A 1.0 kg block is initially at rest on a horizontal frictionless surface when a horizontal force along an `x` axis is applied to the block. The force is given by `vec(F)(x)= (2.5-x^(2)) hat(i)N`, where `x` is in meters and the initial position of the block is `x=0`. (a) What is the kinetic energy of the block as it passes through `x=2.0m` ? (b) What is the maximum kinetic energy of the block between `x=0` and `x=2.0 m` ?

Answer

To solve the problem, we will use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Let's break down the solution step by step.

### Step-by-Step Solution

**Given:**
- Mass of the block, \( m = 1.0 \, \text{kg} \)
- Force as a function of position, \( F(x) = 2.5 - x^2 \, \text{N} \)
- Initial position, \( x_0 = 0 \, \text{m} \)
- Final position for part (a), \( x = 2.0 \, \text{m} \)

#### Part (a): Kinetic Energy at \( x = 2.0 \, \text{m} \)

1. **Calculate the Work Done from \( x = 0 \) to \( x = 2 \):**
The work done \( W \) is given by the integral of force over the displacement:
\[ W = \int_{0}^{2} F(x) \, dx = \int_{0}^{2} (2.5 - x^2) \, dx \]

2. **Evaluate the Integral:**
\[ W = \int_{0}^{2} (2.5 - x^2) \, dx = \left[ 2.5x - \frac{x^3}{3} \right]_{0}^{2} \]
- At \( x = 2 \):
\[ W = 2.5(2) - \frac{(2)^3}{3} = 5 - \frac{8}{3} = 5 - 2.67 = 2.33 \, \text{J} \]

3. **Apply the Work-Energy Theorem:**
Since the block starts from rest, its initial kinetic energy \( KE_0 = 0 \):
\[ W = KE_f - KE_0 \implies KE_f = W \]
Therefore, the kinetic energy at \( x = 2 \):
\[ KE = 2.33 \, \text{J} \]

#### Part (b): Maximum Kinetic Energy between \( x = 0 \) and \( x = 2 \)

1. **Find the Expression for Kinetic Energy:**
From part (a), we have:
\[ KE(x) = 2.5x - \frac{x^3}{3} \]

2. **Differentiate to Find Critical Points:**
To find the maximum kinetic energy, we differentiate \( KE(x) \) with respect to \( x \) and set it to zero:
\[ \frac{d(KE)}{dx} = 2.5 - x^2 = 0 \]
Solving for \( x \):
\[ x^2 = 2.5 \implies x = \sqrt{2.5} \approx 1.58 \, \text{m} \]

3. **Check if \( x = \sqrt{2.5} \) is within the interval [0, 2]:**
Yes, \( 1.58 \, \text{m} \) is within the interval.

4. **Calculate the Kinetic Energy at \( x = \sqrt{2.5} \):**
Substitute \( x = \sqrt{2.5} \) into the kinetic energy expression:
\[ KE\left(\sqrt{2.5}\right) = 2.5(\sqrt{2.5}) - \frac{(\sqrt{2.5})^3}{3} \]
- Calculate \( KE \):
\[ KE\left(\sqrt{2.5}\right) = 2.5 \cdot \sqrt{2.5} - \frac{(2.5)^{3/2}}{3} \]
\[ = 2.5 \cdot 1.58 - \frac{(2.5) \cdot 1.58}{3} \approx 3.95 - 1.32 \approx 2.63 \, \text{J} \]

### Final Answers:
- (a) Kinetic energy at \( x = 2.0 \, \text{m} \) is approximately \( 2.33 \, \text{J} \).
- (b) Maximum kinetic energy between \( x = 0 \) and \( x = 2 \) is approximately \( 2.63 \, \text{J} \).