An object of mass m has a speed `v_(0)` as it passes throgh the origin. Origin. It subjected to a retaeding force given by `F(x) =-Ax`. Here, A is a positive constant. Find its x-coordinate when it stops.

Explanation

To solve the problem step by step, we will use the work-energy theorem which relates the work done by forces acting on an object to its change in kinetic energy.

### Step 1: Understand the Given Information
We have:
- Mass of the object: \( m \)
- Initial speed of the object at the origin: \( v_0 \)
- Retarding force: \( F(x) = -Ax \), where \( A \) is a positive constant.

### Step 2: Apply the Work-Energy Theorem
The work-energy theorem states that the work done by the forces acting on an object is equal to the change in kinetic energy of that object. Mathematically, it can be expressed as:
\[ \Delta KE = W \]
Where:
- \( \Delta KE = KE_{final} - KE_{initial} \)
- \( W = \int F(x) \, dx \)

### Step 3: Calculate the Initial and Final Kinetic Energy
The initial kinetic energy when the object is at the origin is:
\[ KE_{initial} = \frac{1}{2} m v_0^2 \]
The final kinetic energy when the object stops is:
\[ KE_{final} = 0 \]
Thus, the change in kinetic energy is:
\[ \Delta KE = 0 - \frac{1}{2} m v_0^2 = -\frac{1}{2} m v_0^2 \]

### Step 4: Calculate the Work Done by the Retarding Force
The work done by the retarding force from the origin to the point \( x \) is given by:
\[ W = \int_0^x F(x) \, dx = \int_0^x -Ax \, dx \]
Calculating the integral:
\[ W = -A \int_0^x x \, dx = -A \left[ \frac{x^2}{2} \right]_0^x = -A \frac{x^2}{2} \]

### Step 5: Set Up the Equation
According to the work-energy theorem:
\[ -\frac{1}{2} m v_0^2 = -A \frac{x^2}{2} \]
Removing the negative signs and simplifying gives:
\[ \frac{1}{2} m v_0^2 = A \frac{x^2}{2} \]

### Step 6: Solve for \( x \)
Multiplying both sides by 2 to eliminate the fraction:
\[ m v_0^2 = A x^2 \]
Now, rearranging for \( x^2 \):
\[ x^2 = \frac{m v_0^2}{A} \]
Taking the square root of both sides gives:
\[ x = \sqrt{\frac{m v_0^2}{A}} \]

### Final Answer
Thus, the x-coordinate when the object stops is:
\[ x = \sqrt{\frac{m v_0^2}{A}} \]