The potential energy of a particle of mass `m` free to move along the x-axis is given by `U=(1//2)kx^2` for `xlt0` and `U=0` for `xge0` (x denotes the x-coordinate of the particle and k is a positive constant). If the total mechanical energy of the particle is E, then its speed at `x=-sqrt(2E//k)` is

Options

A. (a) Zero
B. (b) `sqrt((2E)/(m))`
C. (c) `sqrt(E/m)`
D. (d) `sqrt((3E)/(2m))`

Explanation

To solve the problem, we need to analyze the potential energy and the total mechanical energy of the particle.

### Step-by-Step Solution:

1. **Understanding the Potential Energy:**
The potential energy \( U \) of the particle is given by:
\[ U = \frac{1}{2} k x^2 \quad \text{for } x < 0 \]
\[ U = 0 \quad \text{for } x \geq 0 \]
Here, \( k \) is a positive constant.

2. **Total Mechanical Energy:**
The total mechanical energy \( E \) of the particle is the sum of its kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]

3. **Kinetic Energy Expression:**
The kinetic energy \( K \) can be expressed as:
\[ K = E - U \]

4. **Substituting for Potential Energy:**
For \( x = -\sqrt{\frac{2E}{k}} \), we substitute this value into the potential energy equation:
\[ U = \frac{1}{2} k \left(-\sqrt{\frac{2E}{k}}\right)^2 \]
Simplifying this:
\[ U = \frac{1}{2} k \left(\frac{2E}{k}\right) = \frac{2E}{2} = E \]

5. **Finding Kinetic Energy:**
Now we substitute \( U \) back into the kinetic energy equation:
\[ K = E - U = E - E = 0 \]

6. **Relating Kinetic Energy to Speed:**
The kinetic energy is also given by the formula:
\[ K = \frac{1}{2} mv^2 \]
Setting this equal to zero:
\[ \frac{1}{2} mv^2 = 0 \]
Since mass \( m \) cannot be zero, we conclude that:
\[ v^2 = 0 \implies v = 0 \]

7. **Conclusion:**
Therefore, the speed of the particle at \( x = -\sqrt{\frac{2E}{k}} \) is:
\[ v = 0 \]

### Final Answer:
The speed of the particle at \( x = -\sqrt{\frac{2E}{k}} \) is \( 0 \).

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