A particle moves along an x axis. Does the kinetic energy of the particle increase, decrease, or reman the same if the particle's velocity changes (a) from `-3 m//s` to `-2m//s` and (b) from `-2 m//s` to `2m//s` ? (c) In each situation, is the work done on the particle positive, negative, or zero?

Answer

To solve the problem, we will analyze the changes in kinetic energy and work done for each scenario step-by-step.

### Step 1: Understanding Kinetic Energy
The kinetic energy (KE) of a particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity. The kinetic energy depends on the square of the velocity, which means it is always positive regardless of the direction of the velocity.

### Step 2: Analyzing Part (a)
In part (a), the velocity changes from \(-3 \, \text{m/s}\) to \(-2 \, \text{m/s}\).

- **Initial Kinetic Energy (KE_initial)**:
\[ KE_{\text{initial}} = \frac{1}{2} m (-3)^2 = \frac{1}{2} m \cdot 9 = \frac{9}{2} m \]

- **Final Kinetic Energy (KE_final)**:
\[ KE_{\text{final}} = \frac{1}{2} m (-2)^2 = \frac{1}{2} m \cdot 4 = 2m \]

- **Comparison**:
Since \(\frac{9}{2} m > 2m\), the kinetic energy decreases.

### Step 3: Work Done in Part (a)
The work done (W) on the particle can be calculated using the change in kinetic energy:
\[ W = KE_{\text{final}} - KE_{\text{initial}} = 2m - \frac{9}{2} m = 2m - 4.5m = -2.5m \]
Since the work done is negative, it indicates that work is done against the motion.

### Step 4: Analyzing Part (b)
In part (b), the velocity changes from \(-2 \, \text{m/s}\) to \(2 \, \text{m/s}\).

- **Initial Kinetic Energy (KE_initial)**:
\[ KE_{\text{initial}} = \frac{1}{2} m (-2)^2 = \frac{1}{2} m \cdot 4 = 2m \]

- **Final Kinetic Energy (KE_final)**:
\[ KE_{\text{final}} = \frac{1}{2} m (2)^2 = \frac{1}{2} m \cdot 4 = 2m \]

- **Comparison**:
Since \(KE_{\text{initial}} = KE_{\text{final}}\), the kinetic energy remains the same.

### Step 5: Work Done in Part (b)
The work done on the particle can be calculated as:
\[ W = KE_{\text{final}} - KE_{\text{initial}} = 2m - 2m = 0 \]
Since the work done is zero, it indicates that there is no net work done on the particle.

### Summary of Results
- (a) Kinetic energy **decreases**; work done is **negative**.
- (b) Kinetic energy **remains the same**; work done is **zero**.