A block weighing 10.0 N is attached to the lower end of a verticle spring `(k = 200.0 N//m)`, the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of 2.00 J as it passes through the point at which the spring is unstretched.
What is the amplitude of the oscillation and the maximum kinetic energy of the block as it oscillates?

Options

A. 0.20 m and 4.50 J
B. 0.30 m and 3.35 J
C. 0.25 m and 3.25 J
D. 0.15 m and 2.25 J

Answer

To solve the problem step by step, we will break it down into two parts: finding the amplitude of the oscillation and the maximum kinetic energy of the block.

### Step 1: Determine the Extension of the Spring at Equilibrium
When the block is attached to the spring, it stretches the spring until it reaches an equilibrium position. At this point, the force exerted by the spring equals the weight of the block.

- Given:
- Weight of the block (W) = 10.0 N
- Spring constant (k) = 200.0 N/m

Using Hooke's Law, the force exerted by the spring can be expressed as:
\[ F_{\text{spring}} = k \cdot x \]
Where \( x \) is the extension of the spring.

At equilibrium:
\[ W = F_{\text{spring}} \]
\[ 10.0 = 200.0 \cdot x \]

Solving for \( x \):
\[ x = \frac{10.0}{200.0} = 0.05 \, \text{m} \]

### Step 2: Calculate Total Energy at the Unstretched Position
When the block passes through the point where the spring is unstretched, it has a kinetic energy of 2.00 J. At this point, the potential energy is zero because the spring is unstretched.

The total mechanical energy (E) in the system is given by:
\[ E = K + U \]
Where \( K \) is the kinetic energy and \( U \) is the potential energy.

At the unstretched position:
\[ E = 2.00 + 0 = 2.00 \, \text{J} \]

### Step 3: Calculate the Total Energy at Maximum Displacement (Amplitude)
At the maximum displacement (amplitude \( A \)), the kinetic energy is zero, and all the energy is potential energy:
\[ E = U = \frac{1}{2} k A^2 \]

Setting the total energy equal to the potential energy at maximum displacement:
\[ 2.00 = \frac{1}{2} \cdot 200.0 \cdot A^2 \]

Solving for \( A^2 \):
\[ 2.00 = 100.0 \cdot A^2 \]
\[ A^2 = \frac{2.00}{100.0} = 0.02 \]
\[ A = \sqrt{0.02} = 0.1414 \, \text{m} \]

### Step 4: Calculate Maximum Kinetic Energy
The maximum kinetic energy occurs when the block is at the equilibrium position. At this point, all the total energy is kinetic energy:
\[ K_{\text{max}} = E = 2.00 \, \text{J} \]

### Final Results
- **Amplitude of the oscillation**: \( A \approx 0.1414 \, \text{m} \)
- **Maximum kinetic energy**: \( K_{\text{max}} = 2.00 \, \text{J} \)