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. zero
B. `sqrt((2E)/(m))`
C. `sqrt(E)/(m)`
D. `sqrt((E)/(2m))`

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy \( E \) of the particle is the sum of its kinetic energy \( K \) and potential energy \( U \).

### Step 1: Write the expression for total mechanical energy
The total mechanical energy \( E \) is given by:
\[ E = K + U \]
where \( K \) is the kinetic energy and \( U \) is the potential energy.

### Step 2: Identify the potential energy at the given position
The potential energy \( U \) is given as:
\[ U = \frac{1}{2} k x^2 \quad \text{for } x < 0 \]
For \( x = -\sqrt{\frac{2E}{k}} \), we substitute this value into the potential energy equation:
\[ U = \frac{1}{2} k \left(-\sqrt{\frac{2E}{k}}\right)^2 \]
Calculating this gives:
\[ U = \frac{1}{2} k \left(\frac{2E}{k}\right) = \frac{E}{k} \cdot k = E \]

### Step 3: Substitute the potential energy into the total energy equation
Now, substituting \( U \) back into the total energy equation:
\[ E = K + E \]
This simplifies to:
\[ K = E - U = E - E = 0 \]

### Step 4: Relate kinetic energy to speed
The kinetic energy \( K \) is also given by the expression:
\[ K = \frac{1}{2} mv^2 \]
Setting this equal to the kinetic energy we found:
\[ 0 = \frac{1}{2} mv^2 \]

### Step 5: Solve for speed
From the equation above, we can see that:
\[ \frac{1}{2} mv^2 = 0 \implies v^2 = 0 \implies v = 0 \]

Thus, the speed of the particle at \( x = -\sqrt{\frac{2E}{k}} \) is:
\[ \boxed{0} \]