed as:
\[ F_{\text{net}} = F_{\text{applied}} - f \]
Substituting the values we have:
\[ ma = \frac{mg}{2} - \mu_0 x \cdot mg \]
Dividing through by \( m \):
\[ a = \frac{g}{2} - \mu_0 x \cdot g \]

### Step 5: Relate acceleration to velocity
Acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} = v \frac{dv}{dx} \]
Thus, we can rewrite the equation:
\[ v \frac{dv}{dx} = \frac{g}{2} - g \mu_0 x \]

### Step 6: Integrate to find the relationship between velocity and distance
We can separate variables and integrate:
\[ \int v \, dv = \int \left( \frac{g}{2} - g \mu_0 x \right) \, dx \]
This gives:
\[ \frac{v^2}{2} = \frac{g}{2} x - \frac{g \mu_0 x^2}{2} + C \]
Assuming the initial velocity \( v = 0 \) when \( x = 0 \), we find that \( C = 0 \):
\[ \frac{v^2}{2} = \frac{g}{2} x - \frac{g \mu_0 x^2}{2} \]

### Step 7: Set the final velocity to zero and solve for \( x \)
To find the distance \( x_0 \) when the box comes to rest again, set \( v = 0 \):
\[ 0 = \frac{g}{2} x_0 - \frac{g \mu_0 x_0^2}{2} \]
Factoring out \( \frac{g}{2} \):
\[ 0 = x_0 - \mu_0 x_0^2 \]
This gives us two solutions:
1. \( x_0 = 0 \) (initial position)
2. \( x_0 = \frac{1}{\mu_0} \)

### Conclusion
The distance the box is pushed until it comes to rest again is:
\[ x_0 = \frac{1}{\mu_0} \]